Answers
Related Questions
Can momentum be negative?
A. No.
B. Yes, if it has a very small mass.
C. Yes, if it is moving very slow.
D. Yes, if it is moving backward.
Answers
Answer:
D
Explanation:
momentum can be negative as it's a vector quantity. positive momentum indicates that the object is traveling in the positive direction as defined by the coordinate system. negative momentum indicates that the object is moving in the opposite direction (backwards). the momentum equation itself is p = mass x velocity.
and only if you consider velocity a a directed vector, can you have a negative momentum. this equation shows that the momentum is in the same direction as velocity.
so, the sign is only for the direction.
the force is the same (going with the absolute value). a negative momentum is not smaller than a positive momentum.
It is possible that momentum can be negative if an object is moving backward. The correct option is D.
A vector quantity called momentum depends on an object's mass and velocity. It is described as the result of the mass and the velocity of an object.
When an item is moving against a selected positive direction, its velocity will be negative since velocity is a vector variable that comprises both magnitude and direction. As a result, the negative velocity will yield a negative value for momentum when it is calculated.
Thus the correct option is D.
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(b) The particle displacement y of air molecules due to a sound wave is given by y 4m and w = = 0.008 cos wt sin kz. Where k - m 50m rad/s. calculate i. The distance between two consecutive nodes ii. The amplitude after 0.56s
Answers
The distance between two consecutive nodes and the amplitude after 0.56s are m/2 and 1.75×10^(-4) m respectively.
What's the distance between consecutive nodes of the displacement of air molecules?Wavelength is the distance between two consecutive nodes or toughs or crests or anti-nodes.So, distance between consecutive nodes = wavelength = 2π÷k= 2π/(4π÷m)
= m/2
What's the amplitude after 0.56s of the displacement of air molecules?Displacement after 0.56 s = 0.008×cos(50π×0.56s)
=1.75×10^(-4) m
Thus, we can conclude that the distance between consecutive nodes and displacement after 0.56 s are m/2 and 1.75×10^(-4) m respectively.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: The particle displacement y of air molecules due to a sound wave is given by y=0.008coswtsinkz where k=4π÷m and w=50π rads/s.
Calculate:
I) the distance between 2 consecutive nodes
ii) the amplitude after 0.565s
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A 850-kg sports car accelerates from rest to 95 km/h in 6.8 s .
What is the average power delivered by the engine?
Express your answer to two significant figures and include the appropriate units.
Answers
From the calculations, the power expended is 43650 W.
What is the power expended?Now we can find the acceleration from;
v = u + at
u = 0 m/s
v = 95 km/h or 26.4 m/s
t = 6.8 s
a = ?
Now
v = at
a = v/t
a = 26.4 m/s/ 6.8 s
a = 3.88 m/s^2
Force = ma = 850-kg * 3.88 m/s^2 = 3298 N
The distance covered is obtained from;
v^2 = u^2 + 2as
v^2 = 2as
s = v^2/2a
s = (26.4)^2/2 * 3.88
s = 696.96/7.76
s = 90 m
Now;
Work = Fs
Work = 3298 N * 90 m = 296820 J
Power = 296820 J/ 6.8 s
= 43650 W
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A 41-kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.5 m/s. Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.
Answer:
___________m
Answers
The altitude of the pole vaulter as she crosses the bar is 5 m.
The altitude of the barv² = u² - 2gh
where;
v is final velocity of the pole vaulteru is the initial velocity of the pole vaulterh is altitude of the barh = (u² - v²)/2g
h = (10² - 1.5²)/(2 x 9.8)
h = 5 m
Thus, the altitude of the pole vaulter as she crosses the bar is 5 m.
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A small object with mass 3.95 kg moves counterclockwise with constant speed 1.65 rad/s in a circle of radius 2.95 m centered at the origin. It starts at the point with position vector 2.95î m. Then it undergoes an angular displacement of 9.10 rad.
(a) What is its new position vector? _____ m
(b) In what quadrant is the particle located and what angle does its position vector make with the positive x-axis? ______ quadrant at _______°
(c) What is its velocity? _____ m/s
(d) In what direction is it moving? _____ ° from the +x direction
(e) What is its acceleration? _____ m/s2
(f) Make a sketch of its position, velocity, and acceleration vectors.
Answers
a)New position vector in vector form= r = 0.233404 î + 2.94056j m
b) Lies in second quadrant at 161.391°
c)Velocity =4.8675 m/s
d)It is moving in a direction making 161.391° with positive x-direction.
e)Acceleration will be centripetal acceleration (8.031 m/s²).
Given:
Mass of the object m = 3.95 kg
ω=1.65 rad/s
Radius of the circle = 2.95 m
a)
new position vector in vector form
=R cos1.65 î + R sin 1.65 j
= 2.95 cos1.65 î +2.95 sin1.65 j
= 2.95 x 0.07912 î + 2.95 x 0.9968 j
r = 0.233404 î + 2.94056j
b)
Angular Displacement = θ₀ = 9.10 rad
9.10 radian = 180/π× 9/10 degree
= 521.391°
=521.391°- 360°
=161.391°
This will lie in second quadrant.
Angle made with positive x-axis
=161.391°
c)
Velocity
v = ω R
= 1.65 x 2.95
=4.8675 m/s
d)
It is moving in a direction making 161.391° with positive x-direction.
e)
Acceleration will be centripetal acceleration.
= v²/R
=(4.8675)² / 2.95
=23.6925562 / 2.95
=8.031 m/s²
f) Position, Velocity and Acceleration graph:
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a) New position vector is r = 0.233404 î + 2.94056j m
b) Lies in second quadrant at 161.391°
c) Velocity =4.8675 m/s
d) It is moving in a direction making 161.391° with positive x-direction.
e) Acceleration will be centripetal acceleration (8.031 m/s²).
Given:
Mass of the object m = 3.95 kg
ω=1.65 rad/s
The radius of the circle = 2.95 m
a) new position vector in vector form
r =R cos1.65 î + R sin 1.65 j
r = 2.95 cos1.65 î +2.95 sin1.65 j
r = 2.95 x 0.07912 î + 2.95 x 0.9968 j
r = 0.233404 î + 2.94056j
b) Angular Displacement = θ₀ = 9.10 rad
9.10 radian = 180/π× 9/10 degree
= 521.391°
= 521.391°- 360°
=161.391°
This will lie in the second quadrant.
Angle made with the positive x-axis =161.391°
c) Velocity
v = ω R
v = 1.65 x 2.95
v = 4.8675 m/s
d) It is moving in a direction making 161.391° with positive x-direction.
e) Acceleration will be centripetal acceleration.= v²/R
=(4.8675)² / 2.95
=23.6925562 / 2.95
=8.031 m/s²
f) Position, Velocity, and Acceleration graph:
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A golf ball is hit so that it leaves the club face at a velocity of 45m/s at an angle of 40° to the horizontal. by ignoring the effect of air resistance and spin on the ball, use the information to answer the question (a) to (d).
a) Calculate the horizontal component of the velocity
b) Determine the vertical component of the velocity.
c) Find the time taken for the ball to reach its maximum height.
d) Calculate the horizontal distance travelled when the ball is at its maximum height.
Answers
Answer:a)45cos40 b)45sin40 c) 2.95 d)102
Explanation:
a) [tex]v_{x} = 45 cos40[/tex]
b) [tex]v_{y}=45sin40[/tex]
c) Since air resistance is negligible, the only force acting on the golf ball is gravity. Thus, its vertical acceleration is -g. We know the final velocity must be 0 m/s, because this will be when the golf ball reaches the maximum height and starts to change direction (it falls back to the ground). We also know initial velocity in the vertical direction (see part b). Thus, we can use this equation: [tex]v_{f} = v_{i} + at[/tex].
[tex]0 m/s = 45sin40 + (-9.8m/s^2)t\\t = 2.95s[/tex]
d) The horizontal distance traveled is dependent on (1) how long the ball is in the air and (2) what the horizontal velocity is. (1) was found in part c, and (2) was found in part a.
[tex]x=vt\\x=45cos40(2.95) =102m[/tex]
A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 4.05 m/s. If the roof is pitched at 34.0° below the horizon and the roof edge is 2.10 m above the ground, find the time the baseball spends in the air and the horizontal distance from the roof edge to the point where the baseball lands on the ground.
(a) the time the baseball spends in the air (in s)
(b) the horizontal distance from the roof edge to the point where the baseball lands on the ground (in m)
Answers
(a) The time the baseball spends in the air is 0.92 s.
(b) The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.
Time spent in air by the baseball
h = vt - ¹/₂gt²
-2.1 = (4.05 x sin 34)t - ¹/₂(9.8)(t²)
-2.1 = 2.26t - 4.9t²
4.9t² - 2.26t - 2.1 = 0
t = 0.92 s
Horizontal distance traveled by the baseballR = Vx(t)
R = (4.05 x cos 34)(0.92)
R = 3.1 m
Thus, the time the baseball spends in the air is 0.92 s.
The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.
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what do u mean by the velocity ratio of lever is four
Answers
Answer:
VR = Velocity of effort / Velocity of Load
Explanation:
The ratio of the out-lever to the in-lever (length of the resistance arm to effort arm) is known as the velocity ratio (VR). It indicates that the distance covered by effort is four times that covered by the load.
The load-to-effort ratio of a machine, or alternatively, the output-to-input ratio of a machine, is known as its mechanical advantage. Another definition of velocity ratio is the ratio of the velocity of effort to the velocity of load.
When a machine has a force ratio of 4 and a velocity ratio of 4, this indicates that the weight moved is multiplied by 4 and the distance moved by the effort is multiplied by 4 at the same time.
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In our solar system, which celestial object is known as the dwarf planet?
Answers
Answer:
unfournatletly
Explanation:
i have no clue sorry to waste ur time ill rather not say a answer that will be incorrect.
Answer:
Pluto
Explanation:
Pluto was part of our solar system till 2006In 2006 international scientific committee removed it from planets listIt's known as dwarf planet nowA thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale.
Evaluate the magnitude of the force on the left hand pole.
Answers
The magnitude of the force on the left-hand pole of the thin flexible gold chain of uniform linear density with a mass of 17.1 and, hangs between two 30.0 cm long vertical sticks, which are a distance of 30.0 cm apart will be, 0.167N.
To find the correct answer, we have to know more about the Basic forces that acts upon a body.
What is force and which are the basic forces that acts upon a body?A push or a pull which changes or tends to change the state or rest, or motion of a body is called Force.Force is a polar vector as it has a point of application.Positive force represents repulsion and the negative force represented attraction.There are 3 main forces acting on a body, such as, weight mg, normal reaction N, and the Tension or pulling force.How to solve the problem?We have given that, the gold chain hangs between the vertical sticks of 30cm and the horizontal distance between then is 30cm.From the given data, we can find the angle (in the free body diagram, it is given as θ).[tex]tan\alpha =\frac{30}{30}\\ \alpha =45^0[/tex]
From the free body diagram given, we can write the balanced equations of total force along y direction as,[tex]y-direction\\T_2sin\alpha =mg\\\\T_2=\frac{mg}{sin\alpha } =\frac{17.1*10^{-3}*9.8}{sin45} \\\\T_2=0.236N[/tex]
From the free body diagram given, we can write the balanced equations of total force along x direction as,[tex]x-direction\\T_1-T_2cos\alpha =0\\T_1=T_2cos\alpha \\T_1=0.236*cos45=0.167N[/tex]
Thus, we can conclude that, the magnitude of force on the left-hand pole will be 0.167N.
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The force acting on the left-hand pole of a thin, flexible gold chain of uniform linear density weighing 17.1 grams that is suspended between two vertical sticks that are 30.0 cm apart will be 0.167 N in magnitude.
We need to learn more about the fundamental forces that affect a body in order to arrive at the right answer.
What exactly is force, and what are the fundamental forces that affect a body?Force is a push or a pull that modifies or tends to modify the condition, rest, or motion of a body.Given that it has a point of application, force is a polar vector.Repulsion is represented by positive force, and attraction by negative force.A body is subject to three main forces: weight mg, normal response N, and tension or pulling force.How can the issue be resolved?We have given that, the gold chain hangs between the vertical sticks of 30cm and the horizontal space between then is 30cm.We can determine the angle (shown as in the free body figure) using the information provided.[tex]\alpha =tan^{-1}(\frac{30}{30})=45^0[/tex]
We may get the balanced equations for the total force in the y direction from the provided free body diagram as follows:[tex]T_2sin\alpha =mg\\T_2=\frac{mg}{sin\alpha } =0.236N[/tex]
We can create the balanced equations for the total force in the x direction using the provided free body diagram:[tex]T_1=T_2cos\alpha \\T_1=0.167N[/tex]
Thus, we may infer that the force acting on the left-hand pole will have a value of 0.167N.
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The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Calculate the speed of the satellite.
Answers
Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
Explanation: To find the answer, we need to know about the different equations of planetary motion.
How to find the initial speed of the rock as it left the astronaut's hand?We have given with the following values,[tex]m_g=7.88*10^{18}kg\\r_g=6.32*10^4 m.\\h_{max}=1.44*10^3m.\\[/tex]
We have the expression for the initial velocity as,[tex]v=\sqrt{2gh}[/tex]
Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,[tex]g_g=\frac{GM_g}{r_g^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2} =0.132m/s^2.[/tex]
Now, the velocity will become,[tex]v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s.[/tex]
How to find the speed of the satellite?As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,[tex]v=\sqrt{ \frac{GM}{r}}=\sqrt{\frac{7.88*10^{18}*6.67*10^{-11}}{1.45*10^5 }} =3.624km/s.[/tex]
Thus, we can conclude that,
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
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1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.
2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.
Understanding the planetary motion equations is necessary in order to determine the solution.
How to determine the rock's original speed when it left the astronaut's hand?The starting velocity's expression is as follows:[tex]v=\sqrt{2gh}[/tex]
So, in order to determine v, we must determine the acceleration of glob caused by gravity. We already have,[tex]a=\frac{GM}{R^2} =0.132m/s^2[/tex]
The velocity will now change to,[tex]v=\sqrt{2*9.8*0.132} =19.46m/s[/tex]
How can I determine the satellite's speed?As we are aware, the centripetal force and gravitational force are equivalent, and thus leads to the following satellite speed equation:[tex]v=\sqrt{\frac{GM}{R} } =3.624km/s\\where, M=7.88*10^{18}kg[/tex]
Consequently, we can say that
1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.
2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.
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Help me please.
A
B
C
D
Answers
I will go to school tomorrow .....is this present tense or past tense or future tense
Answer:
D
Explanation:
The given vector can be broken down into two components (as shown in the figure).
Which results ; vector (s) = 3x + 4y.
A certain force gives mass m1, an acceleration of 12 m/s² and mass m2 an acceleration of 3.3. m/s². What acceleration will this force give to the combined mass?
Answers
Acceleration is the rate of change of velocity. Usually, acceleration means the speed is transforming, but not always. When an object moves in a circular path at a steady speed, it is still accelerating, because the focus of its velocity is changing.
a = 4,552 m / s², b) a = 2,588 m / s²
Newton's second law is
F = ma
a = F / m
in this case the force remains constant
indicate us
* for a mass m₁
a₁ = F/m₁
a₁ = 12, m/ s²
* for a mass m₂
a₂= 3.3 m / s²
acceleration m = m₂-m₁
substitute
[tex]$\begin{aligned}&a=\frac{F}{m_{2}-m_{1}} \\&1 / a=\frac{m_{2}}{F}-\frac{m_{1}}{F}\end{aligned}$[/tex]
let's calculate
1/a=1/3.3 - 1/12 = 0.21969
a = 4,552 m / s²
a= 4,552m/s²
What is speed and acceleration?Speed estimates the rate of movement of an object, that is, the distance traveled per unit of time. Acceleration calculates the rate of change of velocity, that is, the change in velocity between two different moments
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The three displacement vectors have magnitude of A=5.00m,B=5.00m and C=4.00m.Find the resultant (magnitude and directional angle) of the three vectors?
Answers
3.00 m is the magnitude of the resultant vector
137.1° is the directional angle of the resultant vector.
1) To find the magnitude of the resultant
Discover each vector's individual components first.
We must consider the signs of the components because the angles in the figure are measured in various ways.
In this case, both the y component of vector C and the x component of vector A are negative.
The vectors' components are as follows, using a little trigonometry:
Magnitude of A = 5 m
[tex]A_{x}[/tex] = - (5.00m) cos20° = -5 ×0.408 = -4.698 m
[tex]A_{y}[/tex] = + (5.00m)sin20° = +5 × 0.342 = +1.710 m
Magnitude of B = 5m
[tex]B_{x}[/tex] = +(5.00m)cos60° = 5 × 0.5 = +2.5m
[tex]B_{y}[/tex] = +(5.00m)sin60° = 5 × √3/2 = +4.33 m
Cx = 0
Cy = -4.00m
The sum of all three vectors, which we refer to as R, produces components.
Rₓ = Aₓ + Bₓ + Cₓ
= -4.698 + 2.5 + 0
= -2.198 m
[tex]R_{y}[/tex] = [tex]A_{y} + B_{y} + C_{y}[/tex]
= +1.71 +4.33 - 4.00
= 2.040 m
R = [tex]\sqrt{R_{x^{2} }+R_{y^{2} }[/tex]
= [tex]\sqrt{(-2.198)^{2 } + (2.040)^{2} }[/tex]
= 3.00 m
2) To find the directional angle of resultant
tan θ = 2.040/-2.198 = -0.928
θ = -42.9°
Such a vector would be in the so-called "fourth quadrant," as it is well known. However, we discovered that R has a negative x component and a positive y component, indicating that such a vector must reside in the "second quadrant.
The calculator accidentally returned an angle that is 180 degrees off, thus we must add 180 degrees to the naïve angle in order to get the right angle.
Therefore, R's actual direction is determined by-
θ = -42.9° + 180° = 137.1°
Hence, the magnitude of resultant vector is 3.00 m and directional angle is 137.1°
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How many bytes of memory space are there in an 80 GB hard disk and 256 GB card?
Answers
Answer:
80GB= 80000000000 bytes
256GB= 274877906944 bytes
Explanation:80GB= 80000000000 bytes
256GB= 274877906944 bytes
If the voltage between 2 plates is 30 V and the electric field strength is 10 V/m, what is the separation distance between the plates?
A. 10 m
B. 2 m
C. 3 m
D. 30 m
Answers
(These are multiple choice questions)
1) The magnitude of the charge of five electrons is
A. 1.0×10-19 C.
B. 1.6×10-19 C.
C. 6.0×10-19 C.
D. 8.0×10-19 C.
2) Total resistance (in ohms-Ω) between the points A and B of the following circuit is
A. 70 Ω
B. 30 Ω
C. 25 Ω
D. 20 Ω
3) Given the electron configuration of a neutral atom 1s22s22p63s23p5. The atom
A. can become a negative ion easily.
B. is a halogen.
C. has 7 electrons in the outer most shell.
D. all of the above.
4) Identify the number of protons, neutrons and electrons in an atom of
A. 136 protons, 92 neutrons and 92 electrons
B. 92 protons, 136 neutrons and 92 electrons
C. 92 protons, 138 neutrons and 92 electrons
D. 230 protons, 92 neutrons and 92 electrons
5)Which of the following is not an application of Total Internal Reflection?
A. mirage
B. optical fiber
C. prismatic binocular
D. hologram
Answers
The application of total Internal Reflection occurs in a mirage. Option A
What is the charge?1) We know that a charge can be positive or negative. If the charge is negative, we call it an electron. If the charge is positive, we call it a proton. Now we know that the magnitude of charge on each electron is 1.6×10-19 C. Hence, the magnitude of charge on five electrons = 5 * 1.6×10-19 C = 8.0×10-19 C. Option D
2) The details of question 2 are not shown hence the question can not be answered.
3) Looking at the electronic configuration of the element; 1s22s22p63s23p5, it is clear to see that it has the ns2 np5 outermost configuration that is common to halogens thus it;
can become a negative ion easily. is a halogen.has 7 electrons in the outer most shell.4) The atom is composed of the protons, neutrons and electrons. Given the nuclide identified as 230/92U we have 92 protons, 138 neutrons and 92 electrons. Option C
5) The application of total Internal Reflection occurs in a mirage. Option A
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In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.430 m (1.41 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 15.00 millibar. The atmospheric pressure outside is 940 millibar.
1. Calculate the force required to pull the two hemispheres apart. [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
2. Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1450N (i.e., about 326 lbs), what is the minimum number of horses required?
Answers
The force required to pull the two hemispheres apart 53696.25N and the minimum number of horses required is 37 .
( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 )
(1)The contact area between the hemispheres is (π x 0.430²) = 0.5805m²
Pressure difference = (940 - 15) = 925millibars.
(925 x 100) = 92,500N/m^2.
(92500 x 0.5805) = 53696.25N. is the force required to part the hemispheres.
(2)[tex]\frac{53696.25N}{1450N}[/tex] = 37 horses for each side .
37 + 37 = 74 horses will be required.
Force is something which can change the motion of an object, stop it or move it, change its shape or size with it. There are two types of forces, contact forces and non-contact forces. Here, it is a contact force at first, then when the horses come it becomes non-contact force.
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a) Calculate the magnitude of the magnetic field at point P due to the current in the semicircular section of wire shown in the figure (Figure 1). (Hint: Does the current in the long, straight section of the wire produce any field at P?)
Express your answer in terms of the variables I , R , and magnetic constant μ0 .
b) Find the direction of the magnetic field at point P.
Answers
Hello!
a)
To begin, let me first clarify that no section of the wire along the axis of the point 'P' will contribute to the magnetic field (Aka, the straight part of the wire) because the radius vector and current vectors would be parallel.
Now, let's use Biot-Savart's Law:
[tex]dB = \frac{\mu_0}{4\pi }\frac{id\vec{l}\times \hat{r}}{r^2}[/tex]
B = Magnetic field strength (? T)
μ₀ = Permeability of free space (Tm/A)
i = Current in wire ('I' A)
dl = differential length element
r = distance from wire to point P ('R' m, remains constant!)
We can rewrite Biot-Savart as:
[tex]B = \frac{\mu_0}{4\pi } \int \frac{id\vec{l}\times \hat{r}}{r^2}[/tex]
First, let's mind the cross-product.
The angle between the radius vector (Along 'R') and the current vector is ALWAYS 90° since the two vectors are perpendicular along the arc. At a certain point, think about the current as being "tangential" to the differential length and thus perpendicular to the radius.
Therefore, we can disregard the cross-product since sin(90) = 1.
Let's plug in what we already know, replacing 'dl' with 'ds' since this is an arc:
[tex]B = \frac{\mu_0}{4\pi } \int \frac{ids}{R^2}[/tex]
We have a semi-circle, so we are integrating from 0 to π radians.
[tex]B = \frac{\mu_0}{4\pi } \int\limits^{\pi}_{0} {\frac{ids}{R^2}}[/tex]
Simplifying to make the integral easier, we can take constants out of the integral.
[tex]B = \frac{\mu_0i}{4\pi R^2 } \int\limits^{\pi R}_{0} {} \, ds[/tex]
Evaluating:
[tex]B = \frac{\mu_0i}{4\pi R^2} (\pi R- 0) = \boxed{\frac{\mu_0 i}{4R}}[/tex]
b)
Using the current Right-Hand-Rule at the top of the arc, point your thumb to the right. Curl your fingers as if you are gripping the wire over the top and all the way over to the other side of the wire (Where point 'P' would be). Your fingers should point INTO THE PAGE.
A lead weight falls from a height of 6 m onto a muddy surface. It comes to rest after penetrating 0.4 cm into the surface. What was the magnitude of the average acceleration during the impact? How long did it take to stop?
(Also can I have like a little explanation :))
Answers
The the magnitude of the average acceleration during the impact is 9.8 m/s² and the time of motion is 1.11 s.
Average acceleration of the lead weight during the impactThe lead weight will fall under the influence of gravity with magnitude of 9.8 m/s².
Time of motion of the lead weightt = √2h/g
where;
h is the total height of fall, h = 6 m + 0.4 cm = 6.004 mg is acceleration due to gravityt = √((2 x 6.004)/9.8)
t = 1.11 s
Thus, the the magnitude of the average acceleration during the impact is 9.8 m/s² and the time of motion is 1.11 s.
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If you were in a spaceship watching a ball hover at rest (inside the spaceship) in mid-air, and the spaceship suddenly began rapidly accelerating, what would you see happen to the ball? Why?
Answers
Answer:
It will still hover until the spaceship "hits" or exerts a force on it.
Explanation:
Remember, if there is no net force, there is no acceleration or movement.
In this case, our ball is hovering in the spaceship, and in space, we can assume there is no [tex]F_g[/tex], and we can assume there is no [tex]F_N[/tex], nor no forces acting against it.
So, the ball would not move.
However, once the spaceship starts accelerating, the ball would still hover until the spaceship exerts a force on it.
This is because of the same thing as explained above, no forces acting on it, therefore, no acceleration.
Think about it this way.
Imagine you jumped up, then someone threw a ball at you. Now let's imagine you can't move until you hit the floor, meaning that in an ideal situation only [tex]F_g[/tex] is acting on you. Now again, let's imagine time slows really down for you, but not the ball. Before the ball comes and hits you, you are "hovering" like a ball. But after the ball hits you, you move a little because the ball exerted a force on you.
If you did not understand what I meant above, just forget about it, and think about the fact that if there is a Net force (all the force values added up), then there is acceleration and movement.
The following table lists the speed of sound in various materials. Use this table to answer the question.
Substance Speed (m/s)
Glass 5,200
Aluminum 5,100
Iron 4,500
Copper 3,500
Salt water 1,530
Fresh water 1,500
Mercury 1,400
Hydrogen at 0°C 1,284
Ethyl Alcohol 1,125
Helium at 0°C 965
Air at 100°C 387
Air at 0°C 331
Oxygen at 0°C 316
Sound will travel fastest in air at _____.
-5°C
0°C
10°C
15°C
Answers
Sound will travel fastest in air at 15°C.
Speed of sound in air
The speed of sound in air, given in the range of 100 degrees Celsius and 0 degree Celsius include;
Air at 100°C 387 m/s
Air at 0°C 331 m/s
From the date above, the speed of sound in air increases with increases in temperature. Thus, Sound will travel fastest in air at 15°C.
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6. Determine the number of significant figure of:
a. 0.2001 b. 2.0000 c. 243 d. 0.00010300
Answers
Answer:
b
Explanation:
the answer is b because if you do the math it equals to b
in a motor, what do the combined effect of electric currents and magnetic forced turn electrical energy into?
mechanical energy
a battery
electromagnetic energy
a generator
(please explain too)
Answers
A rocket takes off from Earth's surface, accelerating straight up at 69.2 m/s2. Calculate the normal force (in N) acting on an astronaut of mass 87.4 kg, including his space suit. (Assume the rocket's initial motion parallel to the +y-direction. Indicate the direction with the sign of your answer.)
Answers
According to Newton's 3rd law, there will be equal and opposite force on the astronaut which is -6048 N
What does Newton's third law say ?The law state that in every action, there will be equal and opposite reaction.
Given that a rocket takes off from Earth's surface, accelerating straight up at 69.2 m/s2. We are to calculate the normal force (in N) acting on an astronaut of mass 87.4 kg, including his space suit.
Let us first calculate the force involved in the acceleration of the rocket by using the formula
F = ma
Where mass m = 87.4 kg, acceleration a = 69.2 m/s2
Substitute the two parameters into the formula
F = 87.4 x 69.2
F = 6048.08 N
According to the Newton's 3rd law, there will be equal and opposite force on the astronaut.
Therefore, the normal force acting on the astronaut is -6048 N approximately
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What is formed when 2 or more elements combine chemically
Answers
Answer:
compound
Explanation:
a substance made from two or more different elements that have been chemically joined. Examples of compounds include water (H2O), which is made from the elements hydrogen and oxygen, and table salt (NaCl), which is made from the elements sodium and chloride.
The Lunar Module could make a safe landing if its vertical velocity at impact is 2.8 m/s or less. Use conservation of energy to determine h in each case. The acceleration due to gravity at the surface of the Moon is 1.62 m/s2.
Suppose that you want to determine the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is zero.
Express your answer to two significant figures and include the appropriate units.
Suppose that you want to determine the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s downward.
Express your answer to two significant figures and include the appropriate units.
Suppose that you want to determine the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s upward.
Express your answer to two significant figures and include the appropriate units.
Answers
(1) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is zero is 0.4 m.
(2) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s downward is 0.25 m.
(3) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s upward is 0.25 m.
Maximum height
v² = u² - 2gh
where;
v is final velocityu is initial velocitywhen the lander's velocity = 0
0 = u² - 2gh
u² = 2gh
h = u²/2g
h = (2.8²)/(2 x 9.8)
h = 0.4 m
when the velocity of the lander is 1.7 m/s downwardh = (u² - v²)/2g
h = (2.8² - 1.7²)/(2 x 9.8)
h = 0.25 m
when the velocity of the lander is 1.7 m/s upwardh = (u² - v²)/2g
h = (2.8² - 1.7²)/(2 x 9.8)
h = 0.25 m
Thus, the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is zero is 0.4 m.
The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s downward is 0.25 m.
The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s upward is 0.25 m.
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A candle 4.75 cm tall is 40.0 cm to the left of a plane mirror.
a) Where is the image formed by the mirror?
- to the right of the mirror
- to the left of the mirror
b) What is the height of this image?
(Express your answer in centimeters.)
Answers
Answer: A candle 4.75 cm tall is 40.0 cm to the left of a plane mirror. Then, the image formed by the mirror will be to the right of the mirror and the height of this image is same as that of the object.
Explanation: To find the answer, we need to know about the reflection at a plane mirror.
What is the reflection at a plane mirror?There are some points about the reflection at a plane surface.
Distance of image from mirror = distance of object from mirror.Size of the image = size of the object.If object moves with certain velocity, then the image also moves with same velocity in opposite direction.Image formed by a plane mirror will be on the opposite side of the object and which is erect, virtual and laterally inverted.Thus, we can conclude that, a candle 4.75 cm tall is 40.0 cm to the left of a plane mirror. Then, the image formed by the mirror will be to the right of the mirror and the height of this image is same as that of the object.
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In response, a plane mirror is 40.0 cm to the left of a candle that is 4.75 cm tall. The mirror's generated image will then be to the right of the mirror and have the same height as the object.
In order to determine the solution, we must understand how a plane mirror reflects light.
What appears in a plane mirror's reflection?There are a few things to note concerning reflections on flat surfaces.The distance of an object from a plane mirror is equal to the distance of its image.Size of the object equals size of the image.If an object moves at a specific speed, the picture will follow suit and move in the opposite direction at the same speed.The other side of the object will have an erect, virtual, and laterally inverted image created by a plane mirror.As a result, we may say that a flat mirror is 40.0 cm to the left of a candle that is 4.75 cm tall. The mirror's generated image will then be to the right of the mirror and have the same height as the object.
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A jet of water squirts out horizontally from a hole near the bottom of the tank, as seen in the figure.
If the hole has a diameter of 3.13 mm, what is the height, h, of the water level in the tank? Assume that x = 1.33 m and y = 1.72 m.
Answers
The height, h, of the water level in the tank is mathematically given as
h=1.3675
What is the height, h, of the water level in the tank?Generally, the equation for kinematics is mathematically given as
S=v_o t+(1/2)at^2
Therefore
y=0+(1/2)gt^2
Where
t=(2y/g)^{1/2}
t=(2*1.72/9.8)^{1/2}
t=0.5925s
The horizontal exit velocity will be given as
[tex]Vx=x/t[/tex]
Therefore
[tex]Vx=\frac{0.3.13}{0.5925}[/tex]
Vx=0.5283
In conclusion, applying Bernoulli's Law the tank bottom and tank surface
P+(1/2)pv0^2+pgh=P+(1/2)pvx^2+pgh
(1/2)(p)(0m/s)^2+pgh=(1/2)pvx^2+pg(0m)
gh=(1/2)vx^2
h=(0.5283m/s)^2/(2)(9.8)
h=1.3675
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what is fundamental quantity
Answers
Answer:
The Fundamental Quantity is a physical quantity that exists independently and cannot be expressed by any other physical quantity.
Explanation:
Examples of these could be : Length, electric current and mass
Hope this helps! :)