Answer:
3.73m
Explanation:
What we are asked to find is the range covered by the fish. This is given by the following equation (1). Range can simply be defined as the horizontal distance covered by a body whose motion freely under gravity is in two dimensions. The motion of the fish is in two dimensions, the vertical dimension and the horizontal dimension.
[tex]R=\frac{u^2sin2\theta}{g}.........(1)[/tex]
where u = 6.5m/s is the initial velocity, g is acceleration due to gravity which is taken as [tex]9.8m/s^2[/tex] and [tex]\theta=30.0^o[/tex].
Substituting these values into equation (1), we obtain the following;
[tex]R=\frac{6.5^2sin2(3)}{9.8}\\R=\frac{42.25sin60}{9.8}\\\\R=\frac{36.59}{9.8}\\R=3.73m[/tex]
The range should be 3.73m.
Important information:A Chinook salmon can jump out of water with a speed of 6.50 m/s . The initial angle of =30.0°
The range refers to the horizontal distance i.e. covered by a body whose motion freely under gravity should be in two dimensions. It can be the vertical dimension and the horizontal dimension.
calculation of the range:[tex]= 6.5sin^2\div 9.8\\\\= 36.59 \div 9.8[/tex]
= 3.73m
learn more about the direction here: https://brainly.com/question/20827436
A ball thrown downward by initial speed of 3m/s. It hits the ground after 5 second.
How high is throwing point?
Let’s discuss some particle physics. In 1897, physicist J.J. Thomson conducted a seminalexperiment with a cathode ray tube. In the experiment, a beam of an unknown particle of chargeqandmassmis shot through the tube onto a fluorescent screen, at some speedvand in thex-direction. Uponimpact the beam leaves a bright spot on the screen. Sincevis very fast, effects of gravity are negligible andthe beam forms essentially a straight line normal to the screen.
To eliminate this deflection, an external uniform magnetic field of magnitudeBis turned on as well. What should be the direction and magnitudeBof this field such that ∆z= 0?
Answer:
The direction of the magnetic field should be in the x and y coordinates alone, and their should be no change towards the z coordinate. Also the magnitude of the magnetic field in the z direction would be zero if there is no magnetic field in that coordinate.
The coordinate should be:
B = bi + bj
What is the relationship Between frequency and sound?
Answer:
The sensation of a frequency is commonly referred to as the pitch of a sound. A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound wave.
Explanation:
Hope this helps : )
Revlew Millikan's Photoelectric Experiment Robert A. Mlkan (1868 1953). although best known for his "oil-drop experiment," which measured the charge of an electron, also perfomed pioneering research on the photoelectric effect. In experiments on lithium, for example, Millikan observed a maximum kinetic energy of 0.550 eV when electrons were ejected with 433.9-nm light. When light of 253.5 m was used, he observed a maximum kinetic energy of 2.57 eV.
Part A What is the work function,W, for lithium, as determined from Milikan's results? Express your answer to three significant figures and include appropriate units.
Part B What maximum kinetic energy do you expet illikan found when he used light with a wavelength of 362.4 TIm? Express your answer to three significant figures and include appropriate units Value Units
Answer:
A.) Work function = 2.3 eV
B.) Max. K.E observed = 1.1 eV
Explanation:
A.) Millikan observed a maximum kinetic energy of 0.550 eV when electrons were ejected with 433.9-nm light. When light of 253.5 m was used, he observed a maximum kinetic energy of 2.57 eV.
work function (f) is the minimum energy required to remove an electron from the surface of the material.
hf = Ø + K.E (maximum)
Where
h = Plank constant 6.63 x 10-34 J s
Ø = work function
hc/λ = Ø + K.E (max)
(6.63×10^-34 × 3×10^8)/433.9×10^-9 = Ø + 0.550 × 1.6×10^-19
4.58×10^-19 = Ø + 8.8×10^-20
Ø = 4.58×10^-19 - 8.8×10^-20
Ø = 3.7 × 10^-19 J
Converting Joule to eV
Ø = 3.7 × 10^-19/1.6×10^-19
Ø = 2.3 eV
B.) When light of wavelength 362.4 m is used
The maximum K.E observed = incident light K.E - (the work function).
Incident K.E = hf = hc/λ
Incident K.E =
(6.63×10^-34 × 3×10^8)/362.4
Incident K.E = 5.5 × 10^-28J
Let's convert joule to eV
Incident K.E = 5.5×10^-28/1.6×10^-19
Incident K.E = 3.4 × 10^-9
Max. K.E observed = 3.4 - 2.3
Max. K.E observed = 1.1 eV
Thinking about the winter we missed out on this year. Calvin and his tiger go sledding down a snowy hill. There is friction between the snow and the sled, and there is air resistance. Looking at the entire time that they spend on the hill (from when they start from rest at the top of the hill until they reach the bottom moving fast), indicate whether each of the following quantities is positive, negative, or zero, by writing either, -, or 0 for the questions below. We take Calvin, the tiger, and the sled as our system.
a. The work done by the normal force that the snow exerts on the system?
b. The work is done by the frictional force that the snow exerts on the system?
c.The change in the kinetic energy of the system?
d.The change in the gravitational potential energy of the system?
e. The total work is done on the system?
Answer:
a) W=0, b) Work is negative, c) work is positive and scientific energy variation is positive, d) the variation of the potential enrgy is negative,
e) total work is positive
Explanation:
Work in physics is defined by the scalar scalar product of force by displacement
W = F. dx
The bold are vectors; this can be written in the form of the mules of the quantities
W = F dx cos θ
where θ is the angle between force and displacement.
a) The normal force is perpendicular to the inclined plane which is perpendicular to the displacement, therefore the angle is
θ = 90 cos 90 = 0
W=0
In conclusion the work is zero
b) The friction force opposes the displacement whereby the angle is
θ = 180 cos 190 = -1
W = - fr d
Work is negative
c) To calculate the change in kinetic energy we use that the work is equal to the variation of the kinetic energy
m g sin θ L = ΔK
this magnitude is positive since the angle is zero cos 0 = 1
how the system starts from rest ΔK = Kf -K₀= + Kf -0
work is positive and scientific energy variation is positive
d) change in potential energy
The potential energy is is ΔU = Uf -U₀
we fix the reference system in the bases of the plane so Uf = 0
ΔU = -U₀
the variation of the potential enrgy is negative
e) The total work is formed by the work of the weight component, the work of the friction force
W_Total = W_weight - W_roce
as the body moves down
W_Total> 0
Therefore the total work is positive
Which shows evidence of active transport?
A scientist places four identical cells into four different
liquids, each with different concentrations of magnesiuni.
Celll
w
Description of Liquid
Slightly more magnesium than the
cell
The least amount of magnesium
O
O
O
O
cells W and Z
cell W only
cell Y only
cells X and Y
Result
Took in
magnesium
Took in
magnesium
Took in
magnesium
Took in
magnesium
Slightly less magnesium than the
cell
The most amount of magnesium
Answer: D
X and Y
Explanation:
X The least amount of magnesium Took in magnesium
Y Slightly less magnesium than the cell Took in magnesium
Because active transport occurs when ions or molecules move from less concentration region to high concentration region through semi membrane with the help of some energy.
Answer:
Cells X and Y
Explanation:
Active transport occurs when a substance moves across a membrane against its concentration gradient.
Cells W and Z were placed in a liquid containing more magnesium than the cells. Magnesium therefore, diffuses passively down it's concentration gradient into the cells.
However, cells X and Y were placed in a solution containing less magnesium than the cells, yet these cells took in magnesium against this concentration gradient. This, shows that active transport had taken place.
A spring with a 6-kg mass and a damping constant 7 can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons. Suppose the spring is stretched 1 meters beyond its natural length and then released with zero velocity, In the notation of the text, what is the value c2−4mk? m2kg2/sec2 Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t with the general form c1eαtcos(βt)+c2eγtsin(δt)
Answer:
The value for [tex]c^2 - 4mk[/tex] is : [tex]\mathbf{-23 \ m^2kg^2/sec^2}[/tex]
The position of the mass (m) after t seconds is:
[tex]\mathbf{x(t) = e^{\frac{1}{2}t }( cos \frac{\sqrt{36}}{12}t + \frac{6}{\sqrt{36}} sin \frac{\sqrt{36}}{12}t)}[/tex]
Explanation:
The spring constant is :
[tex]F = kx \\ \\ k = \frac{F}{x} \\ \\ k = \frac{1.5}{0.5} \\ \\ k = 3 \ N/m[/tex]
The value for [tex]c^2 - 4mk[/tex] is :
= [tex]7^2 - 4(6)(3)[/tex]
= [tex]49 - 27[/tex]
= [tex]\mathbf{-23 \ m^2kg^2/sec^2}[/tex]
The differential equation for this system is :
[tex]m \frac{d^2x}{dt^2}+c \frac{dx}{dt}+kx = 0[/tex]
[tex]6 \frac{d^2x}{dt^2}+7\frac{dx}{dt}+3x = 0[/tex]
and the auxiliary equation for this differential equation is :
[tex]6m ^2+ 6m + 3 = 0[/tex]
using the quadratic formula :
[tex]\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}[/tex]
[tex]\frac{-6 \pm \sqrt{6^2 - 4(6)(3)} }{2(6)}[/tex]
= [tex]\frac{-6 \pm \sqrt{-36} }{12}[/tex]
= [tex]-\frac{1}{2} \pm \frac{\sqrt{36} }{12}i[/tex]
The general solution is :
[tex]x(t) = e^{-\frac{1}{2}t}}(c_1 cos {\frac{\sqrt{36} }{12}} t + c_2 sin \frac{\sqrt{36} }{12} t})[/tex]
Initial conditions : [tex]x(0) = 1 \ m , x' (0) = 0\\[/tex]
[tex]x(0) = ( c_1 cos 0 + c_2 sin 0)[/tex]
[tex]1 = c_1[/tex]
[tex]x't = e^{- \frac{1}{2}t}}}(- c_1 \frac{\sqrt{36} }{12} sin \frac{\sqrt{36} }{12}t + c_2 \frac{\sqrt{36} }{12} cos \frac{\sqrt{36} }{12}t) - e^{- \frac{1}{2}t}}} (\frac{1}{2}) (c_1cos \frac{\sqrt{36} }{12} t +c_2 sin \frac{\sqrt{36} }{12} t)[/tex]
[tex]x'(0)= e^{- \frac{1}{2}t}}}(- c_1 \frac{\sqrt{36} }{12} sin0 + c_2 \frac{\sqrt{36} }{12} cos 0) - e^{- \frac{1}{2}t}}} (\frac{1}{2}) (c_10+c_2 sin0)[/tex]
[tex]x'(0) = c_2 \frac{\sqrt{36} }{12}-c_1 \frac{1}{2}[/tex]
replacing 1 for [tex]c_1[/tex]
[tex]0 = c_2 \frac{\sqrt{36} }{12} -\frac{1}{2}[/tex]
[tex]c_2 \frac{\sqrt{36} }{12} = \frac{1}{2}[/tex]
[tex]c_2 = \frac{6}{\sqrt{36} }[/tex]
The position of the mass (m) after t seconds is:
[tex]\mathbf{x(t) = e^{\frac{1}{2}t }( cos \frac{\sqrt{36}}{12}t + \frac{6}{\sqrt{36}} sin \frac{\sqrt{36}}{12}t)}[/tex]
Explain why you cannot charge one end of a steel rod and leave the other end uncharged
Answer:
Electrons can be made to move from one object to another. However, protons do not move because they are tightly bound in the nuclei of atoms.Static charge occurs when electrons build up on an object. Static charge:
can only build up on objects which are insulators, eg plastic or wood
cannot build up on objects that act as conductors, eg metals
Conductors allow the electrons to flow away, forming an electric current.
When a static charge on an object is discharged, an electric current flows through the air. This can cause sparks. Lightning is an example of a large amount of static charge being discharged.
Explanation:
Batman and Robin are attempting to escape that dastardly villain, the Joker, by hiding in a large pool of water (refractive index nwater = 1.333). The Joker stands gloating at the edge of the pool. (His makeup is watersoluble.) He holds a powerful laser weapon y1 = 1.49 m above the surface of the water and fires at an angle of θ1 = 27◦ to the horizontal. He hits the Boy Wonder squarely on the letter "R", which is located y2 = 3.77 m below the surface of the water. θ x y y 1 1 2 R J Batplastic surface Mirrored Surface water B How far (horizontal distance) is Robin from the edge of the pool? (Fear not, Batfans. The "R" is made of laser-reflective material.) Answer in units of m.
Answer:
x_total = 4.29m
Explanation:
To solve this exercise we must work in parts. Let's use the law of refraction to find the angle of the refracted ray and trigonometry to find the distances.
Let's start by looking for the angles that the laser refracts
n₁ sin θ₁ = n₂ sin θ₂
where n₁ is the air refraction compensation n₁ = 1, n₂ the water refractive index n₂ = 1,333
θ₂ = sin⁻¹ (n₁ sin θ₁/n₂)
θ₂ = sin⁻¹ (1 sin 27 / 1,333)
θ₂ = sin⁻¹ 0.34057
θ₂ = 19.9º
now let's find the distance from the edge of the pool to the point where the ₂lightning strikes the water
tan θ₁ = y₁ / x₁
x₁ = y₁ / tan θ₁
x₁ = 1.49 / tan 27
x₁ = 2,924 m
Now let's look for the waterfall in the water as far as Robin
tan θ₂₂ = y₂ / x₂
x₂ = y₂ / tan θ₂
x₂ = 3.77 / tan 19.9
x₂ = 1,364
the distance from the edge of the pool to Robin is
x_total = x₁ + x₂
x_total = 2,924 + 1,364
x_total = 4.29m
Tidal forces are gravitational forces exerted on different parts of a body by a second body. Their effects are particularly visible on the earth's surface in the form of tides. To understand the origin of tidal forces, consider the earth-moon system to consist of two spherical bodies, each with a spherical mass distribution. Let re be the radius of the earth, m be the mass of the moon, and G be the gravitational constant.
(a) Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration ae of the earth due to the gravitational pull of the moon? Express your answer in terms of G, m, and r.
(b) Since the gravitational force between two bodies decreases with distance, the acceleration a_near experienced by a unit mass located at the point on the earth's surface closest to the moon is slightly different from the acceleration a_far experienced by a unit mass located at the point on the earth's surface farthest from the moon. Give a general expression for the quantity a_near - a_far. Express your answer in terms of G, m, r, and re.
Answer:
Explanation:
radius of earth = re
mass of the noon = m
mass of the earth = E
distance between earth and moon = r
acceleration of earth ae
force on earth = GMm / r²
acceleration of the earth
ae = force / mass
= GMm / (r² x M )
= Gm / r²
b ) The point on the earth nearest to moon will be at a distance of r - re
a_near = Gm / ( r - re)²
The point farthest on the earth to moon will be at a distance of r + re
a_ far = Gm / ( r + re )²
A car is travelling at 16.7m/s. If the car can slow at a rate of 21.5m/s^2, how much time does the driver need in order to stop the red light?
Answer:
About 0.7767 seconds
Explanation:
[tex]\dfrac{16.7m/s}{21.5m/s^2}\approx 0.7767s[/tex]
Hope this helps!
It has been proposed that we could explore Mars using inflated balloons to hover just above the surface. The buoyancy of the atmosphere would keep the balloon aloft. The density of the Martian atmosphere is 0.0154 (although this varies with temperature). Suppose we construct these balloons of a thin but tough plastic having a density such that each square meter has a mass of 4.60 . We inflate them with a very light gas whose mass we can neglect. So far I found the following: What should be the radius of these balloons so they just hover above the surface of Mars? Radius of the balloon = /896 m What should be the mass of these balloons so they just hover above the surface of Mars? Mass of balloon = 4.64*10^-2 kg If we released one of the balloons from part A on earth, where the atmospheric density is 1.20 , what would be its initial acceleration assuming it was the same size as on Mars? If on Mars these balloons have five times the radius found in part A, how heavy an instrument package could they carry?
Answer:
the radius of the balloon r = 0.896 m and mass m = 4.64 × 10⁻² kg
the initial acceleration is a = 753.47 m/s²
the an instrument package could they carry a required mass of 4.64 kg
Explanation:
a) What should be the radius of these balloons so they just hover above the surface of Mars?
Given that :
The density of the Martian atmosphere, ρ = 0.0154 kg/m³
The volume of the sphere, V = (4/3)πr³
The area of the sphere, A = 4πr²
The mass of the balloon is m = (4.60 g/m²)A
m = (4.60×10⁻³ kg/m²)(4πr²)
The formula for the buoyant force is expressed as :
F = ρVg
m×g = ρ×V×g
m = ρ×V
Now;
(4.60×10⁻³ kg/m²)(4πr²)= ρ(4/3)πr³
r = 3(4.60×10⁻³ kg/m²)/ ρ
r = 3(4.60×10⁻³ kg/m²)/ 0.0154 kg/m³
r = 0.896 m
Thus; the radius of the balloon r = 0.896 m
The mass of the balloon is (4.60×10⁻³ kg/m²)(4πr²)
m = (4.60×10⁻³ kg/m²)(4π×0.896²)
m = 4.64 × 10⁻² kg
b) what would be its initial acceleration assuming it was the same size as on Mars?
The density of the air on earth, ρ = 1.20 kg/m³
The volume of the balloon is V = (4/3)(π)(0.896 m)³
V = 3.01156 m³
Considering the net force acting on the balloon ; we have
ΣF = ρVg - mg = ma
However; making the initial acceleration a of the balloon the subject ; we have:
a = (ρVg - mg)/m
a = (1.20 kg/m³)(3.01156 m³)(9.8 m/s²) - (4.64×10⁻² kg)(9.8 m/s²)]/(4.64×10⁻² kg)
a = 753.47 m/s²
c) If on Mars these balloons have five times the radius found in part A, how heavy an instrument package could they carry?
The volume of the total system is V' = (4/3)π(5r)³
V' = (4/3)π(5)³(0.896 m)³
V' =376.446 m³
The mass of the total system is m = (4.60×10⁻³ kg/m²) (4π(5r)²)
m = [4.60×10⁻³ kg/m²][4π][25](0.896 m)²
m = 1.159587 kg
We can then say that the buoyant force is equals to the weight of the total mass (balloon+load) and is expressed as:
F = (m + m')g
ρV'g = (m + m')g
ρV' = (m + m')
Thus; the required mass m' is = ρV' - m
m' = ρV' - m
m' = (0.0154 kg/m³)(376.446 m³) - (1.159587 kg)
m' = 4.64 kg
Thus; the an instrument package could they carry a required mass of 4.64 kg
A metal ring 5.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.220 T/s.
Required:
What is the magnitude of the electric field induced in the ring?
Answer:
Ein: 2.75*10^-3 N/C
Explanation:
The induced electric field can be calculated by using the following path integral:
[tex]\int E_{in} dl=-\frac{\Phi_B}{dt}[/tex]
Where:
dl: diferencial of circumference of the ring
circumference of the ring = 2πr = 2π(5.00/2)=15.70cm = 0.157 m
ФB: magnetic flux = AB (A: area of the loop = πr^2 = 1.96*10^-3 m^2)
The electric field is always parallel to the dl vector. Then you have:
[tex]E_{in}\int dl=E_{in}(2\pi r)=E_{in}(0.157m)[/tex]
Next, you take into account that the area of the ring is constant and that dB/dt = - 0.220T/s. Thus, you obtain:
[tex]E_{in}(0.157m)=-A\frac{dB}{dt}=-(1.96*10^{-3}m^2)(-0.220T/s)=4.31*10^{-4}m^2T/s\\\\E_{in}=\frac{4.31*10^{-4}m^2T/s}{0.157m}=2.75*10^{-3}\frac{N}{C}[/tex]
hence, the induced electric field is 2.75*10^-3 N/C
I Science Sem 1
1 7.1.2 Exam: Semester 1 Exam
Question 18 of 30
3 Points
How much force is required in Newtons) to accelerate a 12-kg bicycle, along
with its 50-kg rider, at 2 m/s2?
Give your answer as a number.
Answer here
Answer:
Force, F = 124 N
Explanation:
We have,
Mass of bicycle is 12 kg and mass of rider is 50 kg
Total mass of the system is 12 kg + 50 kg = 62 kg
Acceleration of the system is 2 m/s²
It is required to find the force required to accelerate the system. The force acting on an object is given by :
[tex]F=ma\\\\F=62\ kg\times 2\ m/s^2\\\\F=124\ N[/tex]
So, the force of 124 N is acting on the system.
Sonia was experimenting with electric charges. She tied two inflated balloons together, held them next to each other, and rubbed both with a piece of wool.
What did Sonia observe, and why?
Answer: They will repel each other.
Explanation:
Two inflated balloons when rubbed with woolen cloth will lead to repeal each other because of the similar charges on both the balloons.
Rubbing both the balloons together by the woolen cloth will introduce negative charge in the balloons.
As, we know that the same charges repeal each other both of the balloons with be apart from each other.
This is due to the static electricity, the negatively charged particles jump to the positive one. When balloons are rubbed they become negatively charged.
Answer:
Sample Response: Sonia observed that the two balloons repelled each other. This is because both balloons acquired the same charge when she rubbed them with the piece of wool, and like charges repel each other.
Explanation:
Did it on Egde 2020
Which best explains why making a pancake from batter is an example of a chemical change?
-The pancake that forms is a different state of matter.
-The change from batter to pancake can be reversed.
-A new substance forms when the batter is cooked.
-The batter changes shape when it is cooked.
The correct answer is C. A new substance forms when the batter is cooked.
Explanation:
When a chemical change occurs the properties, and composition of substances change. This means atoms in the substance re-arrange to form a new substance. This only occurs when there is a chemical change, but not when physical changes occur, indeed a physical change only affects the state of the matter, shape, size, etc.
In the case of the pancake, this is an example of a chemical change because though the process of cooking the pancake changes its composition. Due to this, the properties of the cooked pancake, and the butter are not the same as a new substance forms. Also, in this and most chemical changes, reversibility is not possible, that is why you cannot reverse the process and make the cooked pancake batter once again.
Answer:
C. on edge
Explanation:
In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors
A =(1,0,−3), B =(−2,5,1), and C =(3,1,1). Calculate the following, and express your answers as ordered triplets of values separated by commas.
a. A- B=______
b. B-C=_______
c. -A +B- C=______
d. 3A- 2C=______
e. -2A+3B- C=_____
f. 2A- 3(B-C)
Answer:
a)A-B =(1,0,-3) +(2,-5,-1) =(3, -5, -4);
b)B-C =(-2-3, 5-1, 1-1) = (-5,4,0);
c)-A +(B –C) = (-1,0,3) +(-5,4,0) =(-6,4,3);
d) =(3,0,-9) –(6,2,2) = (-3, -2, -11);
e) = (-2,0,6) +(-6,15,3) +(-3,-1,-1) =
d) = (2,0,-6) –(-5*3, 4*3, 0) =
Explanation:
1)C=k*A = (k*a1, k*a2, k*a3);
2)C=-A = -1*A;
3)C= A+B = (a1+b1, a2+b2, c2+c2);
4)C=A-B =A +(-B);
5)A+B=B+A;
6)A+B+C =(A+B)+C =A+(B+C
Consider a spinning plate is dropped onto a stationary plate (which is itself at rest on a frictionless surface). Both plates have a radius of 30cm and a mass of 1kg. The spinning plate is initially spinning at a rate of 0.7 revolutions per second. Hint: This is like a totally-inelastic collision.
Required:
a. After a sufficiently long time, what is the angular velocity of the initially-spinning plate? What about the initially-stationary plate?
b. Assume that the period of velocity matching happens over a course of 2 seconds. Further, assume that the torque exerted by each plate on the other is constant over time. In that case, what is the magnitude of the acceleration that each plate feels during those two seconds? Hint: Use the rotational impulse-momentum theorem.
Answer:
The final angular velocity is [tex]w_f = 2.1994 rad/sec[/tex]
The angular acceleration is [tex]\alpha = 1.099 \ rad/sec^2[/tex]
Explanation:
From the question we are told that
The radius of each plate is [tex]r = 30 \ cm = \frac{30}{100} = 0.3 \ m[/tex]
The mass of each plate is [tex]m_p = 1 \ kg[/tex]
The angular speed of the spinning plate is [tex]w = 0.7 \ rev \ per \ sec = 0.7 * 2 \pi = 4.3988 \ rad/sec[/tex]
From the law of conservation of momentum
[tex]L_i = L_f[/tex]
Where [tex]L_i[/tex] is the initial angular momentum of the system (The spinning and stationary plate ) which is mathematically represented as
[tex]L_i = I_1 w + 0[/tex]
here [tex]I_1[/tex] is the moment of inertia of the spinning plate which mathematically represented as
[tex]I_1 = \frac{m_pr^2}{2}[/tex]
and the zero signify that the stationary plate do not have an angular momentum as it is at rest at the initial state
[tex]L_f[/tex] is the final angular momentum of the system (The spinning and stationary plate) , which is mathematically represented as
[tex]L_f = (I_1 + I_2 ) w_f[/tex]
Where
[tex]I_2[/tex] is the moment of inertia of the second plate (This was stationary before but now it spinning due to the first pate ) and is equal to [tex]I_1[/tex]
and [tex]w_f[/tex] is the final angular speed
So we have
[tex]I_1 w = (I_1 + I_2)w_f[/tex]
[tex]\frac{m_p r^2}{2} * w = 2 * \frac{m_p r^2}{2} * w_f[/tex]
[tex]w = 2 * w_f[/tex]
substituting values
[tex]4.3988 = 2 * w_f[/tex]
[tex]w_f = \frac{4.3988 }{2}[/tex]
[tex]w_f = 2.1994 rad/sec[/tex]
The the rotational impulse-momentum theorem can be mathematially represented as
[tex]\tau * \Delta t = 0.09891[/tex]
Where [tex]\tau[/tex] is the torque and [tex]\Delta t[/tex] is the change in time
So at [tex]\Delta t = 2 \ sec[/tex]
[tex]\tau = \frac{0.09891}{2}[/tex]
[tex]\tau = 0.0995 \ Nm[/tex]
now the angular acceleation is mathematically represented as
[tex]\alpha = 2 * \frac{\tau}{m_p * r^2 }[/tex]
substittuting values
[tex]\alpha = 2 * \frac{0.0995}{1 * 0.3^2}[/tex]
[tex]\alpha = 1.099 \ rad/sec^2[/tex]
If the frequency of a wave is tripled, what happens to the period of the wave?
Answer:
if the frequency of the wave if tripled then period of wave gets tripled
When the wave frequency is tripled, the period of the wave becomes one-third of its original.
Relationship between frequency and period of a waveThe frequency (f) of a wave is inversely proportional to the period (T) of the wave.
Mathematically,
[tex]$f \propto \frac{1}{T}$[/tex]
Thus, when frequency increases, the period decrease and when frequency decreases, the period of the wave increases.
So when frequency of a wave increases by three time, the period of the wave decreases by three times.
Hence, when the frequency is tripled, the period of the wave becomes one-third of its original value.
Learn more about "frequency and period" here :
https://brainly.com/question/18657169
Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiempo alcanzará el coche a la furgoneta? ¿A qué distancia se producirá el encuentro?
Answer:
t = 0.33h = 1200s
x = 18.33 km
Explanation:
If the origin of coordinates is at the second car, you can write the following equations for both cars:
car 1:
[tex]x=x_o+v_1t[/tex] (1)
xo = 10 km
v1 = 55km/h
car 2:
[tex]x'=v_2t[/tex] (2)
v2 = 85km/h
For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:
[tex]x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}[/tex]
[tex]t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s[/tex]
The position in which both cars coincides is:
[tex]x=(55km/h)(0.33h)=18.33km[/tex]
A string, fixed at both ends and 2.14 m long, is vibrating in its second harmonic. It excites a 0.5-m long pipe closed on one end into its third harmonic (first overtone). Use a speed of sound in air of 345 m/s if needed. (a) Sketch the normal modes given above for both the string and the closed-open pipe. [8] (b) What is the wave speed on the string? [7] (c) Write down the wave function y(x,t) for the 2nd harmonic wave on the string. Assume an arbitrary amplitude. [10] (d) If the mass density of the string is 0.55 g/m, what is the tension in the string? [5]
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
How much time is needed to push a 5,000 N car 50 meters if you are using a machine with a power of 4,500 W?
Answer:
55.56
Explanation:
5000N * 50 = 250000/4500= 55.55555555 or 55.56
Why will a change in velocity have a greater affect on KE than a change in mass?
Answer: the answer is corect of what everone else said
Explanation:
When you turn down the volume on the television, you reduce the _______ carried by the sound waves, so you also reduce their ________
Answer:
when you turn down the volume on the television, you reduce the intensity carried by the sound waves, so you also reduce their amplitude.
Explanation:
When you turn down the volume of the television, you are actually reducing the intensity of the sound wave, which is directly proportional to the amplitude of the sound. Amplitude is height of the sound wave.
Therefore, when you turn down the volume on the television, you reduce the intensity carried by the sound waves, so you also reduce their amplitude.
Explain the force responsible for the formation of our solar system.
Answer:
gravity
Approximately 4.5 billion years ago, gravity pulled a cloud of dust and gas together to form our solar system.
Explanation:
Which type of power plant uses the movement of air in nature to generate
electricity?
A. Radiant
B. Wind
C. Coal
D. Hydroelectric
Answer:
Wind
Explanation:
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
Answer:
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
b) Determine the magnitude and direction of current in wire 3
Explanation:
a) [tex]F_{net} \text {on wire }3=0[/tex]
[tex]\frac{\mu_0 I_1 I_3}{2 \pi x} = \frac{\mu I_2 I_3}{2 \pi (0.2+x)} \\\\\frac{1.5}{x} =\frac{4}{0.2+x} \\\\0.03+1.5x=4x\\\\x=0.012m\\\\=1.2cm[/tex]
position of wire = 50 - 1.2
= 48.8cm
b) [tex]F_{net} \text {on wire }1=0[/tex]
[tex]\frac{\mu _0 I_1 I_3}{2 \pi (1.2)} = \frac{\mu _0 I_1 I_2}{2 \pi (20)} \\\\\frac{I_3}{1.2} =\frac{4}{20} \\\\I_3=0.24A[/tex]
Direction ⇒ downward
Some students conduct an experiment to find out how long different rubber balls would keep bouncing before they stop, when they dropped from a fixed height. which unit would be the most suitable for measuring the results of their experiment?
A. hours
B. kilograms
C. metres
D. seconds
What does the speed of sound depend on ?
Answer:
As with any wave the speed of sound depends on the medium in which it is propagating.
Explanation:
Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes the moon to orbit the earth. The force of gravitational attraction is represented by the equationF=Gm1m2r2where F is the magnitude of the gravitational attraction on either body, m1 and m2 are the masses of the bodies, r is the distance between them, and G is the gravitational constant. In SI units, the units of force are kgâ‹…m/s2, the units of mass are kg, and the units of distance are m. For this equation to have consistent units, the units of G must be which of the following?Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes the moon to orbit the earth. The force of gravitational attraction is represented by the equation,where is the magnitude of the gravitational attraction on either body, and are the masses of the bodies, is the distance between them, and is the gravitational constant. In SI units, the units of force are , the units of mass are , and the units of distance are . For this equation to have consistent units, the units of must be which of the following?A. kg3mâ‹…s2B. kgâ‹…s2m3C. m3kgâ‹…s2D. mkgâ‹…s2
Answer:
The S.I unit of G must be m³/kg.s² to keep the equation consistent.
Explanation:
We have the equation:
F = Gm₁m₂/r²
where,
F = Gravitational Force between Two Bodies
G = Gravitational Constant
m₁ = Mass of 1st Body
m₂ = Mass of 2nd body
r = Distance between the Bodies
The S.I Units of the quantities are:
F = Newton = kg.m/s²
m₁ = kg
m₂ = kg
r = meter = m
G = ?
Therefore, to find the units of Gravitational Constant (G), we substitute the known units in the formula:
kg.m/s² = G(kg)(kg)/m²
G = m³/kg.s²