Answer:
the correct answer is d
Explanation:
The laws of mechanics are related
F = m a
the acceleration of the body is given by the kinematics
a = [tex]\frac{dv}{dt}[/tex]
v = [tex]\frac{dx}{dt}[/tex]
substituting
a = \frac{d2x}{dt^2}
F = m [tex]\frac{d^2x}{dt^2}[/tex]
Therefore, in order to obtain the force (interaction of a body), continuous curves are needed and derivable from the position and the speed, for which all change in the trajectory of a body must be smooth where smooth is understood to have until the second derived.
Consequently the correct answer is d
A car traveling 85 km/h is 250 m behind a truck
traveling 73 km/h.
Time needed = t = 20.83 s
Further explanationGiven
car speed = 85 km/h
truck speed = 73 km/h
Required
the time it takes for the car to reach the truck
Solution
When the car reaches the truck, the distance between them will be the same
x car - 250 m = x truck
General formula for distance (d) :
d = v.t
So the equation becomes :
85t-250 = 73t
12t=250
t = 20.83 s
Green plants need light in order to survive. Structures in the leaves absorb light, which in turn, helps plants make
their own food
Under which color of light will plants be least likely to make food?
ted
blue
orange
Save and Exit
Answer:
I don't know lol good luck i guess
What is sin(77)?
Α. 0.77
Β. 4.33
Ο Ο Ο
C. 0.22
Ο D. 0.97
Answer:
0.97 id the correct answer
Explanation:
what is efficiency?
Explanation:
the state or quality of being efficient or able to accomplish something
what is the definition of power ? what are the units of power?
Answer:
power is the rate at which energy is transferred or converted
the unit of power is Watts
A car travels 100 km due East in 2 hours. It then travels 50 km South in 1 hour. What is its average velocity?
The average velocity of the car is 37.27 km/h.
The given parameters;
Initial displacement of the car, x = 100 kmTime of motion, t = 2 hoursFinal displacement of the car, y = 50 kmtime of motion, t = 1 hourThe average velocity of the car is calculated as follows;
[tex]average \ velocity = \frac{Total \ displacement}{Total \ time}[/tex]
The total displacement of the car is calculated as follows;
[tex]d = \sqrt{x^2 + y^2} \\\\d = \sqrt{100^2 \ + \ 50^2} \\\\d = 111.803 \ km[/tex]
The average velocity of the car is calculated as follows;
[tex]v = \frac{111.803}{3} \\\\v = 37.27 \ km/h[/tex]
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3. One effective method for coping with change is using:
Recall
Revisitation
Reminiscence
Reminding
Cope up with changes is significant in everyone's life. One of the suitable way is to remind ourselves that we can withstand with the changes.
What is reminder?Reminding a thing that is important to anyone or ourselves until we do it a better way. Thinking past time and wasting time is just ridiculous. Because we must cope up with the current scenario and build up our mental strength.
Our lives are constantly impacted by changes. Change is a constant, uncontrolled aspect of life that has an impact on our work ethics and routines.
Anything that changes may have major or little repercussions. As a result, since the extent of change's first consequences is uncertain, we must refrain from discriminating against it.
Organizations can effectively use change as a tool. Without change, people would be hesitant to form fresh perspectives and ideas. They wouldn't be able to respond to daily obstacles and would become inert in their work and their roles.
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Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 12.0 mg piece of tape held 1.00 cm above another.
Answer:
36.1 nC
Explanation:
The electrostatic force F = kqq'/r² since q = q', F = kq²/r² where q = charge on tape, r = distance between tapes = 1.00 cm = 1 × 10⁻² m and k = 9 × 10⁹ Nm²/C².
Given that F = weight of 12.0 mg piece of tape, F = mg where m = mass of tape = 12.0 mg = 12 × 10⁻³ kg and g = acceleration due to gravity = 9.8 m/s²
So, kq²/r² = mg
q²/r² = mg/k
q² = mgr²/k
taking square-root of both sides,
q = √(mg/k)r
So, substituting the values of the variables into the equation, we have
q = √(mg/k)r
q = √(12 × 10⁻³ kg × 9.8 m/s²/ 9 × 10⁹ Nm²/C²)1 × 10⁻² m
q = √(117.6 × 10⁻³ kgm/s²/9 × 10⁹ Nm²/C²)1 × 10⁻² m
q = √(13.07 × 10⁻¹² C²/m²)1 × 10⁻² m
q = 3.61 × 10⁻⁶ C/m × 1 × 10⁻² m
q = 3.61 × 10⁻⁸ C
q = 36.1 × 10⁻⁹ C
q = 36.1 nC
A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 1.1 m from the axis of rotation, and he rotates with angular speed of 0.64 rad/sec. The moment of inertia of the student plus the stool is 4 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.33 m from the rotation axis.
Required:
a. Find the new angular speed of the student.
b. Find the kinetic energy of the student before and after the objects are pulled in.
Answer:
a) ω₁ = 0.97 rad/sec
b) K₀ = 1.31 J
Kf = 1.99 J
Explanation:
a)
Assuming no external torques acting on the student, total angular momentum must be conserved, as follows:[tex]L_{o} = L_{f} (1)[/tex]
The angular momentum of a rigid body rotating respect an axis of rotation can be written as follows:[tex]L = I*\omega (2)[/tex]
In order to get ωf, since ω₀ = 0.64 rad/sec, we need to find the values of the initial moment of inertia, I₀, and the final one, If:I₀ = 4 kg*m² + 1kg*(1.1 m)² + 1kg*(1.1m)² = 6.42 kg*m² (3)If = 4 kg*m² + 1kg*(0.33m)² + 1kg*(0.33m)² = 4.22 kg*m² (4)Replacing (3), (4) in (1) we can solve for ωf:[tex]\omega_{f} = \frac{I_{o} *\omega_{o} }{I_{f} } = \frac{6.42kgm2*0.64rad/sec}{4.22kgm2} = 0.97 rad/sec (5)[/tex]
b)
Since the student is not translating but he is only rotating, all his kinetic energy is rotational kinetic energy.The expression for the kinetic energy of a rotating rigid body, around an axis of rotation is as follows:[tex]K_{rot} = \frac{1}{2} * I * \omega^{2} (6)[/tex]
The initial kinetic energy of the student, before the objects are pulled in, is as follows:[tex]K_{roto} = \frac{1}{2} *I_{o} * \omega_{o} ^{2} = \frac{1}{2}* 6.42kgm2*(0.64rad/sec)^{2} = 1.32 J (7)[/tex]
The final kinetic energy is given by the following expression:[tex]K_{rotf} = \frac{1}{2} *I_{f} * \omega_{f} ^{2} = \frac{1}{2}* 4.22kg*m2*(0.97rad/sec)^{2} = 1.99 J (8)[/tex]How high will a 1-kilogram ball go if it is thrown straight up into the air with an initial velocity of 30 m/sec?
h max = 45.92 m
Further explanationGiven
1 kg ball
vo=initial velocity = 30 m/s
Required
Height
Solution
We can use the principle of energy conservation or parabolic motion
For parabolic motion
h max = vo²sin²θ/2g(θ= 90° for straight up)
For conservation energy :
KE₁=PE₂
1/2.m.vo² = m.g.h
We choose parabolic motion
h max = vo²sin²θ/2g
h max = 30²sin²90/2 x 9.8
h max = 45.92 m
physical properties for soap
Answer:
there are germs in your hand and soap has chemicals that kill germs
Answer:
dino nuggies
Explanation:
cause whynot
Two soccer players kick a soccer ball back and forth along a straight line. The first player kicks the ball 11 m to the right to the second player. The second player kicks the ball to the left weakly; it only moves 2.7 m before stopping. (Consider the right to be the positive direction. Where applicable, indicate the direction with the sign of your answer.)
(a) What is the total distance that the ball moved?
m
(b) At the end, what is the displacement of the ball (from the first player)?
m
Answer:
a) 13.7 m
b) 8.3 m
Explanation:
a)
The total distance traveled by the ball is just the sum of the distances traveled after being kicked by the first player (11 m.) plus the distance moved after being kicked by the second player (2.7 m), independent of the direction of the movement:[tex]d = d_{1} + d_{2} = 11.0 m + 2.7 m = 13.7 m (1)[/tex]
b)
The total displacement of the ball, is just the difference between the final and initial position, by definition.If we choose the origin to be in the place that the ball is kicked by the first player,[tex]x_{o} = 0 (2)[/tex]
The final position along the x-axis, is just the distance traveled to the right (11 m) minus the distance traveled to the left (considering the right to be the positive direction), 2.7 m:[tex]x_{f} = x_{1} - x_{2} = 11.0 m - 2.7 m = 8.3 m (3)[/tex]
⇒ Δx = xf - x₀ = 8.3 m - 0 m = 8.3 m
Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 2 seconds. Through experiments carried out four centuries ago, Galileo discovered that the distance fallen by any freely falling body is proportional to the square of the time it has been falling. (This model for free fall neglects air resistance.) If the distance fallen after t seconds is denoted by s ( t ) and measured in meters, then Galileo's law is expressed by the equation
Answer:
v = -19.6 m / s, y =y₀ + v₀ t - ½ g t²
Explanation:
This is an exercise in kinetics in one dimension, let's take the upward direction as positive
v = v₀ - g t
in this case as the body is released its initial velocity is zero and the acceleration is -g the sign indicates that it is directed downwards
v = 0 -g t
v = - 9.8 2
v = -19.6 m / s
the sign indicates that the speed is down.
Galileo's equation is
y =y₀ + v₀ t - ½ g t²
where i is the initial height, v₀ the initial velocity and -g the acceleration of the body
A ball is dropped from a top of a 70 m tall building. How fast will the velocity be when it hits the
ground?
Answer:
take gravaiy and times it by 70m to find max v
Explanation:
ps i am olny 10 cool ya
Two insulated wires, each 2.64 m long, are taped together to form a two-wire unit that is 2.64 m long. One wire carries a current of 7.68 A; the other carries a smaller current I in the opposite direction. The two wire unit is placed at an angle of 65.0o relative to a magnetic field whose magnitude is 0.59 T. The magnitude of the net magnetic force experienced by the two-wire unit is 4.11 N. What is the current I
Answer:
[tex]4.77\ \text{A}[/tex]
Explanation:
F = Magnetic force = 4.11 N
[tex]I_n[/tex] = Net current
[tex]I_2[/tex] = Current in one of the wires = 7.68 A
B = Magnetic field = 0.59 T
[tex]\theta[/tex] = Angle between current and magnetic field = [tex]65^{\circ}[/tex]
[tex]l[/tex] = Length of wires = 2.64 m
[tex]I[/tex] = Current in the other wire
Magnetic force is given by
[tex]F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}[/tex]
Net current is given by
[tex]I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}[/tex]
The current I is [tex]4.77\ \text{A}[/tex].
A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.796 m down the incline in 2.00 s. The 0.200 kg block is replaced by a 0.400 kg block. If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s
Answer:
distance travelled by the block is 0.796 m
{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }
Explanation:
Given that;
mass of block m = 0.200 kg
distance travelled d = 0.796 m
time t = 2.00 s
m₂ = 0.400 kg
If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s?
Now, using the second equation of motion;
d = ut + ([tex]\frac{1}{2}[/tex] × at²)
as the object started from rest, u=0
so, we substitute
0.796 = 0×2 + ([tex]\frac{1}{2}[/tex] × a(2)²)
0.796 = 0 + ([tex]\frac{1}{2}[/tex] × 4a)
0.796 = 2a
a = 0.796 / 2
a = 0.398 m/s²
using first equation of motion
[tex]V_{f}[/tex] = u + at
we substitute
[tex]V_{f}[/tex] = 0 + 0.398 × 2
[tex]V_{f}[/tex] = 0.796 m/s
now, average velocity is given as;
[tex]V_{avg}[/tex] = ( 0.796 m/s + 0 ) / 2
[tex]V_{avg}[/tex] = ( 0.796 m/s + 0 ) / 2
now, distance as the block moves in 2s will be;
D = [( 0.796 m/s + 0 ) / 2 ] × 2
D = 0.796 m
Therefore, distance travelled by the block is 0.796 m
{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }
The distance traveled by the object with the uniform motion can be given by the second equation of the motion.
The distance traveled by the big block with weight in 2 seconds is.
What is second equation of motion?
The distance traveled by the object with the uniform motion can be given by the second equation of the motion. It can be given as,
[tex]s=ut+\dfrac{1}{2} at^2[/tex]
Given information-
The mass of the small block is 0.200 kg.
The total distance traveled by the block is 0.796.
Initial velocity of small block is zero.
Total time taken by the block to travel this distance is 2 seconds.
Put the values in the above equation as,
[tex]0.796=0\times 2+\dfrac{1}{2} a\times 2^2\\0.796=2a\\a=0.398[/tex]
Thus the acceleration of the small block is 0.398 meter per second.
Now the mass is doubled which is, 0.400 kg. As the acceleration does not depends on the mass, thus the acceleration for both cases is 0.398.
The velocity of the big block can be given as,
[tex]V=u+0.398\times2\\V=0+0.796\\V=0.796[/tex]
The velocity of the big block is 0.796.
The average velocity of the big block is given by,
[tex]V_{avg}=(\dfrac{0.796+0}{2} })\\V_{avg}=0.398[/tex]
The distance traveled by the object is the ratio of the velocity of the body to the time taken by it. Thus the distance traveled by the big block in 2 seconds is,
[tex]d={{0.398} \times2}\\d=0.796[/tex]
Thus the distance traveled by the big block with weight in 2 seconds is.
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Which two substances are elements?
A. sand and air B. salt and sand
C. iron and helium
D. helium and water
A child on a sled slides (starting from rest) down an icy slope that makes an angle of 15◦ with the horizontal. After sliding 20 m down the slope, the child enters a flat, slushy region, where she slides for 2.0 s with a constant negative acceleration of −1.5 m/s2 with respect to her direction of motion. She then slides up another icy slope that makes a 20◦ angle with the horizontal.
A) How fast was the child going when she reached the bottom of the first slope? How long did it take her to get there?B) How long was the flat stretch at the bottom?C) How fast was the child going as she started up the second slope?D) How far up the second slope did she slide?
Answer:
A) v₁ = 10.1 m/s t₁= 4.0 s
B) x₂= 17.2 m
C) v₂=7.1 m/s
D) x₂=7.5 m
Explanation:
A)
Assuming no friction, total mechanical energy must keep constant, so the following is always true:[tex]\Delta K + \Delta U = (K_{f} - K_{o}) +( U_{f} - U_{o}) = 0 (1)[/tex]
Choosing the ground level as our zero reference level, Uf =0.Since the child starts from rest, K₀ = 0.From (1), ΔU becomes:[tex]\Delta U = 0- m*g*h = -m*g*h (2)[/tex] In the same way, ΔK becomes:[tex]\Delta K = \frac{1}{2}*m*v_{1}^{2} (3)[/tex] Replacing (2) and (3) in (1), and simplifying, we get:[tex]\frac{1}{2}*v_{1}^{2} = g*h (4)[/tex]
In order to find v₁, we need first to find h, the height of the slide.From the definition of sine of an angle, taking the slide as a right triangle, we can find the height h, knowing the distance that the child slides down the slope, x₁, as follows:[tex]h = x_{1} * sin \theta_{1} = 20.0 m * sin 15 = 5.2 m (5)[/tex]
Replacing (5) in (4) and solving for v₁, we get:
[tex]v_{1} = \sqrt{2*g*h} = \sqrt{2*9.8m/s2*5.2m} = 10.1 m/s (6)[/tex]
As this speed is achieved when all the energy is kinetic, i.e. at the bottom of the first slide, this is the answer we were looking for.Now, in order to finish A) we need to find the time that the child used to reach to that point, since she started to slide at the its top.We can do this in more than one way, but a very simple one is using kinematic equations.If we assume that the acceleration is constant (which is true due the child is only accelerated by gravity), we can use the following equation:[tex]v_{1}^{2} - v_{o}^{2} = 2*a* x_{1} (7)[/tex]
Since v₀ = 0 (the child starts from rest) we can solve for a:[tex]a = \frac{v_{1}^{2}}{2*x_{1} } = \frac{(10.1m/s)^{2}}{2* 20.0m} = 2.6 m/s2 (8)[/tex]
Since v₀ = 0, applying the definition of acceleration, if we choose t₀=0, we can find t as follows:[tex]t_{1} =\frac{v_{1} }{a} =\frac{10.1m/s}{2.6m/s2} = 4.0 s (9)[/tex]
B)
Since we know the initial speed for this part, the acceleration, and the time, we can use the kinematic equation for displacement, as follows:[tex]x_{2} = v_{1} * t_{2} + \frac{1}{2} *a_{2}*t_{2}^{2} (10)[/tex]
Replacing the values of v₁ = 10.1 m/s, t₂= 2.0s and a₂=-1.5m/s2 in (10):[tex]x_{2} = 10.1m/s * 2.0s + \frac{1}{2} *(-1.5m/s2)*(2.0s)^{2} = 17.2 m (11)[/tex]
C)
From (6) and (8), applying the definition for acceleration, we can find the speed of the child whem she started up the second slope, as follows:[tex]v_{2} = v_{1} + a_{2} *t_{2} = 10.1m/s - 1.5m/s2*2.0s = 7.1 m/s (12)[/tex]
D)
Assuming no friction, all the kinetic energy when she started to go up the second slope, becomes gravitational potential energy when she reaches to the maximum height (her speed becomes zero at that point), so we can write the following equation:[tex]\frac{1}{2}*v_{2}^{2} = g*h_{2} (13)[/tex]
Replacing from (12) in (13), we can solve for h₂:[tex]h_{2} =\frac{v_{2} ^{2}}{2*g} = \frac{(7.1m/s) ^{2}}{2*9.8m/s2} = 2.57 m (14)[/tex]
Since we know that the slide makes an angle of 20º with the horizontal, we can find the distance traveled up the slope applying the definition of sine of an angle, as follows:[tex]x_{3} = \frac{h_{2} }{sin 20} = \frac{2.57m}{0.342} = 7.5 m (15)[/tex]
It takes 500 W of power to move an object 96 m in 12 s. What force is being applied to the object?
Answer:
Explanation:
Power, by definition, is the amount of work per unit of time. We arent given work in this question, but we can find it because work is how much force per unit of distance.
[tex]P=\frac{W}{t_f-t_i} =\frac{F*d}{t_f-t_i}[/tex]
Plug in all the values, and its algebra at this point
[tex]500=\frac{F*96}{12}[/tex]
6000 = 96F
F = 62.5 Newtons
The force which is being applied to the object with the power of 500 Watt is 62.5 N.
Power can be defined as the rate of doing work. It is the work done in unit time. The SI unit of power is Watt (W) which is equal to joules per second (J/s). Sometimes, the power of motor vehicles and other machines is given in terms of Horsepower (hp), This unit is approximately equal to 745.7 watts of power.
The power of a system can be calculated as the product of force applied and distance travelled by the system per unit time taken.
Therefore, the force of the object with power 500W can be calculated as:
P = f × d/ t
where, P = Power of the object,
f = Force applied,
d = distance travelled,
t = time taken to cover the distance
f = (P × t)/ d
f = (500 × 12) / 96
f = 6000/ 96
f = 62.5 N
Therefore, the force which is being applied to the object is 62.5N (Newton).
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A piece of aluminium with mass 1 kg and density 2700 kg/m3 is suspended from a string and then completely immersed in a container of water. Calculate the tension in the string (a) before and (b) after the metal is immersed.
(with steps please)
A piece of aluminium with mass 1 kg and density 2700 kg/m3 is suspended from a string and then completely immersed in a container of water. Calculate the tension in the string (a) before and (b) after the metal is immersed.
ANSWER: 6.296NI am unable to add the workings out, however please do message me and I will be able to provide you with them :)
A friend claims that as long as he has his seat belt on, he can hold on to a 14 kg child in a 64 mi/h head-on collision with a brick wall in which the car passenger compartment comes to a stop in 0.05 s. Show that the violent force during the collision will tear the child from his arms. A child should always be in a toddler seat secured with a seat belt in the back seat of a car.
Answer:
Force is too large for a person to exert.
Explanation:
m = Mass of child = 14 kg
u = Initial velocity of compartment = 64 mi/h = [tex]\dfrac{64\times 1609.34}{3600}=28.61\ \text{m/s}[/tex]
v = Final velocity of compartment = 0
t = Time taken by the compartment to stop = 0.05 s
Force is given by
[tex]F=ma\\\Rightarrow F=m\dfrac{v-u}{t}\\\Rightarrow F=14\times \dfrac{0-28.61}{0.05}\\\Rightarrow F=-8010.8\ \text{N}[/tex]
The person has to exert a force of [tex]-8010.8\ \text{N}[/tex] in the opposite direction to keep holding the child. This is a huge amount of force and cannot be done by a person's physical strength alone.
The body breaks down or converts most carbohydrates into the
sugar____?*
1.Fructose
2.Galactose
3.Lactose
4.Glucose
The resistivity of a material is 2x10-3Ωm. What is the conductivity\n
Answer:
Conductivity, [tex]\sigma=500\ (\Omega-m)^{-1}[/tex]
Explanation:
Given that,
The resistivity of a material, [tex]\rho=2\times 10^{-3}\ \Omega-m[/tex]
We need to find the conductivity of the material.
We know that the reciprocal of resistivity is called conductivity.
[tex]\sigma=\dfrac{1}{\rho}\\\\\sigma=\dfrac{1}{2\times 10^{-3}}\\\\\sigma=500\ (\Omega-m)^{-1}[/tex]
So, the conductivity of the material is [tex]500\ (\Omega-m)^{-1}[/tex].
Through what potential difference should electrons be accelerated so that their speed is 1.0 % of the speed of light when they hit the target
Answer:
Explanation:
Considering non - relativistic approach : ----
Speed of electron = 1 % of speed of light
= .01 x 3 x 10⁸ m /s
= 3 x 10⁶ m /s
Kinetic energy of electron = 1/2 m v²
= .5 x 9.1 x 10⁻³¹ x ( 3 x 10⁶ )²
= 40.95 x 10⁻¹⁹ J
Kinetic energy in electron comes from lose of electrical energy equal to
Ve where V is potential difference under which electron is accelerated and e is electronic charge .
V x e = kinetic energy of electron
V x 1.6 x 10⁻¹⁹ = 40.95 x 10⁻¹⁹
V = 25.6 Volt .
A ship sets out to sail to a point 141 km due north. An unexpected storm blows the ship to a point 102 km due east of its starting point. (a) How far and (b) in what direction (as an angle from due east, where north of east is a positive angle) must it now sail to reach its original destination
Answer:
Explanation:
The point of destination is 141 north and 102 km east . Vectorially it is represented by unit vector as follows .
D = 102 i + 141 j
magnitude of D = √ ( 102² + 141² )
= √ ( 10404 + 19881)
= √ 30285
= 174 km
To go back to original position , ship should move on the following vector
D = -102 i - 141 j
Tan Ф = 141 / 102 = 1.38
Ф = 54⁰ , direction will be south of west . From north east , angle will be
180 + 54 = 234⁰
Consider a collection of charges in a given region and suppose all other charges are distant and have a negligible effect. Further, the electric potential is taken to be zero at infinity. If the electric potential at a given point in the region is zero, which of the following statements must be true?
A. The electric field is zero at that point.
B. The charges have the same sign and are symmetrically arranged around the given point.
C. The electric potential energy is a minimum at that point.
D. There is no net charge in the region.
E. Some charges in the region are positive, and some are negative.
Answer:
E. Some charges in the region are positive, and some are negative.
Explanation:
Electric potential is given as;
[tex]V = \frac{W}{Q}[/tex]
where;
W is the work done in moving a charge between two points which have a difference in potential
Q is quantity of charge in the given region
If the electric potential at a given point in the region is zero, then sum of the charges in the given region must be equal to zero. For the charges to sum to zero, some will be positive while some will be negative,.
Therefore, the correct statement in the given options is "E"
E. Some charges in the region are positive, and some are negative.
A boat moves up and down as water waves pass under the boat. If the amplitude of the wave gets bigger, then
A)
the boat will rise up higher.
B)
the boat will not rise up as high.
C)
the boat will go up and down more often.
D)
the boat will continue to move the same way.
Answer: The Boat will rise
Explanation: Because high amplitude means high in heights.
A blue ball is thrown upward with an initial speed of 24.1 m/s, from a height of 0.5 meters above the ground. 2.9 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 7.2 m/s from a height of 32 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2. Take up as the positive direction.
1. What is the speed of the blue ball when it reaches its maximum height?.
2. How long does it take the blue ball to reach its maximum height?.
3. What is the maximum height the blue ball reaches?.
4. What is the height of the red ball 3.77 seconds after the blue ball is thrown?.
5. How long after the blue ball is thrown are the two balls in the air at the same height?.
Answer:
1. Speed=0
2. 2.46 s
3.30.1 m
4. 22.0 m
5.1.004 s
Explanation:
We are given that
Initial speed of blue ball, u=24.1 m/s
Height of blue ball from ground y_0=0.5 m
Initial speed of red ball , u'=7.2 m/s
Height of red from ground=y'0=32 m
Gravity, g=[tex]9.81ms^{-2}[/tex]
1.When the ball reaches its maximum height then the speed of the blue ball is zero.
2.v=0
[tex]v=u+at[/tex]
Using the formula and substitute the values
[tex]0=24.1-9.81t[/tex]
Where g is negative because motion of ball is against gravity
[tex]24.1=9.81t[/tex]
[tex]t=\frac{24.1}{9.81}=2.46s[/tex]
3.[tex]y=y_0+ut+\frac{1}{2}at^2[/tex]
Using the formula
[tex]y=0.5+24.1(2.46)-\frac{1}{2}(9.81)(2.46)^2[/tex]
[tex]y=30.1 m[/tex]
4.Time of flight for red ball=3.77-2.9=0.87s
[tex]y'=32-7.2(0.87)-\frac{1}{2}(9.81)(0.87)^2[/tex]
[tex]y'=22.0m[/tex]
Hence, the height of red ball 3.77 s after the blue ball is 22.0 m.
5.According to question
[tex]0.5+24.1(t+2.9)-\frac{1}{2}(9.81)(2.9+t)^2=32-7.2t-\frac{1}{2}(9.81)t^2[/tex]
[tex]0.5+24.1t+69.89-4.905(t^2+5.8t+8.41)=32-7.2t-4.905t^2[/tex]
[tex]0.5+24.1t+69.89-4.905t^2-28.449t-41.25105=32-7.2t-4.905t^2[/tex]
[tex]0.5+69.89-41.25105-32=-24.1t+28.449t-7.2t[/tex]
[tex]-2.86105=-2.851t[/tex]
[tex]t=\frac{2.86105}{2.851}=1.004 s[/tex]
Hence, 1.004 s after the blue ball is thrown are the two balls in the air at the same height.
Fred is doing an experiment with basketballs. He drops one basketball while standing on a chair, and drops another basketball while standing on the roof. Which basketball will probably bounce the highest?
A. The basketball from the chair
B. The basketball from the roof
C. They will bounce the same
D. Neither ball will bounce
A young gazelle is grazing in a beautifully sunlit corner of the savanna. Suddenly, the gazelle raises herhead and spots a lioness in the tall grass 173 m away, so she turns away running at roughly constantspeed. The lioness immediately chases the gazelle, with an explosive acceleration of 2.57 m/s2. Aninformed source tells us that this lioness is capable of enduring her maximum speed of 21.0 m/s for 25.0seconds at the longest. [Assume that both predator and prey never change their direction or motion inthis case, for the sake of simplicity.]
(a) On average, how fast (at least) does that gazelle need to run to survive? [Show all your steps.]
(b) Produce a qualitative graph of animal's position (vertical axis) versus time (horizontal axis), shared for the lioness's motion and as well as the gazelle's. Place graph labels in a way that fits the narration.
Answer:
a) v₂ = 13.20 m / s
Explanation:
To solve this exercise we will use the kinematic relations
Let's start with the Lioness. Let's find the time to reach top speed
v = v₀ + a t
as part of rest, its initial velocity is zero
t = v / a
t₁ = 21.0 / 2.57
t₁ = 8.17 s
the total time is the acceleration time plus the time (t₂ = 25 s) that the maximum speed can withstand
t = t₁ + t₂
t = 8.17 +25.0 = 33.17 s
Now let's find out what distance the lioness travels in these times
during acceleration
x = v₀ t + ½ a t²
x = ½ a t²
x₁ = ½ 2.57 8.17²
x₁ = 85.77 m
during constant speed part
x₂ = v t₂
x₂ = 21.0 25.0
x₂ = 525 m
therefore the total distance traveled is
x = x₁ + x₂
x = 85.77 + 525
x = 610.77 m
a) the average speed of the gazelle
this must be the distance that the lioness travels minus the initial distance that separates the two animals (xo = 173 m) between the time taken
v₂ = [tex]\frac{x -x_o}{t}[/tex]
v₂ = [tex]\frac{610.77 - 173}{33.17}[/tex]
v₂ = 13.20 m / s
b) in the attachment we can see a graph of the displacement of the two animals