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motikmotik
3 years ago
9

How does changing the lengthy but not the height of an inclined plane affect the work done to lift a load? PLZ HELP ME NOW'

Physics
1 answer:
Serhud [2]3 years ago
4 0

Answer: Work Done would remain same.

Let us assume that the velocity is constant while taking the load up the inclined plane. Then, the kinetic energy would remain the same. This is because kinetic energy is dependent on velocity (K.E.=\frac{1}{2}mv^2). If that is constant, the kinetic energy would remain same. The potential energy is dependent on the height(P.E.=mgh). If the height is changed, then potential energy varies. In the question, it is mentioned that without changing the height, the length of the inclined plane is changed. Therefore, the potential energy would be same as before.

We know, work done is equal to potential energy plus kinetic energy. Since there is no change in any of these, the required work done would not change.


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While traveling along a highway a driver slows from 32 m/s and comes to a stop with an acceleration of -6 m/s2. How long did it
diamong [38]

Answer: 6s

Explanation:

Vs=32m/s  speed at beginning of slowing down

Vf=0m/s     stop speed

a= -6 m/s²  acceleration

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Use equation for acceleration :

a=(Vf-Vs)/t

a*t=Vf-Vs

t=(Vf-Vs)/a

t=(0-36)/-6

t=-36/-6

t=6 s

7 0
3 years ago
How is acceleration calculated?
steposvetlana [31]

Answer:

Acceleration is the change in velocity divided by time

Explanation:

This is the correct answer because distance divided by time is the position. Speed multiplied by time is the distance. And acceleration is not just velocity, but the change in velocity over time.

4 0
3 years ago
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A point charge with charge q1 = 4.00 μC is held stationary at the origin. A second point charge with charge q2 = -4.40 μC moves
Bezzdna [24]

Answer:

W=0.94J

Explanation:

Electrostatic potential energy is the energy that results from the position of a charge in an electric field. Therefore, the work done to move a charge from point 1 to point 2 will be the change in electrostatic potential energy between point 1 and point 2.

This energy is given by:

U=\frac{K\left |q_1 \right |\left |q_2 \right |}{r}\\

So, the work done to move the chargue is:

W=U_1-U_2\\W=\frac{K\left |q_1 \right |\left |q_2 \right |}{r_1}-\frac{K\left |q_1 \right |\left |q_2 \right |}{r_2}\\r_1=\sqrt{((0.155 m)^2+0 m)^2}=0.115m\\r_2=\sqrt{((0.245 m)^2+(0.270 m)^2}=0.365m\\W=K\left |q_1 \right |\left |q_2 \right |(\frac{1}{r_1}-\frac{1}{r_2})\\W=8.99*10^9\frac{Nm^2}{c^2}(4.00*10^{-6}C)(4.40*10^{-6}C)(\frac{1}{0.115m}-\frac{1}{0.365})\\W=0.94J

The work is positive since the potential energy in 1 is greater than 2.

5 0
4 years ago
A 50-ω resistor is connected to a 9.0 V battery. How much thermal energy is produced in 7.5 minutes?1.2 * 10^2 J1.3 * 10^3 J3.0
emmasim [6.3K]

In order to calculate the thermal energy, first let's calculate the power, using the formula:

P=\frac{V^2}{R}

For a voltage V = 9 Volts and a resistance R = 50 ohms, we have:

\begin{gathered} P=\frac{9^2}{50} \\ P=\frac{81}{50}=1.62\text{ W} \end{gathered}

Now, multiplying the power by the time (in seconds), we can find the energy:

\begin{gathered} E=P\cdot t \\ E=1.62\cdot7.5\cdot60 \\ E=729\text{ J} \end{gathered}

In scientific notation, we have an energy of 7.3 * 10^2 J, therefore the correct option is the fourth one.

4 0
1 year ago
8. A car moved 20 km East and 70 km West.<br> What is the distance?
SCORPION-xisa [38]
Your answer is.......a car moved 60 km East and 90 km in west.
7 0
4 years ago
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