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algol
6 days ago
15

A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat-top. If the mass of t

he loaded cart is 3.0 kg and the height of the seat top is 0.45 meters, then what is the potential energy of the loaded cart at the height of the seat-top?
Physics
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If a metal wire is 4m long and a force of 5000n causes it to stretch by 1mm, what is the strain?
serg [3582]

Answer:

2.5\cdot 10^{-4}

Explanation:

Strain is defined as the ratio of an object's dimensional change when subjected to a force:

S=\frac{\Delta L}{L_0}

where

\Delta L indicates the alteration in length of the object

L_0 signifies the object's initial length

In this case, we have L_0 = 4 m and \Delta L=1 mm=0.001 m, hence the strain is

S=\frac{\Delta L}{L_0}=\frac{0.001 m}{4 m}=2.5\cdot 10^{-4}


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1 month ago
1. The gravitational pull of the sun on Earth keeps Earth orbiting around the sun. Which statement is correct about the force th
Sav [3153]

Answer:

1. The sun is moved away from the Earth by an equal force exerted by the Earth.

6 0
2 months ago
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Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Which student’s measureme
Ostrovityanka [3204]
Upon comparison, it is evident that the student with the highest percent error is A. Student 4, who measured 9.61 m/s². Four students recorded the acceleration of gravity, with the accepted local value being 9.78 m/s². Now let's find out which student's measurement exhibited the greatest percent error.
4 0
2 months ago
1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r
Softa [3030]

Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle \theta to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

where y_o represents the initial height from ground level, v_{oy} is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

v_y=v_{oy}-gt

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of v_o.

The peak height is attained when v_y=0, which allows us to compute the time to reach that height.

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum height reached is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is equal to 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Continuing with the calculations for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Substituting known values yields

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

y=7.39\ m

5 0
3 months ago
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