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Sauron [17]
3 years ago
14

A worker pushes horizontally on a large crate with a force of 200 N, and the crate is moved 3.5 m. How much work was done in Jou

les?
Physics
1 answer:
sattari [20]3 years ago
6 0

Work = (force) x (distance) =

           (200 N) x (3.5 m)  =  <em>700 joules</em>


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An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the wa
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Answer:

A) 17.7 m/s

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C) Zero

E) 9.8 m/s²

Explanation:

given information

distance, h = - 34 m

time, t = 5 s

A) What is the initial speed of the egg?

h - h₀ = v₀t - \frac{1}{2} gt², h₀ = 0

- 34 = v₀ 5 - \frac{1}{2} 9.8 5²

- 34 = 5 v₀ - 122.5

v₀ = 122.5 - 34/5

    = 17.7 m/s

B) How high does it rise above its starting point?

v² = v₀² - 2gh

v = 0 (highest point)

2gh = v₀²

h = v₀²/2g

  = 17.7²/2 (9.8)

  = 15.98 m

C) What is the magnitude of its velocity at the highest point?

v = 0 (at highest point)

E) What are the magnitude and direction of its acceleration at the highest point?

g= 9.8 m/s², since the egg is moved vertically, the acceleration is the same as the gravitational acceleration.

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One end of a horizontal spring with force constant 130.0 N/m is attached to a vertical wall. A 3.00 kg block sitting on the floo
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Answer:

a) v = 0

b) The aceleration is 1.41 m/s^{2}

c) The block is accelerating away from the wall.

Explanation:

First, you need to think about the effect this constant force is causing in the spring: it causes a displacement in the equilibrium point of the system, therefore we need to know where it sits now:

At equilibrium no movement is present reducing friction to 0:

\sum{F} = 0 = F_{spring} - F_{external}

F_{spring} = F_{external}

Kx = F_{external}

x = \frac{F_{external}}{K}=\frac{88}{130}=0.68m=68cm

This means that the spring can be compressed with the single force up to 68 cm, Any further compression will cause an unbalanced system and the occilation of the mass.

The spring can't be compressed by the given force to 80 cm, therefore it must have been compressed by another force and then released.

In this case, the instantanous speed is 0, since the block has just been released.

In the same instant we can stimate the free body diagram of forces by the next two equations:

\sum_y{F}={F_N-W}=0\\\sum_x{F}={F_{spring}-F_{external}-F_{friction}}=ma

For the y axis:

F_N = W = mg = 3*9.8 = 29.4N

To calculate the force of friction:

F_{friction} = \mu_k F_N=0.4*29.4 = 11.76N

Therefore for x axis:

{Kx-F_{external}-F_{friction}}=ma

a = \frac{130*0.8-88-11.76}{3} = \frac{104-88-11.76}{3}=\frac{4.24}{3}=1.41\frac{m}{s^2}

7 0
3 years ago
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