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Rzqust [24]
3 years ago
8

1500 kg wrecking ball traveling at a speed of 3.5 m/s hits a wall that does not crumble but is pushed back 75 cm. If the wreckin

g ball comes to rest in the wall, what is the size of the force that pushes the wall
Physics
1 answer:
Rudiy273 years ago
7 0

Answer:

The size of the force that pushes the wall is 12,250 N.

Explanation:

Given;

mass of the wrecking ball, m = 1500 kg

speed of the wrecking ball, v = 3.5 m/s

distance the ball moved the wall, d = 75 cm = 0.75 m

Apply the principle of work-energy theorem;

Kinetic energy of the wrecking ball = work done by the ball on the wall

¹/₂mv² = F x d

where;

F is the size of the force that pushes the wall

¹/₂mv² = F x d

¹/₂ x 1500 x 3.5² = F x 0.75

9187.5 = 0.75F

F = 9187.5 / 0.75

F = 12,250 N

Therefore, the size of the force that pushes the wall is 12,250 N.

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A man pushes a 35.2 kg box across a frictionless floor with a force of 128 N. What is the acceleration of the box
vodka [1.7K]
The formula for Force is F = MA, or Force is equivalent to the product of Mass and Acceleration. 

F = 128N.
M = 35.2kg.

128 = 35.2A
Divide both sides by 35.2 to solve for the acceleration.
A = ~3.636

The acceleration is 3.636 m/s^2.

I hope this helps!
4 0
3 years ago
Read 2 more answers
An energy source forces a constant current of 2A to flow through a light bulbfilament for twenty seconds. If 4.6 kJ is given off
QveST [7]

Answer:

The voltage drop across the bulb is 115 V

Explanation:

The voltage drop equation is given by:

V=\frac{\Delta W}{\Delta q}

Where:

ΔW is the total work done (4.6kJ)

Δq is the total charge

We need to use the definition of electric current to find Δq

I=\frac{\Delta q}{\Delta t}

Where:

I is the current (2 A)

Δt is the time (20 s)

2=\frac{\Delta q}{20}

q=40 C

Then, we can put this value of charge in the voltage equation.

V=\frac{4600}{40}=115 V

Therefore, the voltage drop across the bulb is 115 V.

I hope it helps you!

6 0
3 years ago
Which electromagnetic waves have the shortest wavelength and the highest frequency?
ra1l [238]
It’s supposed to be gamma, what are your other options
5 0
3 years ago
compare your previous result to the present hrf result are there any changes if yes explain your answer​
garri49 [273]

Answer:

<em>I </em><em>don't</em><em> know</em><em> </em><em>what</em><em> </em><em>are </em><em>you </em><em>saying</em><em> </em><em>but </em><em>I </em><em>don't</em><em> </em><em>have </em><em>any</em><em> </em><em>results</em><em> </em>

Explanation:

kai6417

4 0
2 years ago
An electron with a speed of 1.2 × 107 m/s moves horizontally into a region where a constant vertical force of 5.2 × 10-16 N acts
Aliun [14]

Answer: 0.642mm

Explanation: F= force = 5.2×10^-16 N,

v = velocity of electron = 1.2×10^7 m/s,

m = mass of electron = 9.11×10^-31 kg.

We will assume the motion of the object to be of a constant acceleration, hence newton's laws of motion is applicable.

Recall that f = ma.

Where a = acceleration

This acceleration of vertical because it occurred when the object deflected.

5.2×10^-16 = 9.11×10^-31 (ay)

ay = 5.2×10^-16 / 9.11×10^-31

ay = 5.71×10^14 m/s²

For the horizontal motion, x = vt

Where x = horizontal distance = 0.019m and v is the velocity = 1.2×10^7 m/s,

By substituting the parameters, we have that

0.019 = 1.27×10^7 × t

t = 0.019 / 1.27 × 10^7

t = 1.5×10^-9 s

The vertical distance (y) is gotten by using the formulae below

y = ut + at²/2

but u = 0

y = at²/2

y = 5.71×10^14 × (1.5×10^-9)²/2

y = 0.00128475/2

y = 0.000642m = 0.642mm

7 0
3 years ago
Read 2 more answers
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