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zloy xaker [14]
3 years ago
9

Suppose you exert a 400-N force on a wall but the wall does not move. The work you are doing on the wall is : unknown, because t

he mass of the wall is not given
zero, because the wall is not moving
3920 J
400 J
Physics
1 answer:
Ilya [14]3 years ago
6 0

Answer:

W = 0 :The work done on the wall is zero,because the wall is not moving

Explanation:

Work theory

Work is the product of a force applied to a body and the displacement of the body in the direction of this force.

W= F*d Formula (1)

W: Work (Joules) (J)

F: force applied (N)

d=displacement of the body (m)

The work is positive (W+) if the force goes in the same direction of movement.

The work is negative (W-)if the force goes in the opposite direction to the movement

Data

F= 400-N

d= 0

Problem development

We apply formula (1) to calculate the work done on the wall:

W= 400*0

W=0

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Elena-2011 [213]

Answer:

I = PRT

18 = 150(R)(2)

18 = 300R

R = 0.06

So A is correct.

Let me know if this helps!

7 0
3 years ago
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You walk exactly 250 steps North, turn around, and then walk exactly 400 steps South. How far are you from your starting
german

Answer:

150 steps south

Explanation:

250 north 250 back to start then continue south for remainder of 400 steps. 150 south

3 0
3 years ago
The driver of a car traveling at 31.3 m/s applies the brakes and undergoes a constant deceleration of 1.6 m/s2.How many revoluti
lisov135 [29]

Answer:

R=156.99\operatorname{Re}vs

Explanation: The equations used are as follows:

\begin{gathered} x(t)=x_o+v_ot+\frac{1}{2}at^2\Rightarrow(1) \\ v(t)=v_o+at\Rightarrow(2) \end{gathered}

By using equation (2), the time needed for the car to come to rest is calculated as follows:

\begin{gathered} v(t)=(31.3ms^{-1})_{}+(-1.6ms^{-2})t=0 \\ t=\frac{31.3ms^{-1}}{1.6ms^{-2}}=19.56s \\ t=19.563s \end{gathered}

By using equation (1), The total distance traveled in that time would be as:

\begin{gathered} x(19.563s)=_{}(31.3ms^{-1})\cdot(19.563s)+\frac{1}{2}(-1.6ms^{-2})\cdot(19.563s)^2\Rightarrow(1) \\ x(19.563s)=612.31-306.17=306.14m \\ \therefore\Rightarrow \\ x(19.563s)=306.14m \end{gathered}

The revolutions taken by the tire before the car comes to rest would be:

\begin{gathered} C=2\pi\cdot(0.31m)=1.95m \\ R=\frac{x(19.563s)}{C}=\frac{306.14m}{1.95m}=156.99\operatorname{Re}v \\ R=156.99\operatorname{Re}vs \end{gathered}

3 0
2 years ago
At night, large bodies of water release heat to the atmosphere quickly. True False
Nikolay [14]

Answer:

false I hope. have a good day!

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2 years ago
A 60 kg student is standing atop a spring in an elevator that is accelerating upward at 3.0 m/s2. The spring constant is 2.5 x 1
dimaraw [331]

the spring will be compressed by 0.3072 m

Explanation:

acceleration of elevator=3 m/s²

mass of student= 60 Kg

spring constant=2.5 x 10³ N/m

the force on the student is given by F = m ( g +a)

F=60 (9.8+3)

F=768 N

now the formula for spring force is given by

F= k x

768= 2.5 x 10³ (x)

x=0.3072 m

6 0
3 years ago
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