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2 years ago

Are you doing work holding a heavy piece of furniture in place

Physics

1 answer:

Crazy boy [7]2 years ago

8 0

Answer:

Explanation:

Work Done is Force x Distance.

You are exerting a force to hold the piece of furniture in place, however work is not being done as you are not moving it. (as you are holding it in place)

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PLEASE HELP!! ITS URGENT!!!

Natasha2012 [34]

Answer:

F = 800 [N]

Explanation:

To be able to calculate this problem we must use the principle of momentum before and after the impact of the hammer.

We must summarize that after the impact the hammer does not move, therefore its speed is zero. In this way, we can propose the following equation.

ΣPbefore = ΣPafter

$(m_{1}*v_{1}) - F*t = (m_{1}*v_{2})$

where:

m₁ = mass of the hammer = 0.15 [m/s]

v₁ = velocity of the hammer = 8 [m/s]

F = force [N] (units of Newtons)

t = time = 0.0015 [s]

v₂ = velocity of the hammer after the impact = 0

$(0.15*8)-(F*0.0015) = (0.15*0)\\F*0.0015 = 0.15*8\\F = 1.2/(0.0015)\\F = 800 [N]$

Note: The force is taken as negative since it is exerted by the nail on the hammer and this force is directed in the opposite direction to the movement of the hammer.

6 0

2 years ago

HELP PLS i need this right now

N76 [4]

Answer:

V = 44.4 units.

Explanation:

In order to solve this problem, we must use the Pythagorean theorem. Which is defined by the following expression.

$V = \sqrt{(V_{x})^{2} +(V_{y})^{2} }$

where:

Vx = - 43 units

Vy = 11.1 units

Now replacing:

$V=\sqrt{(-43)^{2} +(11.1)^{2} }\\V=44.4 units$

7 0

3 years ago

What causes thunder?

LUCKY_DIMON [66]

I believe the answer is B.

4 0

3 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be: