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2 years ago

9

Levers pivot on the A. fulcrum B. resistance arm C. effort arm D.moon

1 answer:

AVprozaik [17]2 years ago

6 0

Answer: it’s A

Explanation: hope this is right!!!!

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Which force can affect an object without direct physical contact?

denis-greek [22]

Gravity affet everything and it touches nothing.
Hope this helps!

8 0

3 years ago

Read 2 more answers

Please help ASAP!!

bixtya [17]

There is kinetic energy when it is sitting at the top, then as it goes towards the bottom, the kinetic energy is transformed into potential energy.

6 0

3 years ago

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coeffi

Flura [38]

The given data is incomplete. The complete question is as follows.

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36. Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)

Explanation:

Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and $\mu$ be the friction coefficient and m be the mass of car.

Hence, the given data is as follows.

v = 0, s = 84 m, $\mu$ = 0.36

According to Newton's law of second motion the expression for acceleration is as follows.

F = ma

$-\mu N$ = ma

$-\mu mg$ = ma

a = $-\mu g$

Also,

$v^{2} = u^{2} + 2as$

$(0)^{2} = u^{2} + 2(-\mu g)s$

$u^{2} = 2(\mu g)s$

= $\sqrt{2(0.36)(9.81 m/s^{2})(84 m)}$

= 24.36 m/s

Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.

4 0

2 years ago

Luis wants to compare the density of three solid objects that are different shapes and sizes. What information does he need to m

tatyana61 [14]

Answer:

u need to make sure that comparison is = to shapes and then find the shapes sizes and add them

7 0

2 years ago

On a vacation flight, you look out the window of the jet and wonder about the forces exerted on the window. Suppose the air outs

user100 [1]

Answer:

A) $\Delta P = 14512.5 Pa = 14.512 kPa$

B) F = 1632.65 N

Explanation:

Given details

outside air speed is given as $v_2 = 150 m/s$

since inside air is atmospheric , $v_1 = 0 m/s$

a) By using bernoulli equation between outside and inside of flight

$P_1 + \frac{1}{2} \rho v_1^2 + \rho gh = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh$

$\Delta P = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2$

$\Delta P = \frac{1}{2} \rho[ v_2^2 -v_1^2]$

$\Delta P = \frac{1}{2} 1.29 [ 150^2 - 0^2]$

$\Delta P = 14512.5 Pa = 14.512 kPa$

b) force exerted on window

Area of window $= 25\times 45 = 1125 cm^2 = 0.1125 m^2$

We know that force is given as

$F = P\times A$

$F = 14512.5 \times 0.1125$

F = 1632.65 N

5 0

3 years ago