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Andrews [41]
1 year ago
8

a box weighing 155 N is pushed horizontally down the hall at constant velocity. the applied force is 83 n what is the coefficien

t of friction between the box and the floor
Physics
1 answer:
meriva1 year ago
8 0

Answer:

μ = 0.535

Explanation:

On a level floor, normal force = weight.

N = W

Friction force = normal force × coefficient of friction.

F = Nμ

Substitute:

F = Wμ

83 = 155μ

μ = 0.535

Round as needed.

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Imagine you want to get 1 kcal of energy from a cow. How much energy would the cow need to get from plants? Why?
ZanzabumX [31]
1000 kcal because you only get 10% of the energy of the thing you eat
7 0
1 year ago
A helicopter travels west at 80 mph. It is moving above a car traveling on a highway at 80 mph. Given this information, you can
gavmur [86]

Answer:

d. at the same velocity

Explanation:

I will assume the car is also travelling westward because it was stated that the helicopter was moving above the car. In that case, it depends where the observer is. If the observer is in the car, the helicopter would look like it is standing still ( because both objects have the same velocity). If the observer is on the side of the road, both objects would be travelling at the same velocity. Also recall that, velocity is a vector quantity; it is direction-aware. Velocity is the rate at which the position changes but speed is the rate at which object covers distance and it is not direction wise. Hence velocity is the best option.

5 0
1 year ago
Imagine that a small boy and his much larger father are enjoying a winter morning, sliding along the ice on a nearby playground.
kramer
You can write an hypothesis such as this:
The weight of an object has effects on the operating frictional force, the greater the weight, the higher the operating frictional force.
The father is the one with the higher weight while the son has the lower weight. The operating frictional force is the friction that their weights exert.
8 0
2 years ago
Read 2 more answers
The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode al
SashulF [63]

Answer:

7.9 \frac{m}{s^{2} }

Explanation:

Take the fact that mass is inversely proportional to accelertation:

m ∝ a

Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:

\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\

Using algebra, we can rearrange our equation to find the final acceleration, a_{2}:

a_{2}  = \frac{a_{1}*m_{1}  }{m_{2} } \\

Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:

0.85g * 9.8 \frac{m }{s^{2} } = 8.33 \frac{m }{s^{2} }

Plug everything in:

7.9 \frac{m }{s^{2} } = \frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}

(1590kg the initial weight plus the weight of the added passenger)

8 0
1 year ago
A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .
nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

v is image distance

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm

Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3

So, the magnification of the lens is 3.

3 0
2 years ago
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