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andrew11 [14]
2 years ago
14

1. Force of impact is the force ________________.

Physics
1 answer:
siniylev [52]2 years ago
6 0
Force of impact is the force generated when objects meet.
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The static frictional force between a 95-kilogram object and the floor is 45 Newtons. The kinetic frictional force is only 22 Ne
Lisa [10]

Answer:

F = 69.5 [N]

Explanation:

We must remember that the friction force is defined as the product of the normal force by the coefficient of friction, and it can be calculated by the following expression.

f=N*miu

where:

N = normal force [N]

miu = friction coefficient

f = friction force = 22 [N]

Now we must calculate the force exerted by means of Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

F - f = m*a

where:

F = force exerted [N]

f = friction force [N]

m = mass = 95 [kg]

a = acceleration = 0.5 [m/s²]

Now replacing:

F - 22 = 95*0.5\\F = 47.5 + 22\\F = 69.5 [N]

6 0
3 years ago
What is 1113.28 in a sig fig?
Afina-wow [57]

Answer:

6 significant figure

Explanation:

The digits 111328 all are 6 figures with no figure being zero, neither zero after the other digits. In this case, all the numbers are significant and since they are only six numbers, then this is a six significant figure. In case we add another zero after digit 8, the zero is not significant but if added either infront of 8 or 2, the zero becomes significant.

4 0
3 years ago
What is the Kinetic Energy of the glass when it is
poizon [28]

Answer: 197

Explanation:

Because mechanical energy and mass it speeds up

7 0
2 years ago
Toy car in a science experiment covers 1.6 meters in half a second. If a the car travels at a steady speed, how far will it go i
Tanzania [10]
The answer is D. 32 m.

The simple equation that connects speed (v), time (t), and distance (d) can be expressed as:
v= \frac{d}{t}         ⇒ d=v*t

It is given:
v =  \frac{1.6m}{0.5s} = \frac{1.6m*2}{0.5s*2}= \frac{3.2m}{1s}  = 3.2 m/s
t = 10 s
d = ?

So:
d= v*t=3.2m/s*10s = 32m
3 0
2 years ago
The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of
natita [175]

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

\mu = Coefficient of friction = 0.4

Energy stored in spring is given by

U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J

As the energy in the system is conserved we have

\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s

The speed of the 8 kg block just before collision is 3.258 m/s

7 0
3 years ago
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