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3 years ago

Example of Newton’s law of universal gravitation…

Physics

1 answer:

insens350 [35]3 years ago

4 0

The force that holds the gases in the sun. The force that causes a ball you throw in the air to come down again

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The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps an

maksim [4K]

Answer:

The approximate spring constant is $k = 55533.33 \ N/m$

Explanation:

From the question we are told that

The mass of the person is $m = 68 \ kg$

The dip of the car is $x = 1.2 \ cm = 0.012 \ m$

Generally according to hooks law

$F = k * x$

here the force F is the weight of the person which is mathematically represented as

$F = m * g$

=> $m * g = k * x$

=> $k = \frac{m * g }{x }$

=> $k = \frac{68 * 9.8}{ 0.012}$

=> $k = 55533.33 \ N/m$

8 0

3 years ago

Which has more KE a baseball traveling at 50 mph or a baseball travelling at 100 mph? Explain.

Trava [24]

A baseball traveling at 100 mph has more kinetic energy than a baseball traveling at 50 mph because the kinetic energy = 1/2 x mass x velocity. Since the baseballs should have the same mass, the velocity is what will determine which ball has more kinetic energy. Since the 100 mph baseball has a higher velocity than the 50 mph baseball, it has more kinetic energy.

6 0

2 years ago

Given the following lens combination:

natulia [17]

Given:

Lens.........diameter ...fl#

eyepiece...2cm............5

objective...40cm........15

focal length of eyepiece = 2*5 = 10cm

focal length of objective = 40*15 = 600cm

magnification = FL obj / FL eyp = 600/10 = 60x

7 0

3 years ago

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You create a ramp using two text books and a 0.50m board. Using a timer you determine that a cart can roll down the ramp in 0.55

ahrayia [7]

Answer:

The velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.

Explanation:

Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:

$v=v_0+at\\\\v^2=v_0^2+2ad$

Dividing the second equation by the first one, we obtain:

$v=\frac{v_0^2+2ad}{v_0+at}$

And, since $v_0=0$ , then:

$v=\frac{2ad}{at}\\\\v=\frac{2d}{t}\\\\v=\frac{2(0.50m)}{0.55s}\\\\v=1.81m/s$

It means that the velocity at the bottom of the ramp is 1.81m/s.

We could use this data, plus any of the two initial equations, to determine the acceleration:

$v=v_0+at\\\\\implies a=\frac{v}{t}\\\\a=\frac{1.81m/s}{0.55s}\\\\a=3.30m/s^2$

So the acceleration is 3.30m/s^2.

7 0

3 years ago

Is the expression "The bigger they are, the harder they fall" a generally true statement since, in the absence of air resistance

lions [1.4K]

No, because in oxygen depraved rooms, if you drop a feather and a bowling ball at the same height and time, they will fall at the same speed and have the same amount of impact.

6 0

3 years ago

Read 2 more answers