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3 years ago

The tides on earth are caused mainly by earths gravitational interactions with which of the following?

Physics

1 answer:

Llana [10]3 years ago

3 0

<span>The tides on earth are caused mainly by earths gravitational interactions with the sun and the moon.</span>

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A mass m neutron has elastic collision with a mass m'

hoa [83]

Answer:

The neutron loses all of its kinetic energy to nucleus.

Explanation:

Given:

Mass of neutron is 'm' and mass of nucleus is 'm'.

The type of collision is elastic collision.

In elastic collision, there is no loss in kinetic energy of the system. So, total kinetic energy is conserved. Also, the total momentum of the system is conserved.

Here, the nucleus is still. So, its initial kinetic energy is 0. So, the total initial kinetic energy will be equal to kinetic energy of the neutron only.

Now, final kinetic energy of the system will be equal to the initial kinetic energy.

Now, as the nucleus was at rest initially, so the final kinetic energy of the nucleus will be equal to the initial kinetic energy of the neutron.

Thus, all the kinetic energy of the neutron will be transferred to the nucleus and the neutron will come to rest after collision.

Therefore, the neutron loses all of its kinetic energy to nucleus.

5 0

3 years ago

a train leaves a station heading east at 50 mph from that same station a car drives north at a speed of 30mph after 3 hours how

Alecsey [184]

Answer:

d = 175 miles

Explanation:

Train is moving towards East with constant speed of 50 mph

While car is moving at speed of 30 mph

so after t = 3 hours

the distance moved by the train is given as

$d_1 = vt$

$d_1 = (50)(3) = 150 miles$

at the same time the distance moved by the car is given as

$d_2 = 30(3) = 90 miles$

now we know that both car and train is moving perpendicular to each other

So the distance between them after t = 3 hours is given as

$d= \sqrt{d_1^2 + d_2^2}$

$d = \sqrt{150^2 + 90^2}$

$d = 175 miles$

8 0

3 years ago

An 1800 kg helicopter rises with an upward acceleration of 2.0 m/s?. What lifting force is supplied by its rotating blades?

Viktor [21]

Answer:

Lifting force, F = 21240 N

Explanation:

It is given that,

Mass of the helicopter, m = 1800 kg

It rises with an upward acceleration of 2 m/s². We need to find the lifting force supplied by its rotating blades. It is given by :

F = mg + ma

Where

mg is its weight

and "ma" is an additional acceleration when it is moving upwards.

So, $F=1800\ kg(9.8\ m/s^2+2\ m/s^2)$

F = 21240 N

So, the lifting force supplied by its rotating blades is 21240 N. Hence, this is the required solution.

4 0

3 years ago

A 66-foot-tall woman walks at 55 ft/s toward a street light that is 2424 ft above the ground. What is the rate of change of th

I am Lyosha [343]

Answer:

a. $\frac{dx}{dt}=20ft/s$

b. $\frac{d(x+L)}{dt}==25ft/s$

Explanation:

Using the triangle theorem both triangle the woman makes between the light so the rate of change of length can use geometry first

$tan(\beta)=\frac{24ft}{L+x}=\frac{6ft}{x}$

Solve to find the rate relation

$x=\frac{24}{6}*L$

$x=4*L$

Now the rate of the change rate

$\frac{dx}{dt}=4*\frac{dL}{dt}$

$\frac{dx}{dt}=4*5ft/s=20ft/s$

Finally the rate of her shadow moving

$\frac{d(x+L)}{dt}=\frac{dx}{dt}+\frac{dL}{dt}$

$\frac{d(x+L)}{dt}=20ft/s+5ft/s=25ft/s$

5 0

3 years ago

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