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3 years ago

using mass and distance, identify and compare the sun's and moon's contribution to the formation of tides on earth

Physics

1 answer:

vekshin13 years ago

7 0

Answer:

Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth.

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A child at the top of a slide has a gravatational store of 1800j what is the childs maximum kinetic store as he slides down expl

Mrac [35]

Answer:

1800 J

Explanation:

Energy is conserved, so the maximum kinetic energy equals the change in gravitational energy.

7 0

3 years ago

The average speed of a car that travels 500 km in 5 hours is

ASHA 777 [7]

Answer:

100km/h

Explanation:

500km/5 = 100km per hour

5 0

4 years ago

A small wooden block with mass 0.750 kg is suspended from the lower end of a light cord that is 1.72 m long. The block is initia

Ierofanga [76]

Answer:

$v_{0}=319.2 m/s$

Explanation:

We need to use the momentum and energy conservation.

$p_{0}}=p_{f}$

$mv_{0}=(m+M)V_{1}$

Where:

m is the mass of bullet (m=0.01 kg)
M is the mass of the wooden (M=0.75 kg)
v(0) initial velocity of bullet
V(1) is the velocity of the combined object in the moment the bullet hist the block

Conservation of energy.

We have kinetic energy at first and kinetic and potential energy at the end.

$(1/2)(m+M)V_{1}^{2}=(1/2)(m+M)V_{2}^{2}+(m+M)gh$

Here:

V(1) is the velocity of the combined at the initial position
h is the vertical height ( h = 0.800 m)

We can find V(2) using the definition of force at this point:

$\Sigma F=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)$

$T-(m+M)gcos(\theta)=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)$

$cos(\theta) =(L-h)/L=(1.72-0.8)/1.72$

$\theta =57.66$

Now, we can solve the equation to find V(2)

$V_{2}=\sqrt{\frac{R*(T-(m+M)*g*cos(\theta))}{(m+M)}}$

$V_{2}=\sqrt{\frac{1.72*(4.86-(0.01+0.75)*9.81*cos(57.66))}{(0.01+0.75)}}$

$V_{2}=1.40 m/s$

Now we can find V(1) using the conservation of energy equation

$(1/2)V_{1}^{2}=(1/2)V_{2}^{2}+gh$

$V_{1}=\sqrt{V_{2}^{2}+2gh}$

$V_{1}=\sqrt{1.40^{2}+2*9.81*0.8}$

$V_{1}=4.20 m/s$

Finally, using the momentum equation we find v(0)

$v_{0}=\frac{(m+M)V_{1}}{m}$

$v_{0}=\frac{(0.01+0.75)*4.20}{0.01}$

$v_{0}=319.2 m/s$

I hope it helps you!

7 0

4 years ago

N discussing engines, the ratio of output work to input work expressed as a percentage is called

djyliett [7]

The ration of output work to input work expressed as a percentage is called <u>Efficiency</u>.

5 0

3 years ago

If a car has a mass of 200 kg and produces a force of 500 N how fast was the car accelerate

Crank

Answer:

2.5

Explanation:

force/weight

5 0

4 years ago

Read 2 more answers