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Yakvenalex [24]
3 years ago
13

A 1500 kg car has the same momentum as a 2500 kg truck moving at 25 m/s. How fast is the car moving in m/s

Physics
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

41.6m/s

Explanation:

P=mv

2500kg × 25m/s = 62500kgm/s

62500kgm/s ÷ 1500 kg = 41.6m/s

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4 (JAMB) The diagram below shows a light see-saw, which is balanced horizontally by the weights W₁, W2, W3, W4 in the positions
damaskus [11]

The equation that represents the principle of the lever balance is:

  • W₁ + W₂ = W3 + W4; option A.

<h3>What is the principle of moments?</h3>

The principle of moments states when a body is in equilibrium, the sum of the clockwise moment about a point equals the sum of anticlockwise moment about that point.

A see-saw represents a balanced system of moments.

The sum of clockwise moment = The sum of anticlockwise moments.

Assuming W1 and W2 are clockwise moments and W3 and W4 are anticlockwise moments.

The equation will b:  W₁ + W₂ = W3 + W4

In conclusion, a balanced see-saw illustrates the principle of the lever balance.

Learn more about principle of moments at: brainly.com/question/20519177

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6 0
2 years ago
Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6
Nana76 [90]

Answer:

The answer to the question is as follows

The  acceleration due to gravity for low for orbit is  9.231 m/s²

Explanation:

The gravitational force is given as

F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }

Where F_{G} = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹\frac{Nm^{2} }{kg^{2} }

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by  F_{G} and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and  F_{G} = \frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} } = 9.231 N

Therefore the acceleration due to gravity =  F_{G} /mass  

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is  9.231 m/s²

4 0
3 years ago
Read 2 more answers
3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a
serious [3.7K]

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as \mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}.

Where, {V}_{i} is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

{V}_{i} =3 m/s, V = 0 m/s, and  ∆s = 2 m

Substitute the values in the above formula,

0=3^{2}-2 \times 2 \times a

0 = 9 - 4a

4a = 9

a=2.25 \mathrm{m} / \mathrm{s}^{2}

a=2.25 \mathrm{m} / \mathrm{s}^{2} is the acceleration.

Calculating Coefficient of friction:

\mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}

Compare the above equation

\mu \times m \times g=m \times a

Cancel "m" common term in both L.H.S and R.H.S

\text { Equation becomes, } \mu \times g=a

\text { Coefficient of friction } \mu=\frac{a}{g}

\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}

\mu=\frac{2.25}{9.8}

\mu=0.229

Hence coefficient of friction is 0.229.

calculating force:

\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

7 0
3 years ago
Need Help with this physics question, just a written response
Nadusha1986 [10]

Answer:

Explanation:

According to Newton's third law, every action has an equal and opposite reaction

so it tells us that the force exerted by the earth on the spacecraft is equal to the force exerted by the spacecraft on the earth. But we do not see the earth moving towards the spacecraft because the inertia of the spacecraft is very less than the inertia of the earth.

3 0
3 years ago
What are 2 uses for this tiny electromagnet?
Ray Of Light [21]

Electromagnets are used in all kinds of electric devices, including hard disk drives, speakers, motors, and generators, as well as in scrap yards to pick up heavy scrap metal. They're even used in MRI machines, which utilize magnets to take photos of your insides!

6 0
4 years ago
Read 2 more answers
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