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Marta_Voda [28]
3 years ago
6

If a car with a mass of 5Kg, has a momentum of 25kgm/s, how fast is it travelling?

Physics
2 answers:
Fantom [35]3 years ago
8 0

Answer:

5m/s

Explanation:

p=mv, or momentum (p) is equal to mass (m) times velocity (v).

so:

m=5Kg

p=25Kgm/s

v=p÷m

v=25÷5

v=5m/s

hoped this helped :)

Anna35 [415]3 years ago
6 0

Answer:

5hrs

Explanation:

Mass•Accelleration=Momentum (5kg•5hrs=25Mps)

Mass÷Momentum=Accelleration (25Mps÷5kg=5hrs)

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An archer puts a 0.30-kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m. Assuming th
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Answer:

41.74 m/s

Explanation:

The energy used to draw the bowstring = the kinetic energy of the arrow.

Fd = 1/2mv²................................ Equation 1

Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.

make v the subject of the equation

v = √(2Fd/m)...................... Equation 2

Given: F = 201 N, m = 0.3 kg, d = 1.3 m.

Substitute into equation 2

v = √(2×201×1.3/0.3)

v = √(1742)

v = 41.74 m/s.

Hence the arrow leave the bow with a speed of 41.74 m/s

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3 years ago
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What must be true about a surface in order for diffuse reflection to occur?
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Answer:

carpet

Explanation:

Diffuse reflection is the reflection of light from a surface such that an incident ray is reflected at many angles rather than at just one angle as in the case of specular reflection.

The structure of carpet's surface is as shown. Thus it shows large amount of diffuse reflection.

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3 years ago
What contributions did J.J. Thomson make to atomic history?
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J.J. Thomson discovered electrons and noticed that an atom can be divided. Also, he concluded atoms are made of positive cores and negatively charged particles within it.
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A charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a s
topjm [15]

Answer:

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Explanation:

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U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Now, we will apply this formula to both cases:

U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}

So, the change in the potential energy is

\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

7 0
3 years ago
A 0.050 kg bullet strikes a 5.0 kg wooden block with a velocity of 909 m/s and embeds itself in the block which fies off its sta
serg [7]

Answer:

The final velocity of the bullet is 9 m/s.

Explanation:

We have,

Mass of a bullet is, m = 0.05 kg

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The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,

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So, the final velocity of the bullet is 9 m/s.

5 0
3 years ago
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