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2 years ago

If 2 objects had the same momentum, what must be true about the mass of the object that traveled the fastest?

Physics

1 answer:

julsineya [31]2 years ago

4 0

Yes, the above-given statement is true

<u>Explanation:</u>

The product of the mass x the velocity will be the same for both. Momentum is the action of a body with a particular mass through space and there is the conservation of momentum.
Momentum is described as the mass of the object multiplied by its velocity.
<u>Momentum (p) = Mass (M) * Velocity (v)</u>
Therefore for two objects with many masses to have a similar momentum, then the lighter one has to be moving quicker than the heavier object.

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Answer:

Explanation:

Given the following data;

Mass = 35 kg

Velocity = 3 m/s

To find the kinetic energy of the child;

K.E = ½mv²

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3 years ago

A small sandbag is dropped from rest from a hovering hot-air balloon. (assume the positive direction is upward.) (a) after 1.5 s

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solution:

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2 years ago

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Explanation:

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2 years ago

In 1911, Ernest Rutherford and his assistants Geiger and Marsden conducted an experiment in which they scattered alpha particles

alexandr402 [8]

Answer:

Explanation:

We shall apply conservation of mechanical energy

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1/2 mv² = k q₁q₂/d , d is closest distance

d = 2kq₁q₂ / mv²

= 2 x 9 x 10⁹ x 79e x 2e / 4mv²

= 1422 x2x (1.6 x 10⁻¹⁹)² x 10⁹ /4x 1.67 x 10⁻²⁷ x (1.5 x 10⁷)²

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3 years ago

A nylon guitar string is fixed between two lab posts 2.00 m apart. The string has a linear mass density of μ=7.20 g/m\mu=7.20~\t

vladimir2022 [97]

Answer:

4.6 m

Explanation:

First of all, we can find the frequency of the wave in the string with the formula:

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where we have

L = 2.00 m is the length of the string

T = 160.00 N is the tension

$\mu =7.20 g/m = 0.0072 kg/m$ is the mass linear density

Solving the equation,

$f=\frac{1}{2(2.00 m)}\sqrt{\frac{160.00 N}{0.0072 kg/m}}=37.3 Hz$

The frequency of the wave in the string is transmitted into the tube, which oscillates resonating at same frequency.

The n=1 mode (fundamental frequency) of an open-open tube is given by

$f=\frac{v}{2L}$

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v = 343 m/s is the speed of sound

Using f = 37.3 Hz and re-arranging the equation, we find L, the length of the tube:

$L=\frac{v}{2f}=\frac{343 m/s}{2(37.3 Hz)}=4.6 m$

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3 years ago