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Pachacha [2.7K]
3 years ago
8

A force F produces an acceleration a on an object of mass m. A force 3F is exerted on a second object, and an acceleration 8a re

sults. What is the mass of the second object in terms of m?
a. 3m
b. 9m
c. 24m
d. (3/8)m
e. (8/3)m
Physics
1 answer:
Pie3 years ago
3 0

To solve this problem we will proceed to use Newton's second law for which the mass (m) multiplied by the acceleration (a) is defined as the Force (F) applied on a body, mathematically that is,

F = ma

According to the statement for the first object, the acceleration is,

a = \frac{F}{m} \rightarrow 1^{st} Object

For the second object the acceleration is,

8a = \frac{3F}{m_2} \rightarrow 2^{nd} Object

Solving for the mass of the second object,

m_2 = \frac{3F}{8a}

m_2 = \frac{3F}{8(F/m)}

m_2 = \frac{3}{8}m

Therefore the correct answer is D.

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Calculate the recoil speed of a 1.4 kg rifle shooting 0.006 kg bullets with muzzle speed of 800 m/a.3.43 m/s
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Answer:

a) 3.43 m/s

Explanation:

Due to the law of conservation of momentum, the total momentum of the bullet - rifle system must be conserved.

The total momentum before the bullet is shot is zero, because they are both at rest, so:

p_i = 0

Instead the total momentum of the system after the shot is:

p_f = mv+MV

where:

m = 0.006 kg is the mass of the bullet

M = 1.4 kg is the mass of the rifle

v = 800 m/s is the velocity of the bullet

V is the recoil velocity of the rifle

The total momentum is conserved, therefore we can write:

p_i = p_f

Which means:

0=mv+MV

Solving for V, we can find the recoil velocity of the rifle:

V=-\frac{mv}{M}=-\frac{(0.006)(800)}{1.4}=-3.43 m/s

where the negative sign indicates that the velocity is opposite to direction of the bullet: so the recoil speed is

a) 3.43 m/s

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What is a deadly diseases with no cure
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Two cars collide at an intersection. Car A , with a mass of 2000kg , is going from west to east, while car B , of mass 1400kg ,
ahrayia [7]

Complete Question:

Two cars collide at an intersection. Car A , with a mass of 2000 kg, is going from west to east, while car B, of mass 1500 kg, is going from north to south at 15 m/s. As a result, the two cars become enmeshed and move as one. As an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65∘ south of east from the point of impact. (a) How fast were the enmeshed cars moving just after the collision? (b) How fast was car A going just before the collision?

Answer:

a) 6.36 m/s b) 4.57 m/s

Explanation:

a) Assuming no external forces acting during the collision, total momentum must be conserved.

As momentum is a vector, we can decompose it along two directions perpendicular each other.

Just for convenience, we choose as our x-axis to the W-E direction, and as our y-axis, the direction N-S.

If we know that total momentum must be conserved, same must be true for both components, px and py.

Applying the information provided (both cars become enmeshed after the collision, moving at an angle of 65º south of east from the point of impact), we have:

px = ma * va = (ma+mb) * vab * cos 65º  (1)

py = mb * vb = (ma + mb) * vab * sin 65º (2)

Replacing by the values of ma, mb, and sin 65º, we can solve for vab, as follows:

vab = 1,400 kg* 14.0 m/s / (3,400 Kg * sin 65º) = 6.36 m/s

b) Replacing vab from above in (1), and solving for va, we have:

va = 3,400 kg* 6.36 m/s* cos 65º / 2,000 Kg = 4.57 m/s

7 0
3 years ago
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