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STatiana [176]
3 years ago
11

A 14-kg object moving with a constant velocity

Physics
1 answer:
natta225 [31]3 years ago
4 0

The final velocity of the 14 kg object is 1.6 m/s in the same direction

Explanation:

We can solve this problem by using the law of conservation of momentum: the total momentum of the system must be conserved before and after the collision. Therefore, we can write

p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

m_1 = 14 kg is the mass of the first object

u_1 = 5.0 m/s is the initial velocity of the first object

v_1 is the final velocity of the first object

m_2 = 8.0 kg is the mass of the second object

u_2 = 3.0 m/s is the initial velocity of the second object

v_2 = 9.0 m/s is the final velocity of the second object

Re-arranging the equation and substituting the values, we find:

v_1 = \frac{m_1 u_1 + m_2 u_2 - m_2 v_2}{m_1}=\frac{(14)(5.0)+(8.0)(3.0)-(8.0)(9.0)}{14}=1.6 m/s

And the direction is the same as the initial direction, since it has the same sign.

Learn more about conservation of momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

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In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x=
swat32

In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x=20.0 cm to x=60.0cm.

<h3>Where is the electric potential, when the particle moved?</h3>

The charge field system's electric potential energy rose. The particle experiences an electric force that is directed against the x-axis. It is pushed uphill by an outside force, which raises the potential energy.

When a charge to be moved against an applied electric field, electric potential energy is needed. A charge must be moved through a stronger electric field with more energy than it would require to carry it via a weaker electric field.

In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x=20.0 cm to x=60.0cm.

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The correct option is a).

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7 0
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A nail gets magnetized when kept close to a magnet
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A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a hal
Karolina [17]

Answer:

d = 44.64 m

Explanation:

Given that,

Net force acting on the car, F = -8750 N

The mass of the car, m = 1250 kg

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Final speed, v = 0 (it stops)

The formula for the net force is :

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a is acceleration of the car

a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2

Let d be the breaking distance. It can be calculated using third equation of motion as :

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m

So, the required distance covered by the car is 44.64 m.

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