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2 years ago

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a car accelerates uniformly from rest to a speed of 51.9mi/h in 9.37s. find the constant acceleration of the car. answer in unit

s of m/s^2

1 answer:

Darina [25.2K]2 years ago

3 0

Here is my step-by-step-work. Let me know if you have any questions! :)

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Is this right?? please help me. IT IS SOCIOLOGY!!

Westkost [7]

Answer:

Yes, it is correct : )

Explanation:

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2 years ago

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The transfer of thermal energy is a major aspect of which branch of physics?

scZoUnD [109]

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The motion of a pendulum for which the maximum displacement from equilibrium does not change is an example of simple harmonic mo

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2 years ago

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After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg

kaheart [24]

Answer:

a) v₂ = 4.2 m/s

b) v₂ = 5 m/s

Explanation:

a)

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s$

<u>v₂ = 4.2 m/s</u>

<u></u>

b)

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)$

<u>v₂ = 5 m/s</u>

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3 years ago

In an electricity experiment, a 1.10 g plastic ball is suspended on a 56.0 cm long string and given an electric charge. A charge

Shalnov [3]

Answer:

Tension, T = 0.0115 N

Explanation:

Given that,

Mass of the plastic ball, m = 1.1 g

Length of the string, l = 56 cm

A charged rod brought near the ball exerts a horizontal electrical force F on it, causing the ball to swing out to a 21.0 degree angle and remain there. According to attached figure :

$T\cos\theta=mg$

T is tension in the string

$T=\dfrac{mg}{\cos\theta}\\\\T=\dfrac{1.1\times 10^{-3}\times 9.8}{\cos(21)}\\\\T=0.0115\ N$

So, the tension in the string is 0.0115 N.

8 0

3 years ago