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NikAS [45]
3 years ago
5

A car is moving at a constant speed of 14 m/s when the driver presses down on the gas pedal and accelerates for 14 s with an acc

eleration of 1.8 m/s 2 . what is the average speed of the car during the period?
Physics
1 answer:
Vinil7 [7]3 years ago
7 0

acceleration of car is 1.8 m/s^2

time = 14 s

initial speed = 14 m/s

so the final speed is calculated by

v_f = v_i + at

v_f = 14 + 14 * 1.8

v_f = 39.2 m/s

so the total distance moved in this interval of time is

d = \frac{v_f + v_i}{2}* t

d = \frac{39.2 + 14}{2}* 14

d = 372.4 m

now the average speed is given as

v = \frac{d}{t}

v = \frac{372.4}{14}

v = 26.6 m/s

so the average speed will be 26.6 m/s

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An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.  

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

\sum F_x = \sum F_y = 0

\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }

  • So; On the L.H.S

\mathbf{mg sin \theta =f_s}

\mathbf{mg sin \theta =\mu_s N}

  • On the R.H.S

N = mg cos θ

Equating both force component together, we have:

\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}

\mathbf{sin \theta =\mu_s \ \ cos \theta}

\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}

From trigonometry rule:

\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}

∴

\mathbf{\mu_s =\tan \theta}}

\mathbf{\mu_s =\tan 35^0}}

\mathbf{\mu_s = 0.700}}

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

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An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.50 m above the ground. The railcar
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<span>An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.5 m above the ground. The railcar has a mass of 38,500 kg and is moving to the right at a constant speed of 8.7 m/s on a frictionless rail...
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A ball traveling at 15 m/s hits a bat with a force of 200N. How much force does the bat (moving at 20m/s)
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Answer:

200 N

Explanation:

Given that,

A ball traveling at 15 m/s hits a bat with a force of 200 N.

We need to find the force that the bat moving at 20 m/s hit the ball with.

We know that, this probelm is based on Newton's third law of motion. The force that the ball exerting on bat should be equal to the force that the bat exerting in the ball but in opposite direction.

It would mean that the ball hits the ball with a force of 200 N. Hence, the correct option is (a).

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A certain atom has only three energy levels. From lowest to highest energy, these levels are denoted n = 1, n = 2, and n = 3. Wh
shtirl [24]

Answer:

D. 160 nm

Explanation:

The energy released from n = 3 to n = 1 must be equal to the sum of energies released from n = 3 to n = 2 and from n = 2 to n = 1. Therefore,

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\frac{hc}{\lambda} = \frac{hc}{\lambda_{1}} + \frac{hc}{\lambda_{2}}\\\\\frac{1}{\lambda} = \frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}}

where,

λ = wavelength of photon released from 3 to 1 = ?

λ₁ = wavelength of photon released from 3 to 2 = 800 nm

λ₂ = wavelength of photon released from 2 to 1 = 200 nm

Therefore,

\frac{1}{\lambda} = \frac{1}{800\ nm} + \frac{1}{200\ nm}\\\\\frac{1}{\lambda} = 0.00625 nm^{-1}\\\\\lambda = \frac{1}{0.00625 nm^{-1}}\\\\\lambda = 160 nm

Therefore, the correct option is:

<u>D. 160 nm</u>

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3 years ago
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