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attashe74 [19]
3 years ago
6

A person pushes horizontally with a force of 220. N on a 61.0 kg crate to move it across a level floor. The coefficient of kinet

ic friction is 0.270. (a) What is the magnitude of the frictional force? (b) What is the magnitude of the crate's acceleration? Use g=9.81 m/s2.
Physics
1 answer:
Ede4ka [16]3 years ago
7 0

Answer:

(a) 161.57 N

(b) 0.958 m/s^2

Explanation:

Force applied, F = 220 N

mass of crate, m = 61 kg

μ = 0.27

(a) The magnitude of the frictional force,

f = μ N

where, N is the normal reaction

N = m x g = 61 x 9.81 = 598.41 N

So, the frictional force, f = 0.27 x 598.41

f = 161.57 N

(b) Let a be the acceleration of the crate.

Fnet = F - f = 220 - 161.57

Fnet = 58.43 N

According to newton's second law

Fnet = mass x acceleration

58.43 = 61 x a

a = 0.958 m/s^2

Thus, the acceleration of the crate is 0.958 m/s^2.  

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Answer:

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As a train accelerates away from a station, it reaches a speed of 4.9 m/s in 5.4 s. if the train's acceleration remains constant
romanna [79]

The speed of the train after an additional 5 s has elapsed is 9.45 m/s

<h3>What is acceleration? </h3>

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

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  • v is the final velocity
  • u is the initial velocity
  • t is the time

<h3>How to determine the acceleration in 5.4 s</h3>
  • Initial velocity (u) = 0 m/s
  • Final velocity (v) = 4.9 m/s
  • Time (t) = 5.4 s
  • Acceleration (a) =?

a = (v – u) / t

a = (4.9 – 0) / 5.4

a = 4.9 / 5.4

a = 0.91 m/s²

<h3>How to determine the final velocity after 5 s</h3>
  • Initial velocity (u) = 4.9 m/s
  • Acceleration (a) = 0.91 m/s²
  • Time (t) = 5 s
  • Final velocity (v) = ?

a = (v – u) / t

0.91 = (v – 4.9) / 5

Cross multiply

v – 4.9 = 0.91 × 5

v – 4.9 = 4.55

Collect like terms

v = 4.55 + 4.9

v = 9.45 m/s

Learn more about acceleration:

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3 0
2 years ago
Two balls have masses 18 kg and 47 kg. The 18 kg ball has an initial velocity of 76 m/s (to the right) along a line joining the
BigorU [14]

Answer:

The final velocity of 18 kg ball is V_{2} = 42.09 \frac{m}{sec}

Explanation:

Mass of first ball m_{1} = 18 kg

Mass of second ball m_{2} = 47 kg

Initial velocity of 18 kg ball V_{1} = 76 \frac{m}{sec}

Initial velocity of 47 kg ball = 0

Final velocity of 18 kg ball V_{2} = ??

Final velocity of 18 kg ball  is given by the formula

V_{2} = \frac{2 m_{1} V_{1} }{m_{1} + m_{2}  }

Put all the values in above formula we get

V_{2} = 2 × 18 × \frac{76}{65}

V_{2} = 42.09 \frac{m}{sec}

Thus, the final velocity of 18 kg ball is V_{2} = 42.09 \frac{m}{sec}

3 0
3 years ago
When does resonance occur?
bija089 [108]
Resonance occurs<span> when the amplitude of an object's oscillations are increased by the matching vibrations of another object.

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4 0
4 years ago
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A automotive test driver travels due north in a prototype hybrid vehicle at 30 mi/h
Nikolay [14]

Answer:

Option A. 40 mi/h

Explanation:

To obtain the average speed of the vehicle, we'll begin by calculating the distance travelled by the vehicle in each case. This is illustrated below:

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Distance =...?

Speed = Distance /Time

30 = Distance /2

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Total distance travelled = 120 mi

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Average speed = Total Distance travelled /Total time

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Therefore, the average speed of the vehicle is 40 mi/h

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