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zvonat [6]
2 years ago
7

A person riding a bike has an initial velocity of 10 meters per second and accelerates at a constant rate of 2 meters per second

^2. After 5 seconds, the person would travel ___ meters.
Physics
1 answer:
Montano1993 [528]2 years ago
7 0

After 5 seconds, the person would have travel 75 m

<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

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

<h3>How to determine the final velocity</h3>

We'll begin by obtaining the final velocity of the person. This can be obtained as follow:

  • Initial velocity (u) = 10 m/s
  • Acceleration (a) = 2 m/s²
  • Time (t) = 5 s
  • Final velocity (v) = ?

a = (v – u) / t

2 = (v – 10) / 5

Cross multiply

v – 10 = 2 × 5

v – 10 = 10

Collect like terms

v = 10 + 10

v = 20 m/s

<h3>How to determine the distance travelled</h3>

The distance travelled can be obtained as follow:

  • Initial velocity (u) = 10 m/s
  • Final velocity (v) = 20 m/s
  • Acceleration (a) = 2 m/s²
  • Distance (s) = ?

v² = u² + 2as

Collect like terms

2as = v² - u²

Divide both sides by 2a

s = (v² - u²) / 2a

s = (20² - 10²) / (2 × 2)

s = 75 m

Thus, the distance travelled by the person is 75 m

Learn more about velocity:

brainly.com/question/3411682

Learn more about acceleration:

brainly.com/question/491732

#SPJ1

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m_a_m_a [10]

Answer:

C) W = - 190 J

Explanation:

Notation

Wf = work done by the friction force (unknown)

Ff = force of the friction

d = distance travelled by the box = (2 pi 1.82 m) = 11.435 m

6 0
3 years ago
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A vehicle moving at 5m/s, i . what should be the constant declaration in order to stop it within 15m? ii. How long it takes to s
Flura [38]

Answer:

Refer to the attachment for solution (1).

<h3><u>Calculating time taken by it to stop (t) :</u></h3>

By using the second equation of motion,

→ v = u + at

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  • u denotes initial velocity
  • t denotes time
  • a denotes acceleration

→ 0 = 5 + (-5/6)t

→ 0 = 5 - (5/6)t

→ 0 + (5/6)t = 5

→ (5/6)t = 5

→ t = 5 ÷ (5/6)

→ t = 5 × (6/5)

→ t = 6 seconds

→ Time taken to stop = 6 seconds

7 0
3 years ago
A wheel of mass 4kg is pulled up a plane inclined at 30° to the horizontal by a force of 45N applied to the axle and parallel to
Len [333]

Answer:

v = 10 m/s

Explanation:

Let's assume the wheel does not slip as it accelerates.

Energy theory is more straightforward than kinematics in my opinion.

Work done on the wheel

W = Fd = 45(12) = 540 J

Some is converted to potential energy

PE = mgh = 4(9.8)12sin30 = 235.2 J

As there is no friction mentioned, the remainder is kinetic energy

KE = 540 - 235.2 = 304.8 J

KE = ½mv² + ½Iω²

ω = v/R

KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²

v = √(2KE / (m + I/R²))

v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6

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A typical electric refrigerator has a power rating of 500 Watts, which is the rate (J/s) at which electrical energy is supplied
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Answer:

The rate of heat removed from inside the refrigerator is 300 watts.

Explanation:

By the First Law of Thermodynamics and the definition of a Refrigeration Cycle, we have the following formula to determine the rate of heat removed from inside the refrigerator (\dot Q_{L}), in watts:

\dot Q_{L} = \dot Q_{H}-\dot W (1)

Where:

\dot Q_{H} - Rate of heat released to the room, in watts.

\dot W - Rate of electric energy needed by the refrigerator, in watts.

If we know that \dot Q_{H} = 800\,W and \dot W = 500\,W, then the rate of heat removed from inside the refrigerator is:

\dot Q_{L} = \dot Q_{H}-\dot W

\dot Q_{L} = 300\,W

The rate of heat removed from inside the refrigerator is 300 watts.

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