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never [62]
2 years ago
13

A 1700kg rhino charges at a speed of 50.0km/h. what is the magnitude of the average force needed to bring the rhino to a stop in

0.50s? express your answer to two significant figures.

Physics
2 answers:
klio [65]2 years ago
7 0
First, let's convert the initial speed of the rhino:
v_i = 50 km/h = 13.9 m/s
The acceleration of the rhino is given by:
a= \frac{v_f-v_i}{t}= \frac{0-13.9 m/s}{0.5 s}=-27.8m/s^2
where the negative sign means it is a deceleration. We can neglect the negative sign since we are interested only in the magnitude of the force applied, which is given by Netwon's second law:
F=ma = (1700 kg)(27.8 m/s^2)=4.7 \cdot 10^4 N
katrin [286]2 years ago
3 0

The magnitude of the average force needed to bring the rhino to a stop in 0.50s is about 4.7 × 10⁴ Newton

\texttt{ }

<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

\large {\boxed {F = ma }

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

mass of rhino = m = 1700 kg

initial speed of rhino = u = 50.0 km/h = 13⁸/₉ m/s

final speed of rhino = v = 0 m/s

time taken = t = 0.50 s

<u>Asked:</u>

average force = F = ?

<u>Solution:</u>

F = ma

F = m \times \frac{v - u}{t}

F = 1700 \times \frac{0 - 13\frac{8}{9}}{0.50}

F = 1700 \times \frac{-250}{9}

F \approx -4.7 \times 10^4 \texttt{ Newton}

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441
  • Newton's Law of Motion: brainly.com/question/10431582
  • Example of Newton's Law: brainly.com/question/498822

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

\texttt{ }

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

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Answer:

m = 0.111 kg

Explanation:

Heat required to release from the body of the person when his temperature cool down by 1 degree C is given as

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here we know that

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7 0
2 years ago
A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa
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The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

\texttt{ }

<h3>Further explanation</h3>

Let's recall Impulse formula as follows:

\boxed {I = \Sigma F \times t}

<em>where:</em>

<em>I = impulse on the object ( kg m/s )</em>

<em>∑F = net force acting on object ( kg m /s² = Newton )</em>

<em>t = elapsed time ( s )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of ball = m = 0.500 kg

initial speed of ball = vo = 20.0 m/s

final kinetic energy = Ek = 70% Eko

<u>Asked:</u>

magnitude of the change of momentum of the stone = Δp = ?

<u>Solution:</u>

<em>Firstly, we will calculate the final speed of the ball as follows:</em>

Ek = 70\% \ Ek_o

\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )

v^2 = 70 \% \ (v_o)^2

v = - v_o \sqrt{70 \%} → <em>negative sign due to ball rebounds</em>

v = - v_o \sqrt{0.7} \texttt{ m/s}

\texttt{ }

<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

\Delta p_{stone} = - \Delta p_{ball}

\Delta p_{stone} = - [ mv - mv_o ]

\Delta p_{stone} = m[ v_o - v ]

\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]

\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]

\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]

\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}

\texttt{ }

<h3>Learn more</h3>
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\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

8 0
2 years ago
The speed of a bus increases uniformly from 15 ms per second to 60 ms per second in 20 seconds. calculate 1. the average speed 2
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vs = 15m/s
t=20s
a = vf-vs/t
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Answer:

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Explanation:

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