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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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Yakvenalex [24]
T = √(h)/(0.5)(9.81)
t = √(25)/(4.905)
t = √5.1
t = 2.26 seconds

hope this helps and have a great day :)



3 0
2 years ago
The drawing shows three identical springs hanging from the ceiling. Nothing is attached to the first spring, whereas a 4.50-N bl
Pavlova-9 [17]

Answer:

a. 30 N / m

b. 9.0 N

Explanation:

Given that

Unstretched length of the spring, L_o = 20.0cm = 0.2m

a) When the mass of 4.5N is hanging from the second spring, then extended length Is

L_1 = 35.0cm =  0.35m

So, the change in spring length when mass hangs is

x =  L_1 - L_o

= (0.35 - 0.20) m

= 0.15m

As spring are identical

Let us assume that the spring constant be "k", so at equilibrium

Restoring Force on spring = Block weightage

kx =  W =  4.50

k= \frac{4.50}{x} = \frac{4.50}{0.15}

= 30 N / m

b)  Now for the third spring, stretched the length of spring is

L_2 = 50cm = 0.5m

So, the change in spring length is

x'= L_2 - L_o

= (0.5-0.20)m

=  0.30m

At equilibrium,

Restoring Force on spring = Block weightage

Now using all mentioned and computed values in above,

W'= kx'

= 30(0.3)

= 9.0 N

4 0
2 years ago
A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by
Airida [17]

Answer:

Mobility of the minority carriers, \mu_{n} =1184.21 cm^{2} /V-sec

Diffusion coefficient for minority carriers,D_{n} = 29.20 cm^2 /s

Verified from Einstein relation as  \frac{D_{n} }{\mu_{n} }  = 25 mV

Explanation:

Length of sample, l_{s} = 2 cm

Separation between the two probes, L = 1.8 cm

Drift time, t_{d} = 0.608 ms

Applied voltage, V = 5 V

Mobility of the minority carriers ( electrons), \mu_{n} = \frac{V_{d} }{E}

Where the drift velocity, V_{d} = \frac{L}{t_{d} }

V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s

and the Electric field strength, E = \frac{V}{l_{s} }

E = 5/2

E = 2.5 V/cm

Mobility of the minority carriers:

\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec

The electron diffusion coefficient, D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }

\triangle x = (\triangle t )V_{d}, where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)

\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3}  }\\\triangle x =0.533 cm

D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s

For the Einstein equation to be satisfied, \frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V

\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV

Verified.

4 0
2 years ago
A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end tex
love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

\Delta x = v_{i}.t + \frac{a}{2}.t^{2}

v_{i} is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

\Delta x = 5(6) - \frac{4}{2}.6^{2}

\Delta x = 30 - 2.36

\Delta x = - 42

Displacement of the boat for t=6.0s interval is \Delta x = - 42m, i.e., 42 m to the left.

8 0
2 years ago
A container explodes and breaks into three fragments that fly off 120° apart from each other, with mass ratios 1: 4: 2. If the f
RSB [31]

Answer:

V₂ = 1.5 m/s

Explanation:

given,

speed of the first piece = 6 m/s

speed of the third piece = 3 m/s

speed of the second fragment = ?

mass ratios = 1 : 4 : 2

fragment break  fly off = 120°

α = β = γ  = 120°

sin α = sin β = sin γ = 0.866

using lammi's theorem

\dfrac{A}{sin\alpha}=\dfrac{B}{sin\beta}=\dfrac{C}{sin\gamma}

A,B and C is momentum of the fragments

\dfrac{m\times 6}{0.866}=\dfrac{4m\times v_2}{0.866}=\dfrac{2m\times 3}{0.866}

4 x V₂ = 2 x 3

V₂ = 1.5 m/s

3 0
2 years ago
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