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expeople1 [14]
3 years ago
5

If your speed triples, you need __________ times the distance to stop

Physics
3 answers:
aivan3 [116]3 years ago
5 0
Speed is a scalar quantity given by the rate of change in distance. Thus, it is given by distance covered divided by the time taken to cover the distance.
 Speed= distance/time
therefore, speed is directly proportional to the distance covered if time taken is kept constant, such that an increase in distance causes a corresponding increase in speed and vice versa, hence if the speed triples then you will need thrice times the distance to stop. 
butalik [34]3 years ago
5 0

Answer:

9 Times

Explanation:

The stopping distance of a car (or any traveling object) is proportional to the square of the speed of the car.

This is a consequence of the work-kinetic energy theorem, which states that the work done on the car is equal to its loss of kinetic energy:

W=K_i-K_f

Since the final speed of the car is zero, its final kinetic energy, so we can write:

W=K_i\\Fd=\frac{1}{2}mv^2

where

F is the force that stops the car (the force of friction)

d is the stopping distance

m is the mass of the car

v is the initial speed of the car

As we see from the equation, the stopping distance (d) depends on the square of the speed (v^2). Therefore, it the speed is tripled, the stopping distance will acquire a factor 3^2 = 9, so we will need 9 times the distance to stop.

phy2 years ago
0 0

9 times. Because distance is directly proportional to the square of time.
which means if time is 3sec distance will be 3(3)=9.

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A ball of mass 0.5kg is thrown at a man of mass 59.5kg standing on roller skates, at a speed of 5 m/s. It is stopped by the man.
Akimi4 [234]

The final velocity of the man standing on roller skates with the ball caught, after it is thrown at him at a speed of 5 m/s, is 0.042 m/s.                        

 

The velocity of the man with the ball can be calculated by conservation of linear momentum:

p_{i} = p_{f}

m_{b}v_{b_{i}} + m_{m}v_{m_{i}} = m_{b}v_{b_{f}} + m_{m}v_{m_{f}}

Where:

m_{b}: is the mass of the ball = 0.5 kg

m_{m}: is the <u>mass</u> of the man = 59.5 kg

v_{b_{i}}: is the initial velocity of the <u>ball</u> = 5 m/s

v_{m_{i}}: is the <u>initial velocity</u> if the <u>man</u> = 0 (he is standing still)

v_{b_{f}}: is the final velocity of the <u>ball</u> =?

v_{m_{f}}: is the <u>final velocity</u> of the <u>man</u> =?

Since the man catches the ball, his final velocity is the same that the final ball's velocity, so:

m_{b}v_{b_{i}} + m_{m}v_{m_{i}} = v_{f}(m_{b} + m_{m})

v_{f} = \frac{m_{b}v_{b_{i}} + m_{m}v_{m_{i}}}{m_{b} + m_{m}} = \frac{0.5 kg*5 m/s + 0}{0.5 kg + 59.5 kg} = 0.042 m/s

Therefore, the velocity of the man with the ball is 0.042 m/s.

Learn more about the conservation of linear momentum here:  

  • brainly.com/question/2141713?referrer=searchResults
  • brainly.com/question/14283213?referrer=searchResults

I hope it helps you!  

4 0
2 years ago
Air at 30°C and 2MPa flows at steady state in a horizontal pipeline with a velocity of 25 m/s. It passes through a throttle valv
erastovalidia [21]

Answer:

V_2=159.9\ m/s

T_2=290.6K

Explanation:

At initial condition

P=2 MPa

T=30°C

V=25 m/s

At final condition

P=0.3 MPa

Now from first law for open system

h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}

We know that for air

h= 1.010 x T  KJ/kg

1.01\times 303+\dfrac{25^2}{2000}=1.01\times T+\dfrac{V_2^2}{2000}                               -----1

Now from mass balance

\rho_1A_1V_1=\rho_2A_2V_2

We also know that

\rho=\dfrac{P}{RT}

\dfrac{P_1}{RT_1}V_1=\dfrac{P_2}{RT_2}V_2

\dfrac{2}{R\times 303}\times 25=\dfrac{0.3}{RT_2}V_2

T_2=1.81V_2                  ----------2                                                                                                

Now from equation 1 and 2

V_2^2+3673.749V-612916.49=0

So we can say that

V_2=159.9\ m/s

This is the outlet velocity.

Now by putting the values in equation 2

T_2=1.81\times 159.9  

T_2=290.6K

This is the outlet temperature.

4 0
4 years ago
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