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Montano1993 [528]
3 years ago
12

A car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver n

otices a red light ahead and slows down with constant acceleration −a0. Just as the car comes to a full stop, the light immediately turns green, and the car then accelerates back to its original speed v0 with constant acceleration a0. During the same time interval, the train continues to travel at the constant speed v0.Part A :- How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0Part B:- How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.Part C:- The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.
Physics
1 answer:
satela [25.4K]3 years ago
8 0

Answer:

a) t1 = v0/a0

b) t2 = v0/a0

c) v0^2/a0

Explanation:

A)

How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Vf = 0

Vf = v0 - a0*t

0 = v0 - a0*t

a0*t = v0

t1 = v0/a0

B)

How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

at this point

U = 0

v0 = u + a0*t

v0 = 0 + a0*t

v0 = a0*t

t2 = v0/a0

C)

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

t1 = t2 = t

Distance covered by the train = v0 (2t) = 2v0t

and we know t = v0/a0

so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0

now distance covered by car before coming to full stop

Vf2 = v0^2- 2a0s1

2a0s1 = v0^2

s1 = v0^2 / 2a0

After the full stop;

V0^2 = 2a0s2

s2 = v0^2/2a0

Snet = 2v0^2 /2a0 = v0^2/a0

Now the separation between train and car

= (2v0^2)/a0 - v0^2/a0

= v0^2/a0

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Gre4nikov [31]

Answer:

a) h_m=74625\ m

b) t=265.55\ s

Explanation:

Given:

  • mass of rocket, m_r=200\ kg
  • mass of fuel, m_f=100\ kg
  • acceleration of the rocket consuming fuel, a=30\ m.s^{-2}
  • time after which the fuel exhaust, t_f=35\ s

<u>During the phase of fuel exhaustion:</u>

<u>velocity attained by the rocket just as the fuel ends:</u>

v_f=u+a.t_f

where:

u= initial velocity of the rocket = 0

v_f=0+30\times 35

v_f=1050\ m.s^{-1} this will be the initial velocity for the phase of ascending of the rocket's height under the influence of gravity.

<u>height at which the fuel finishes:</u>

v_f^2=u^2+2a.h_f

1050^2=0^2+2\times 30\times h_f

h_f=18375\ m

<u>During the phase of ascend in height of rocket after the fuel is over:</u>

<u>Time taken to reach the top height after the fuel is over:</u>

v=v_f+g.t'

at top v = (final velocity during this course of motion )= 0 m.s^{-1}

0=1050-9.8\times t'

t'=107.1429\ s

<u>Height ascended by the rocket after the fuel is over:</u>

v^2=v_f^2+2g.h'

at the top height the velocity is zero

0^2=1050^2-2\times 9.8\times h' (-ve sign denotes that the direction of motion is opposite to that of acceleration)

h'=56250\ m

<u>Therefore the maximum altitude attained by the rocket:</u>

h_m=h_f+h'

h_m=18375+56250

h_m=74625\ m

b)

time taken by the rocket to fall back to the earth:

h_m=v.t_m+\frac{1}{2} g.t_m^2

where:

v= initial velocity of the rocket during the course of free fall from the top height.

74625=0+4.9\times t_m^2

t_m=123.41\ s

Now the total time for which the rocket is in the air:

t=t_f+t'+t_m

t=35+107.1429+123.41

t=265.55\ s

4 0
3 years ago
An electron passes location &lt; 0.02, 0.04, -0.06 &gt; m and 5 us later is detected at location &lt; 0.02, 1.62,-0.79 &gt; m (1
hram777 [196]

Answer:

(a)

Average velocity = < 0, 316000, 146000> m/s

(b) < 0, 2.844, 1.314 > m

Explanation:

r1 = < 0.02, 0.04, - 0.06 > m

r2 = < 0.02, 1.62, - 0.79 > m

time, t = 5 micro second = 5 x 10^-6 s

(a) Average velocity is defined as the ratio of total displacement to the total time taken.

Displacement = r = r2 - r1

r = < 0.02 - 0.02, 1.62 - 0.04, - 0.79 + 0.06 > m

r = < 0, 1.58, - 0.73 > m

So. Average velocity =  \frac{< 0, 1.58, - 0.73 >}{5 \times 10^{-6}}

Average velocity = < 0, 316000, 146000> m/s

Average velocity = < 0, 316000, 146000> m/s

(b)

Distance = velocity x time

Here time, t = 9 micro second

d =  < 0, 316000, 146000>  x 9 x 10^-6 m

d = < 0, 2.844, 1.314 > m

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what is the mechanical advantage of a crowbar when a worker uses 10N of force to pry open a window that has a resistance of 500N
Oksana_A [137]

Answer:

50

Explanation:

The mechanical advantage of a machine is given by

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where

F_{out} is the output force

F_{in} is the input force

For the crowbar in this problem,

F_{in}=10 N is the force in input applied by the worker

F_{out}=500 N is the force that the machine must apply in output to overcome the resistance of the window and to open it

Substituting into the equation, we find

MA=\frac{500}{10}=50

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