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Elina [12.6K]
3 years ago
13

7. And 8. Please help, I would prefer someone to explain how to do it please

Physics
1 answer:
natta225 [31]3 years ago
6 0

#7

Given that an electron revolves in a circle and the essential centripetal force is given as

F = 4.60 \times 10^{-14}N

now we need to find its speed

so we know that this force is given as

F = \frac{mv^2}{R}

4.60 \times 10^{-14} = \frac{9.11 \times 10^{-31} v^2}{2 \times 10^{-2}}

v^2 = 1.01 \times 10^{15}

v = 3.18 \times 10^7 m/s

#8

Here the force due to earth on satellite is the centripetal force to revolve around earth

so we can say that force due to earth is given as

F = \frac{mv^2}{R}

we know that

m = 2.7 \times 10^3 kg

v = 4.7 \times 10^3 m/s

R = 1.8 \times 10^7 m

now from above equation we have

F = \frac{(2.7 \times 10^3)(4.7 \times 10^3)^2}{1.8 \times 10^7}

F = 3313.5 N

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a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. t_2=2.5714\ s is the time taken to stop after braking

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  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
  • distance between the cars, \Delta s=45\ m
  • deceleration of the leading car after braking, a_1=-2\ m.s^{-2}

a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

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v_2^2=u_2^2+2.a_2.\Delta s

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v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

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a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

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