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Lina20 [59]
3 years ago
9

A man claims that he can hold onto a 13.0-kg child in a head-on collision as long as he has his seat belt on. Consider this man

in a collision in which he is in one of two identical cars each traveling toward the other at 59.0 mi/h relative to the ground. The car in which he rides is brought to rest in 0.11 s.
Physics
1 answer:
Softa [21]3 years ago
4 0

Answer:

F = 3116.49 N

Explanation:

given,                          

mass of the child = 15 kg

speed = 59 mi/hr                    

time taken to come to rest = 0.11 s

force on the boy = ?                    

1 mi/hr = 0.447 m/s                    

59 mi/hr = 59 × 0.447 = 26.37 m/s

a = \dfrac{dv}{dt}

a = \dfrac{26.37}{0.11}

a=239.73 m/s²

                                                 

the calculation of force

Force = mass × acceleration

F = m × a

F = 13 × 239.73

F = 3116.49 N

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A 0.0450-kg golf ball initially at rest is given a speed of 25.2 m/s when a club strikes. part a part complete if the club and b
Ksenya-84 [330]
We are given information:
m = 0.0450 kg
Δv = 25.2 m/s
Δt = 1.95 ms = 0.00195s

To find force we use formula:
F = m * a

a is acceleration. To find it we use formula:
a = Δv / Δt 
a = 25.2 / 0.00195
a = 12923.1 m/s^2

Now we can find force:
F = 0.0450 * 12923.1
F = 581.5 N 


To check the effect of the ball's weight on this movement we need to calculate it and then compare it to this force.
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3 years ago
Two application of heat energy​
densk [106]

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4 0
3 years ago
Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.4 m from the slits. The center of the 3
VARVARA [1.3K]

Answer:

Part a)

f = 4.76 \times 10^{14} Hz

Part b)

d = 3.48 \times 10^{-4} m

Part c)

\theta = 0.311 degree

Explanation:

Part a)

As we know that the speed of light is given as

c = 3 \times 10^8 m/s

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now the frequency of the light is given as

f = \frac{c}{\lambda}

so we have

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Part b)

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so here we know for 3rd order maximum intensity

y_3 = 0.76 cm

n = 3

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0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}

d = 3.48 \times 10^{-4} m

Part c)

angle of third order maximum is given as

d sin\theta = 3 \lambda

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8 0
2 years ago
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