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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 wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in t
Margarita [4]

Answer: 996m/s

Explanation:

Formula for calculating velocity of wave in a stretched string is

V = √T/M where;

V is the velocity of wave

T is tension

M is the mass per unit length of the wire(m/L)

Since the second wire is twice as far apart as the first, it will be L2 = 2L1

Let V1 and V2 be the speed of the shorter and longer wire respectively

V1 = √T/M1... 1

V2 = √T/M2... 2

Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1

The equations will now become

249 = √T/(m/L1) ... 3

V2 = √T/(m/2L1)... 4

From 3,

249² = TL1/m...5

From 4,

V2²= 2TL1/m... 6

Dividing equation 5 by 6 we have;

249²/V2² = TL1/m×m/2TL1

{249/V2}² = 1/2

249/V2 = (1/2)²

249/V2 = 1/4

V2 = 249×4

V2 = 996m/s

Therefore the speed of the wave on the longer wire is 996m/s

3 0
3 years ago
A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav
Vinvika [58]
<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

g=1.62\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of the moon

As we know y_{o}=0 , equation (2) is rewritten as:

y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

4 0
3 years ago
What is a good example of netwon 3rd law​
satela [25.4K]

A fish pushes water backwards in order to move forward is a good example of Newton's 3rd Law.

8 0
3 years ago
A car with a mass of 1.1 × 103 kilograms hits a stationary truck with a mass of 2.3 × 103 kilograms from the rear end. The initi
snow_lady [41]

Answer:

The velocity of the truck after this elastic collision is 15.7 m/s            

Explanation:

It is given that,

Mass of the car, m_1=1.1\times 10^3\ kg

Mass of the truck, m_2=2.3\times 10^3\ kg

Initial velocity of the car, u_1=22\ m/s

Initial velocity of the truck, u₂ = 0

After the collision the velocity of the car is, v₁ = -11 m/s

Let v₂ is the velocity of the truck after this elastic collision. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

1.1\times 10^3\times 22+0=1.1\times 10^3\times (11)+2.3\times 10^3\times v_2    

v_2=15.7\ m/s

So, the velocity of the truck after this elastic collision is 15.7 m/s. Hence, the correct option is (c).

4 0
3 years ago
Adam is teeing off on hole number two. The hole is 390 yards away. It is a par four hole. What club should he use to tee off? Ex
kkurt [141]

Answer:

A driver.

Explanation:

Using a driver while at least 350 yds away is better than using a iron, because it will be a waste of the par 4 as it is not as powerful as the driver.

3 0
3 years ago
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