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anyanavicka [17]
2 years ago
4

A ski jumper starts with a horizontal take-off velocity of 25 m/s and lands on a straight landing hill inclined at 30°. determi

ne (a) the time between take-off and landing, (b) the length d of the jump, (c) the maximum vertical distance between the jumper and the landing hill.

Physics
1 answer:
nlexa [21]2 years ago
6 0
Refer to the diagram shown below.

The time between take off and landing is
t= \frac{dcos(30^{o}) \, m}{25 \, m/s} =0.0346d \, s

The vertical drop is
h=dsin(30^{o})= \frac{1}{2}gt^{2} \\ 0.5d = 0.5(9.8)(0.0346d)^{2} \\0.5=0.00588d\\d=85.034 \, m

Part (a)
Therefore,
t = 0.034685.034 = 2.946 s

Answer:  2.95 s (nearest hundredth)

Part (b)
The length of the jump is d = 85.034 m

Answer:  85.0 m (nearest tenth)

Part (c)
The vertical distance is
h = d sin(30) = 42.52 m

Answer: 42.5 m (nearest tenth)

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hoa [83]

Answer:

<h2>187,500N/m</h2>

Explanation:

From the question, the kinectic energy of the train will be equal to the energy stored in the spring.

Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².

Equating both we will have;

1/2 mv² = 1/2ke²

mv² = ke²

m is the mass of the train

v is the velocity of then train

k is the spring constant

e is the extension caused by the spring.

Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m

Substituting this values into the formula will give;

30000*4² =  k*1.6²

k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}

The value of the spring constant is 187,500N/m

7 0
2 years ago
A girl rolls a ball up an incline and allows it to return to her. For the angle ! and ball involved, the acceleration of the bal
Musya8 [376]

Answer:

The distance the ball moves up the incline before reversing its direction is 3.2653 m.

The total time required for the ball to return to the child’s hand is 3.2654 s.

Explanation:

When the girl is moving up:

The final velocity (v) = 0 m/s

Initial velocity (u) = 4 m/s

a = -0.25g = -0.25*9.8 = -2.45 m/s². (Negative because it is in opposite of the velocity and also it deaccelerates while going up).

Let time be t  to reach the top.

Using

v = u + a×t

0 = 4 - 2.45*t

t = 1.6327 s

Since, this is the same time the ball will come back. So,

<u>Total time to go and come back = 2* 1.6327 = 3.2654 s </u>

To find the distance, using:

v² = u² + 2×a×s

0² = 4² + 2×(-2.45)×s

s = 3.2653 m

<u>Thus, the distance the ball moves up the incline before reversing its direction is 3.2653 m.</u>

5 0
2 years ago
if you apply a Force of F1 to area A1 on one side of a hydraulic jack, and the second side of the jack has an area that is twice
Firlakuza [10]

Answer:

2F_{1}

Explanation:

F₁ = Force on one side of the jack

A₁ = Area of cross-section of one side of the jack

F₂ = Force on second side of the jack

A₂ = Area of cross-section of second side of the jack = 2 A₁

Using pascal's law

\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{A_{_{2}}}

\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{2A_{_{1}}}

F_{1}= \frac{F_{2}}{2}\\

F_{2}= 2F_{1}

7 0
2 years ago
A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.
Bumek [7]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

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extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver  will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

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hence, number of car of additional car length  n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

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Answer:

c

Explanation:

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4 0
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