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Rzqust [24]
2 years ago
11

An Olympic diver is on a diving platform 3.80 m above the water. To start her dive, she runs off of the platform with a speed of

1.24 m/s in the horizontal direction. What is the diver's speed just before she enters the water
Physics
1 answer:
AVprozaik [17]2 years ago
3 0

Answer:

the diver's speed just before she enters the water is 8.718 m/s.

Explanation:

Given;

height of the diving platform, h = 3.8 m

initial velocity of the diver, u = 1.24 m/s

the diver's speed just before she enters the water is her final velocity, = v

Apply the following kinematic equation to determine the final velocity of the girl;

v² = u² + 2gh

v² = (1.24)² + 2(9.8 x 3.8)

v² = 76.018

v = √76.018

v = 8.718 m/s

Therefore, the diver's speed just before she enters the water is 8.718 m/s.

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What does kinetic energy do when it go up a hill?what do potential energy do when it does up a hill?
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Answer:

Kinetic energy decreases as you go up hill

Potential energy increases as you go up hill

6 0
2 years ago
Two billiard balls move toward each other on a table. The mass of the number three ball, m1, is 5 g with a velocity of 3 m/s. Th
Ymorist [56]

This question involves the concepts of the law of conservation of momentum and velocity.

The velocity of the eight ball is "5.7 m/s".

According to the law of conservation of momentum:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m₁ = mass of number three ball = 5 g

m₂ = mass of the eight ball = 6 g

u₁ = velocity of the number three ball = 3 m/s

u₂ = velocity of the eight ball = - 1 m/s (negative sign due to opposite direction)

v₁ = final velocity of the three number ball = - 5 m/s

v₂ = final velocity of the eight ball = ?

Therefore,

(5 g)(3 m/s) + (6 g)(- 1 m/s) = (5 g)(- 5 m/s) + (6 g)(v₂)

v_2=\frac{34\ g.m/s}{6\ g}\\\\

<u>v₂ = 5.7 m/s</u>

<u></u>

Learn more about the law of conservation of momentum here:

brainly.com/question/1113396?referrer=searchResults

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2 years ago
To completely describe the motion of an object, you need
notsponge [240]

Answer:

D

Explanation:

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3 years ago
Critics of the electoral college argue that the system can result in the selection of —
Romashka [77]

Answer:

President who did not win the popular vote

Explanation:

Critics of the Electoral college argue that the system can result in a selection of what?

6 0
2 years ago
Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



7 0
3 years ago
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