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cluponka [151]
3 years ago
8

When you drop an object from a certain height, it takes time T to reach the ground with no air resistance. If you dropped it fro

m three times that height, how long (in terms of T) would it take to reach the ground?
Physics
1 answer:
alexandr402 [8]3 years ago
5 0

Answer:

Explanation:

Given

When we drop an object from height , suppose h

it takes time T

using equation of motion

h=ut+\frac{1}{2}at^2

where  

h=displacement

u=initial\ velocity

a=acceleration

t=time

here u=0 because it dropped from a certain height

h=\frac{1}{2}gT^2

T=\sqrt{\frac{2h}{g}}

When height is increases to three times of original height

i.e. h'=3 h

then time period becomes

T'=\sqrt{\frac{2\times 3h}{g}}

T'=\sqrt{3}T

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A ball is thrown so that its initial vertical and horizontal components of velocity are 30 m/s and 15 m/s, respectively. Estimat
mihalych1998 [28]

Answer:

H = 45 m

Explanation:

First we find the launch velocity of the ball by using the following formula:

v₀ = √(v₀ₓ² + v₀y²)

where,

v₀ = launching velocity = ?

v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s

v₀y = Vertical Component of Launch Velocity = 30 m/s

Therefore,

v₀ = √[(15 m/s)² + (30 m/s)²]

v₀ = 33.54 m/s

Now, we find the launch angle of the ball by using the following formula:

θ = tan⁻¹ (v₀y/v₀ₓ)

θ = tan⁻¹ (30/15)

θ = tan⁻¹ (2)

θ = 63.43°

Now, the maximum height attained by the ball is given by the formula:

H = (v₀² Sin² θ)/2g

H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)

<u>H = 45 m</u>

6 0
3 years ago
Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's
mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

4 0
3 years ago
What two things are necessary for work to be done on an object?
NemiM [27]

Answer: Force and Movement

Explanation:

The first is that the object moves. The second is that a force must act on the object in the direction the object moves.

6 0
3 years ago
The moon’s period of revolution is 27 1/3 days, and its period of rotation about its axis is _____.
mestny [16]
It lasts 29 1/2 days.
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-the difference of 2 days between the synodic and sidereal cycles is due to the Earth- moon system also moving in an orbit around the sun.
5 0
3 years ago
Read 2 more answers
All rules to find work in changing force in physics ​
saw5 [17]

Answer:

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WORK - product of force and displacement, ie [velocity may be constant or variable but change in position with certain force is of more concern] JOULES

I hope you understood from this..

7 0
2 years ago
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