The table shows the mass of and acceleration due to gravity for several

From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.10 m/s and angle

GarryVolchara [31]

Answer:

Part a)

$y = 88.5 m$

Part b)

$v_x = 7.7 m/s$

$v_y = 2.5 m/s$

Part c)

equation of position in x direction is given as

$x = v_x t$

equation of position in y direction is given as

$y = v_y t + \frac{1}{2}gt^2$

Part d)

$x = 30.8 m$

Part e)

H = 88.5 m

Part f)

t = 1.2 s

Explanation:

As we know that ball is projected with speed

v = 8.10 m/s at an angle 18 degree below the horizontal

so we will have

$v_x = 8.10 cos18$

$v_x = 7.7 m/s$

$v_y = 8.10 sin18$

$v_y = 2.5 m/s$

Part a)

Since it took t =4 s to reach the ground

so its initial y coordinate is given as

$y = v_y t + \frac{1}{2}a_y t^2$

$y = 2.5(4) + \frac{1}{2}(9.81)(4^2)$

$y = 88.5 m$

Part b)

components of the velocity is given as

$v_x = 8.10 cos18$

$v_x = 7.7 m/s$

$v_y = 8.10 sin18$

$v_y = 2.5 m/s$

Part c)

equation of position in x direction is given as

$x = v_x t$

equation of position in y direction is given as

$y = v_y t + \frac{1}{2}gt^2$

Part d)

distance where it will strike the floor is given as

$x = v_x t$

$x = 7.7 \times 4$

$x = 30.8 m$

Part e)

Height from which it is thrown is same as initial y coordinate of the ball

so it is given as

H = 88.5 m

Part f)

time taken by ball to reach 10 m below is given as

$y = v_y t + \frac{1}{2}gt^2$

$10 = 2.5t + \frac{1}{2}(9.81) t^2$

t = 1.2 s

7 0

3 years ago

What are the movements of boiling water? I'm looking for if they speed up, like that sort of thing. Not how, not why, just what

fgiga [73]

When a substance is heated, the molecules begin to vibrate faster and the space between molecules increases. So it would be speeding up.

6 0

3 years ago

An ultracentrifuge accelerates from rest to 99300 rpm in 1.95 min. Randomized Variables f = 99300 rpm t = 1.95 min l = 9.4 cm

Levart [38]

Answer: i'm pretty sure its part B

7 0

4 years ago

Which of the following is represented by the letter A in the diagram below?

Kruka [31]

We have that the letter A in the diagram below given as

Amplitude

Option A

<h3>Amplitude</h3>

Question Parameters:

Amplitude

Crest

Trough

Wavelength

Generally, the amplitude of a wave is the maximum displacement of the wave in the medium from its initial position.

Amplitude is denoted with the letter A

Therefore,Amplitude

Option A

For more information on displacement visit

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6 0

2 years ago

A newly discovered planet has the same mass as the Earth, but a person standing on the surface of the planet experiences a force

Dafna11 [192]

This question deals with the acceleration due to gravity on the surface of the planet. The value of the acceleration due to gravity on the surface of two planets, that is Earth and on the surface of another planet are compared to find out the relation between the radii of both planets, provided their masses are the same.

The correct option for the radius of the new planet is "Option 3: Square Root of 3 R"

The value of the acceleration due to gravity on the surface of a planet is given by the following formula:

$g = \frac{GM}{R^2}$ ----------- eqn(1)

where,

g = acceleration due to gravity on Earth

G = Universal Gravitational Constant

M = mass of the Earth

R = Radius of the Earth

Now, the same formula for the acceleration due to gravity on the surface of the other planet can be written as follows:

$g' = \frac{GM'}{R'}\\\\$ ------------- eqn(2)

where,

g' = acceleration due to gravity on the new planet

G = Universal Gravitational Constant

M' = mass of the new planet

R' = Radius of the new planet

According to the given condition, the new planet has the same mass as the Earth's mass but the acceleration due to gravity on the surface of the new planet is one-third of the acceleration due to gravity on the surface of Earth.

$M'=M\\\\g'=\frac{1}{3}g$

using eqn (1) and eqn (2):

$\frac{GM}{R'^2}=\frac{1}{3}\frac{GM}{R^2}\\\\R'^2=3R^2\\\\R' = \sqrt{3}R$

Hence, the radius of the new planet is found to be equal to the square root of 3 times the radius of the Earth (Square Root of 3 R).

Learn more about acceleration due to gravity on the surface of a planet here:

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7 0

3 years ago