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3 years ago

If a ball is tossed straight up into the air, at what position is its potential energy the greatest?

2 answers:

4 0

A. when it reaches the top of its flight

potential energy increases as the height of the object does, and the highest point of a ball's flight is when it reaches the top.

abruzzese [7]3 years ago

3 0

Answer:

A.when it reaches the top of its flight

Explanation:

The potential energy of an object is given by:

$U=mgh$

where

m is the mass of the object

g is the acceleration of gravity

h is the heigth at which the object is located above the ground

Therefore, we see that the higher the object, the larger its potential energy: this means that the ball will have the greatest potential energy at the top of its trajectory.

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An object is propelled along a straight-line path by a force. If the new force were doubled, is acceleration would

Aneli [31]

Still go straight but would obviously go up in speed!!

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6 0

3 years ago

Read 2 more answers

the presence of which magnetic feature best explains why a magnet can act a distance on other magnets or on objects containing c

katen-ka-za [31]

Magnetic fields

Explanation:

The presence of magnetic fields best explains why a magnet can act a distance on other magnets or on objects containing certain metals.

Magnetic fields are lines of forces around a bar magnet.
These lines of forces attracts and repels other magnetic bodies and metallic bodies round it.
Magnetic lines of forces originates at the north pole and enters in the south pole.
Areas around a magnetic body are bounded by force fields.
A magnet has permanent magnetic fields round it.

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8 0

3 years ago

a car initially at rest move with the constant accerates along straght line read after it's spread increase and finally related

nasty-shy [4]

Answer:

32km per hour

Explanation:

Explanation:

In first case v = a t

==> a t = 40 km p h

Now distance covered S1 + S2 + S3

S1 = 1/2 a t^2 and S3 = 1/2 a t^2

But S2 = 3t * 40 = 120 t km

Hence total distance = at^2 + 120 t

Time taken (total) = t + 3t + t = 5 t

Hence average speed = at^2 + 120 t / 5 t

Cancelling t we have at + 120 / 5 = 40 + 120 / 5 = 160/5 = 32 km per hour

8 0

3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

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3 years ago

According to newton's second law of motion,the acceleration of an object equals the net force acting on the object divided by th

Shtirlitz [24]

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5 0

3 years ago