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3 years ago

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Imagine that two balls, a basketball and a much larger exercise ball, are dropped from a parking garage. If both the mass and ra

dius of the exercise ball are 3 times larger than the mass and radius of the basketball, which of the two balls do you think will reach the ground first?

1 answer:

pashok25 [27]3 years ago

3 0

Here if we assume that there is no air friction on both balls then we can say

$F = mg$

now the acceleration is given as

$F = ma = mg$

$a = g$

so here both the balls will have same acceleration irrespective of size and mass

so we can say that to find out the time of fall of ball we can use

$y = \frac{1}{2}gt^2$

$t = \sqrt{\frac{2y}{g}}$

now from above equation we can say that time taken to hit the ground will be same for both balls and it is irrespective of its mass and size

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A ski lift is used to transport people from the base of a hill to the top. If the lift leaves the

Liono4ka [1.6K]

The energy of the ski lift at the base is kinetic energy:
$K= \frac{1}{2}mv^2$
where m is the mass of the ski lift+the people carried, and $v=15.5 m/s$ is velocity at the base.
As long as the ski lift goes upward, its velocity decreases and its kinetic energy converts into potential energy. Eventually, when it reaches the top, its final velocity is v=0, so no kinetic energy is left and it has all converted into gravitational potential energy, which is
$U=mgh$
where $g=9.81 m/s^2$ and h is the height at the top of the hill.

So, since the total energy must conserve, we have
$U=K$
and so
$mgh = \frac{1}{2}mv^2$
from which we find the height:
$h= \frac{v^2}{2g}= \frac{(15.5m/s)^2}{2\cdot 9.81 m/s^2}=12 m$

8 0

3 years ago

How much energy was released if rocket hydrogen fuel was burnt ?

suter [353]

When ignited, the gas mixture converts to water vapor and releases energy, which sustains the reaction: 241.8 kJ of energy (LHV) for every mole of H2 burned.” A mole of hydrogen weighs 2 grams. So, this is a LHV (lower heating value) of 120.9 kJ/gram of hydrogen when heat of vaporization is subtracted.

6 0

1 year ago

Read 2 more answers

What is the velocity of a proton that has been accelerated by a potential difference of 15 kV? (i)\:\:\:\:\:9.5\:\times\:10^5\:m

andre [41]

Answer:

Velocity of a proton, $v=1.7\times 10^6\ m/s$

Explanation:

It is given that,

Potential difference, $V=15\ kV=15\times 10^3\ V$

Let v is the velocity of a proton that has been accelerated by a potential difference of 15 kV.

Using the conservation of energy as :

$\dfrac{1}{2}mv^2=qV$

q is the charge of proton

m is the mass of proton

$v=\sqrt{\dfrac{2qV}{m}}$

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 15\times 10^3\ V}{1.67\times 10^{-27}\ kg}$

$v=1695361.75\ m/s$

$v=1.69\times 10^6\ m/s$

or

$v=1.7\times 10^6\ m/s$

So, the velocity of a proton is $1.7\times 10^6\ m/s$ . Hence, this is the required solution.

8 0

3 years ago

it is about 384,750 kilometers from earth to the moon. it took the apollo astronauts about 2 days and 19.5 hours to fly to the m

Jet001 [13]

We know, speed = Distance / Time
d = 384,750 Km
t = 2 days, 19.5 hours = 48+19.5 = 67.5 hour

Substitute their values,
s = 384,750 / 67.5
s = 5700 Km/h

In short, Your Answer would be 5700 Km/h

Hope this helps!

7 0

3 years ago

Read 2 more answers

1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/

mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

8 0

2 years ago