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3 years ago

Can someone help me with these two? Pls

Physics

1 answer:

Gelneren [198K]3 years ago

8 0

For the first one: A falling object has higher gravitational potential energy and as it falls this decreases but kinetic energy increases. As the object hits the floor, energy is transferred to heat and sound.

42*9.8*12= 4939.2
Kinetic energy = 4939.2 OR 4939

To find the final speed use the kinetic energy formula;

Ek = 1/2 MV2

4939=1/2*42*v2
V2= 4939 divided by (1/2 * 42)
V2= 4939 / 21
V2= 235.1904761904762
Find square root
V= 15.3359211067
= 15.

Reach the ground at 15m s -¹

Note:
Formula To calculate kinetic energy:
Ek = 1/2 MV2
Ek= kinetic energy of an object
M= mass of object
V= velocity/speed of object

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A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th

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Answer:

The height is $h_c = 42.857$

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

From the question we are told that

The height is $h_s = 30 \ cm$

The angle of the slope is $\theta = 15^o$

According to the law of conservation of energy

The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

$mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2$

Where I is the moment of inertia which is mathematically represented as this for a sphere

$I = \frac{2}{5} mr^2$

The angular velocity $w$ is mathematically represented as

$w = \frac{v}{r}$

So the equation for conservation of energy becomes

$mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2$

$mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]$

$mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

$gh_s =[\frac{7}{10} ] v^2$

$v^2 = \frac{10gh_s}{7}$

Considering a circular hoop

The moment of inertial is different for circle and it is mathematically represented as

$I = mr^2$

Substituting this into the conservation equation above

$mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

$mgh_c = mv^2$

$gh_c = v^2$

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

Recall that $v^2 = \frac{10gh_s}{7}$

$h_c= \frac{\frac{10gh_s}{7} }{g}$

$= \frac{10h_s}{7}$

Substituting values

$h_c = \frac{10(30)}{7}$

$h_c = 42.86 \ cm$

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