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3 years ago

An object is dropped near the earth's surface. At the end of 4.0 s, how fast is it falling in m/s? (Ignore air resistance.)

2 answers:

7 0

For free falling bodies, the final velocity may be calculated through the equation,
Vf = gt
Where g is the acceleration due to gravity (9.8 m/s²) and t is the time elapsed. Substituting the known values,
Vf = (9.8 m/s²) x (4 s) = 39.2 m/s
Therefore, the object's velocity is approximately 39.2 m/s.

DochEvi [55]3 years ago

4 0

The answer would be 39.2 m/s

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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

photoshop1234 [79]

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

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

$h_c = 42.86 \ cm$

8 0

2 years ago

A skier (m=59.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h = 3

marissa [1.9K]

Answer:

35.20 m

Explanation:

By the law of conservation of energy we have,

$mg(H-h)=\frac{1}{2}mv^2$

$g(H-h)=\frac{1}{2}v^2$

$\Rightarrow H=\frac{v^2}{2g}+h$

where m= mass of the skier, h= 3.00 m

D= horizontal distance=13.9 m

H= maximum height attained

Also, the horizontal distance covered by the skier is

D=vt

$=v\frac{2g}{h}$

$\Rightarrow v^2=\frac{gD^2}{2h}$

thus, height H in terms of D is given by

$H=\frac{D^2}{2h}+h$

$H=\frac{13.9^2}{2\times3}+3$

H=35.20 m

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Dmitriy789 [7]

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Answer:

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Explanation:

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Write down the effect of humidity and temperature on the speed of sound.

finlep [7]

Answer:

The speed of sound is affected by temperature and humidity. Because it is less dense, sound passes through hot air faster than it passes through cold air. ... The attenuation of sound in air is affected by the relative humidity. Dry air absorbs far more acoustical energy than does moist air.

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3 years ago