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3 years ago

2) If the initial velocity of an object is equal to a final velocity, what is the acceleration of the object?

Physics

1 answer:

Arlecino [84]3 years ago

8 0

Answer:

The acceleration is 0.

Explanation:

The formula for acceleration is:

a = (v-u)/t

If v and u are equal, thus it would be 0 when subtracted, and anything divided by 0 is 0.

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What increases when the frequency of a sound increases

andrew-mc [135]

The sound is basically just frequency. Because the sound becomes higher in pitch.

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3 years ago

Why the Earth is dependent on the Sun?

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The Earth revolves (orbits) around the Sun in one year. The Earth's rotation axis is tilted relative to the plane of its orbit around the Sun. This tilt of the Earth is responsible for the seasons as the Earth orbits the Sun. The Sun provides energy that sustains all life on Earth.

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3 years ago

A basketball player throws a chall -1 kg up with an initial speed of his hand at shoulder height = 2.15 m Le gravitational poten

Talja [164]

Complete Question:

A basketball player tosses a basketball m=1kg straight up with an initial speed of v=7.5 m/s. He releases the ball at shoulder height h= 2.15m. Let gravitational potential energy be zero at ground level

a) Give the total mechanical energy of the ball E in terms of maximum height hn it reaches, the mass m, and the gravitational acceleration g.

b) What is the height, hn in meters?

Answer:

a) Energy = mghₙ

b) Height, hₙ = 5.02 m

Explanation:

a) Total energy in terms of maximum height

Let maximum height be hₙ

At maximum height, velocity, V=0

Total mechanical energy , E = mgh + 1/2 mV^2

Since V=0 at maximum height, the total energy in terms of maximum height becomes

Energy = mghₙ

b) Height, hₙ in meters

mghₙ = mgh + 1/2 mV^2

mghₙ = m(gh + 1/2 V^2)

Divide both sides by mg

hₙ = h + 0.5 (V^2)/g

h = 2.15m

g = 9.8 m/s^2

V = 7.5 m/s

hₙ = 2.15 + 0.5(7.5^2)/9.8

hₙ = 2.15 + 2.87

hₙ = 5.02 m

6 0

3 years ago

A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de

EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/π $r^{2}$

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / π $r^{2}$ ex

E = $\frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6} } = 17.368 * 10^{9} Pa = 17.4 GPa$

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = $\frac{1}{E}$ (бy - v (бx + бz)) = $-\frac{v}{E}$ бx

= $\frac{vFx}{Epir^{2} }$ = $\frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9} } = -0.63 *10^{-6}$

Finally

ey = Δr / r

Δr = ey * r = 10 * $-0.63* 10^{-6} mm = -6.3 * 10^{-6} mm$

Δd = 2Δr = $-12.6 * 10^{-6} mm$

Explanation:

5 0

3 years ago

The maximum tension that a 0.80 m string can tolerate is 15 N. A 0.35-kg ball attached to this string is being whirled in a vert

zimovet [89]

Answer:

v=5.86 m/s

Explanation:

Given that,

Length of the string, l = 0.8 m

Maximum tension tolerated by the string, F = 15 N

Mass of the ball, m = 0.35 kg

We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

$F=\dfrac{mv^2}{r}$

v is the maximum speed

$v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{15\times 0.8}{0.35}} \\\\v=5.86\ m/s$

Hence, the maximum speed of the ball is 5.86 m/s.

3 0

3 years ago