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Agata [3.3K]
3 years ago
9

a volleyball is hit upward with an initial velocity of 7.5 m/s. calculate the displacement of the volleyball when its final velo

city is 2.2m/s upwad
Physics
1 answer:
Luden [163]3 years ago
7 0

Answer:

The displacement of the volleyball is 2.62 m

Explanation:

Given;

initial velocity of the volleyball, u = 7.5 m/s

final velocity of the volleyball, v = 2.2 m/s

displacement of the volleyball, d = ?

Apply the following kinematic equation;

v² = u² - 2gd

2gd = u² - v²

d = \frac{u^{2}-v^{2}  }{2g}\\\\d = \frac{7.5^{2}-2.2^{2}  }{2*9.8}\\\\d = 2.62 \ m

Therefore, the displacement of the volleyball is 2.62 m

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A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0
3 years ago
PROVE THAT G = GM/R² WHERE THE SYMBOLS HAS THEIR USUAL MEANINGS<br>​
butalik [34]

Answer:

g=GM/R^2

Universal Gravutation Constant:

f=GM×m/R^2

Force can be also expressed as

f=m×g

so,

mg=GMxm/R^2

The m gets cancelled so

g=GM/R^2

8 0
4 years ago
What is the energy difference between the second excited state and first excited state of an electron in the "box" of size L=1nm
jolli1 [7]
I thinks it’s A, tell me if you get it right
4 0
3 years ago
A. How long (in ns) does it take light to travel 1.0 m in a vacuum?
suter [353]

Answer:

a) 3.33 ns

b) Water distance = 0.75 m

Glass distance = 0.66 m

Diamond distance = 0.41 m

Explanation:

We take the speed of light, c = m/s.

Speed = distance/time

Time = distance/speed

a)

t=\dfrac{1}{3.0\times10^{8}}=3.33\times10^{-9}

t = 3.33 ns

b)

Refractive index, n = speed of light in vacuum / speed of light in medium

n=\dfrac{c}{s}

s=\dfrac{c}{n}

d=s\times t

d=\dfrac{c}{n}\times \dfrac{1}{c}

d=\dfrac{1}{n}

Thus, the distance traveled in the same time is numerically equal to the reciprocal of the refractive index.

For water n = 1.333

d = 1/1.333 = 0.75 m

For glass n = 1.517

d = 0.66 m

For diamond n = 2.417

d = 0.41 m

4 0
3 years ago
A person throws a ball straight up. He releases the ball at a height of 1.75 m above the ground and with a velocity of 12.0 m/s.
Lyrx [107]

Answer:

a) 1.22 s

b) 9.089 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-12}{-9.81}\\\Rightarrow t=1.22\ s

Time taken by the ball to reach the highest point is 1.22 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=12\times 1.22+\frac{1}{2}\times -9.81\times 1.22^2\\\Rightarrow s=7.339\ m

The maximum height the ball will reach above the ground is 1.75+7.339 = 9.089 m

8 0
3 years ago
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