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Valentin [98]
3 years ago
8

A small mass is released from rest from the edge of a frictionless, hemispherical bowl. As the mass passes through the bottom of

the bowl its acceleration vector is:
Physics
1 answer:
Greeley [361]3 years ago
4 0

Answer:

a=2g

Explanation:

Let the radius of the hemisphere is R. The mass is released from rest so the initial speed is equal to zero and acceleration is equal to the g.

By the Newton's 3rd equation,

v^2=u^2+2as\\v^2=2gR\\v=\sqrt{2gR}

Now, at the bottom hemisphere the centripetal acceleration will act on the mass and the direction of this force is towards center,

a_c=\frac{v^2}{R}\\a_c=\frac{(\sqrt{2gR} )^2}{R}\\a_c=2g

Hence, the acceleration on the bottom of the hemisphere is equal to the 2g.

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the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity
NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

\int\limits^s_0 {a} \, ds  = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds  = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\

Remember that

v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t=  (5\sqrt{2} ) ln  \frac{| [s + 300 + \sqrt{(s^{2}  + 600s)} ] |}{300} .......2

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0
3 years ago
A 75kg hockey player is skating across the ice at a speed of 6.0m/s. What is the magnitude of the average force required to stop
liq [111]

Answer:

692.31 N

Explanation:

Applying,

F = ma............... Equation 1

Where F = Average force required to stop the player, m = mass of the player, a = acceleration of the player

But,

a = (v-u)/t............ Equation 2

Where v = final velocity, u = initial velocity, t = time.

Substitute equation 2 into equation 1

F = m(v-u)/t............ Equation 3

From the question,

Given: m = 75 kg, u = 6.0 m/s, v = 0 m/s (to stop), t = 0.65 s

Substitute these values into equation 3

F = 75(0-6)/0.65

F = -692.31 N

Hence the average force required to stop the player is 692.31 N

6 0
2 years ago
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Anit [1.1K]

Answer:

A collision in which both total momentum and total kinetic energy are conserved

Explanation:

In classical physics, we have two types of collisions:

- Elastic collision: elastic collision is a collision in which both the total momentum of the objects involved and the total kinetic energy of the objects involved are conserved

- Inelastic collision: in an inelastic collision, the total momentum of the objects involved is conserved, while the total kinetic energy is not. In this type of collisions, part of the total kinetic energy is converted into heat or other forms of energy due to the presence of frictional forces. When the objects stick together after the collision, the collisions is called 'perfectly inelastic collision'

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3 years ago
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Answer:

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