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eduard
3 years ago
7

Identify whether the following statements describe a change in acceleration. Explain your response answer them all plz

Physics
2 answers:
weqwewe [10]3 years ago
5 0

Answer:c

Explanation:it picking up speed

stepladder [879]3 years ago
3 0
I got C
Hoped it was help byee:)
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Distinguish between resultant and equilibrant forces<br><br><br>​
SIZIF [17.4K]

Answer:

resultant is a single force that can replace the of a number of forces , equililbrant is a force that is exactly opposite to resultant

Explanation:

4 0
3 years ago
A cue ball of mass m1 = 0.325 kg is shot at another billiard ball, with mass m2 = 0.59 kg, which is at rest. The cue ball has an
Roman55 [17]

Answer:

v_{2f} = \frac{2vm_1}{m_2 + m_1}

Explanation:

If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls

Before the collision

P_i = m_1v

After the collision

P_f = m_1v_{1f} + m_2v_{2f}

So using the law of momentum conservation

P_i = P_f

m_1v = m_1v_{1f} + m_2v_{2f}

We can solve for the speed of ball 1 post collision in terms of others:

v_{1f} = v - v_{2f}\frac{m_2}{m_1}

Their kinetic energy is also conserved before and after collision

m_1v^2/2 = m_1v_{1f}^2/2 + m_2v_{2f}^2/2

m_1v^2 = m_1v_{1f}^2 + m_2v_{2f}^2

From here we can plug in v_{1f} = v - v_{2f}\frac{m_2}{m_1}

m_1v^2 = m_1\left(v - v_{2f}\frac{m_2}{m_1}\right)^2 + m_2v_{2f}^2

m_1v^2 = m_1\left(v^2 - 2vv_{2f}\frac{m_2}{m_1} + v_{2f}^2\frac{m_2^2}{m_1^2}\right) + m_2v_{2f}^2

m_1v^2 = m_1v^2 - 2vv_{2f}m_2 + v_{2f}^2\frac{m_2^2}{m_1} + m_2v_{2f}^2

v_{2f}^2(m_2 + \frac{m_2^2}{m_1}) - 2vm_2v_{2f} = 0

v_{2f}(1 + \frac{m_2}{m_1}) = 2v

v_{2f} = \frac{2v}{1 + \frac{m_2}{m_1}} = \frac{2v}{\frac{m_1 + m_2}{m_1}} = \frac{2vm_1}{m_2 + m_1}

8 0
3 years ago
Read 2 more answers
A 23.5 kg object travels a distance of 85 m in 30 s. Find the momentum. (Hint: you must find speed first)
VLD [36.1K]
Yup, I think you add all of them
7 0
2 years ago
A 4kg object has a momentum of 12 kg*m/s, what is the objects velocity?
Bezzdna [24]
Momentum = mass x velocity
12 = 4 x v | ÷ both sides by 4
12 ÷ 4 =v
v= 3 m/s
8 0
2 years ago
A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul
GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

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

here we know that for spherical shell and pure rolling conditions

v = R \omega

I = \frac{2}{3}mR^2

\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh

\frac{5}{6}mv^2 = mgh

h = \frac{5v^2}{6g}

h = \frac{5(8^2)}{6(9.81)} = 5.44 m

Part b)

If ball is not rolling and just sliding over the hill then in that case

\frac{1}{2}mv^2 = mgh

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

h = \frac{8^2}{2(9.81)} = 3.16 m

3 0
3 years ago
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