PLEASE HELP!! 13 POINTS! SCIENCE!

A very elastic rubber ball is dropped from a certain height and hits the floor with a (2pts) downward speed v. Since it is so el

artcher [175]

Answer:

3.True. The magnitude of momentum is the same

Explanation:

Let's propose the solution of the problem

The initial moment is

p₀ = m v

The final moment

$p_{f}$ = m (-v)

p₀ = - $p_{f}$

Now we can review the claims

1. False. We see that the moment module is the same, but its direction changes

2. False. The impulse is a vector

3.True. The magnitude of momentum is the same

7 0

3 years ago

A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7

maxonik [38]

First, let's find the speed $v_i$ of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
$p_i = m_1 v_1$
After the collision, the two blocks stick together and so now they have mass $m_1 +m_2$ and they are moving with speed $v_i$ :
$p_f = (m_1 + m_2)v_i$
For conservation of momentum
$p_i=p_f$
So we can write
$m_1 v_1 = (m_1 +m_2)v_i$
From which we find
$v_i = \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s$

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force $\mu (m_1+m_2)g$ . The work done by the frictional force to stop the two blocks is
$\mu (m_1+m_2)g d$
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
$\frac{1}{2} (m_1+m_2)v_i^2$
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
$\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g d$
From which we can find the value of the coefficient of kinetic friction:
$\mu = \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49$

3 0

3 years ago

Which statement accurately explains why the kinetic energy of the first rider is greater?

Dvinal [7]

Hey

The formula of kinetic energy is 1/2mv^2

So it depends on mass and velocity

As mass increases , kinetic energy increase .

So option b , the first rider had more mass is correct z

5 0

2 years ago

Read 2 more answers

An ideal monatomic gas at temperature T is held in a container. If the gas is compressed isothermally, that is at constant tempe

OlgaM077 [116]

Answer:

a) 0 J

b) W = nRTln(Vf/Vi)

c) ΔQ = nRTln(Vf/Vi)

d) ΔQ = W

Explanation:

a) To find the change in the internal energy you use the 1st law of thermodynamics:

$\Delta U=\Delta Q-W$

Q: heat transfer

W: work done by the gas

The gas is compressed isothermally, then, there is no change in the internal energy and you have

ΔU = 0 J

b) The work is done by the gas, not over the gas.

The work is given by the following formula:

$\\W=nRTln(\frac{V_f}{V_i})$

n: moles

R: ideal gas constant

T: constant temperature

Vf: final volume

Vi: initial volume

Vf < Vi, then W < 0 and the work is done on the gas

c) The gas has been compressed. Thus, its temperature increases and heat has been transferred to the gas.

The amount of heat is equal to the work done W

d)

$\Delta U = \Delta Q-W\\\\0=\Delta Q-W\\\\\Delta Q=W=nRTln(\frac{V_f}{V_i})$

5 0

3 years ago

The sound level near a noisy air conditioner is 70 dB. If two such units operate side by side, the sound level near them would b

Schach [20]

Answer:

73 db

Explanation:

A single air conditioner is equivalent to 70 dB frequency. An extra air conditioning unit would therefore double the sound frequency. It does not, however, double the decibels to 140 dB. Instead, it adds only 3 dB to the 70 dB, making the total decibels of two air conditioning units equal to 73 db

Hence the correct option is b that is 73 db

7 0

3 years ago