This when a satellite orbits in an oval-shaped path around a central object.

In a lab, four balls have the same velocities but different masses.

olya-2409 [2.1K]

Answer:

New Momentum of Ball B $=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$

<u>Explanation:</u>

Given:

Mass of Ball A=1kg

Mass of Ball B= 2kg

Mass of Ball C=5kg

Mass of Ball D=7kg

Velocities of A=B=C=D=2.2 $\frac{m}{s}$

Momentum of Ball A=2.2 $\frac{k g m}{s}$

Momentum of Ball B=4.4 $\frac{k g m}{s}$

Momentum of Ball C=11 $\frac{k g m}{s}$

Momentum of Ball D=15 $\frac{k g m}{s}$

To Find:

Change in Momentum When of Ball B gets tripled

Solution:

Though all balls have same velocity, thus we get

Velocities of A=B=C=D=2.2 $\frac{m}{s}$

Initial Momentum of Ball B=4.4 $\frac{k g m}{s}$

If the Mass of Ball B gets tripled;

We get New Mass of Ball B=3×Actual Mass of the ball

=3×2=6kg

Thus we get Mass of Ball B=6kg

According to the formula,

Change in momentum of Ball B $\Delta p=m \times \Delta v$

Where $\Delta p$ =change in momentum

m=mass of the ball B

$\Delta v$ =change in velocity ball B

And $\Delta v=v,$ since all balls, have same velocity

Thus the above equation, changes to

$\Delta p=m \times v$

Substitute all the values in the above equation we get

$\Delta p=6 \times 2.2$

$=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$

Result:

Thus the New Momentum of ball B $=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$

3 0

3 years ago

Read 2 more answers

How to answer this question?

laila [671]

Answer:

measure the vector diagram first

5 0

2 years ago

Help me pls???!!

rosijanka [135]

Answer:

A. I = V / R = 12 / 252 = .048 amps

V = I * R = .048 * 252 = 12 V

V is also the reading the voltage across the battery (12 Volts)

4 0

2 years ago

If s is the specific heat capacity of 4kg water, what is the specific heat capacity of 16kg

uranmaximum [27]

Answer: d

Explanation: divide it by 4

6 0

3 years ago

A rigid, insulated tank contains steam at 3 MPa, 400 Celsius. A valve on the tank is opened, allowing steam to escape. The overa

enot [183]

Answer:

x=0.8

Explanation:

Defining our values we have

$p_1=3Mpa\\T_1=400\° c\\q_{1-2}=0$

Where $p_1$ is the pressure of the super heated steam

$T_1$ is the Temperature of the super heated steam

and $q_{1-2}$ is the pressure in an adiabatic process.

State 1

$v_1 = 0.09936m^3/Kg\\s_1=6.9213kJ/kg.K$

State 2

$s_2=s_1=6.9212$

Note that here in state 2 the process is Reversible.

At the value where $T_2 = 141\°c$ we have

$v_2=(v_g)_{T_2}=0.4972m^3/kg$

Through this values we can calculate the Fraction of steam,

$x=\frac{m_1-m_2}{m_1}\\x=1-\frac{m_2}{m_1}\\x=1-\frac{v_1}{v_2}\\x=1-\frac{0.9936}{0.4972}\\x=0.8$

4 0

3 years ago