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3 years ago

A bowling ball has a mass of 6 kg. What happens to its momentum when its speed increases from 2m/s to 4 m/s?

2 answers:

6 0

Here, Initial momentum = mu = 6*2 = 12 Kg m/s
Final momentum = mv = 6*4 = 24 Kg m/s

In short, Your Answer would be Option C

Hope this helps!

olganol [36]3 years ago

5 0

Answer: The initial momentum is 12 kg m/s, and the final momentum is 24 kg m/s

Explanation:

Momentum is defined as the force that keeps the object moving. It is also defined as the product of mass and velocity of an object.

Mathematically,

$p=m\times v$ .......(1)

where,

p = momentum of the bowling ball

m = mass of the bowling ball

v = velocity of the bowling ball

Calculating the initial momentum:

We are given:

$m=6kg\\u=\text{Initial speed}=2m/s$

Putting values in equation 1, we get:

$p=6kg\times 2m/s\\\\p=12kg.m/s$

Calculating the final momentum:

We are given:

$m=6kg\\u=\text{Final speed}=4m/s$

Putting values in equation 1, we get:

$p=6kg\times 4m/s\\\\p=24kg.m/s$

Hence, the initial momentum is 12 kg m/s, and the final momentum is 24 kg m/s

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3.27 moles of an ideal gas in a 50.0 L tank has a pressure of 171000 Pa. What is the temperature of the gas? (Unit=degrees C)

laiz [17]

The temperature of the gas is 41.3 °C.

Answer:

The temperature of the gas is 41.3 °C.

Explanation:

So on combining the Boyle's and Charles law, we get the ideal law of gas that is PV=nRT. Here P is the pressure, V is the volume, n is the number of moles, R is gas constant and T is the temperature. The SI unit of pressure is atm. So we need to convert 1 Pa to 1 atm, that is 1 Pa = 9.86923× $10^{-6}$ atm. Thus, 171000 Pa = 1.6876 atm.

We know that the gas constant R = 0.0821 atmLMol–¹K-¹. Then the volume of the gas is given as 50 L and moles are given as 3.27 moles.

Then substituting all the values in ideal gas equation ,we get

1.6876×50=3.27×0.0821×T

Temperature = $\frac{84.38}{0.268467} =314.3 K$

So the temperature is obtained to be 314.3 K. As 0°C = 273 K,

Then 314.3 K = 314.3-273 °C=41.3 °C.

Thus, the temperature is 41.3 °C.

3 0

3 years ago

PLEASE HELP ASAP...TIMED TEST. PLEASE ANSWER WITH AT LEAST A PARAGRAPH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

charle [14.2K]

Answer: I am pretty sure that you should pick radio waves.

Explanation: The scientist should use radio waves. I think this because you can use the radio waves to analyze the signals from outer space. This will work much better than anything there, to analyze it the best possible.

The best I could do.

8 0

3 years ago

A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

Where, R = resistor

C = capacitor

Put the value into the formula

$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

8 0

3 years ago

20. In a parallel RL circuit, 10 mA flows through the resistor and 4 mA flows through the inductor. What phase angle separates v

stellarik [79]

Answer:

Option D: 21.8 degrees

Explanation:

In a parallel RL circuit, the current in the resistor R and that in the inductor L are separated among themselves 90 degrees as illustrated in the attached image. In the image the current in the resistor is represented in orange, that of the inductor in blue, and the total current (vector addition of the previous two) is represented in red, forming a certain angle (theta) with respect to the current in the resistor. The output voltage is the same as the input voltage as measured over the resistor R.

Therefore, the phase angle that separated output voltage and total current can be obtained using the fact that tan(phase angle) = $\frac{I_l}{I_R} = \frac{x}{y} \frac{4}{10}$ , therefore the angle is the arctangent of 4/10: