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Alchen [17]
3 years ago
6

What is the equation for an inelastic collision

Physics
1 answer:
abruzzese [7]3 years ago
7 0
M1 v1 = (m1 + m2)v2.

All of the exponents should be lowered to the bottom right of the letters.
You might be interested in
Help with 5-10 please
Zina [86]

Answer:

explained in attachment

Explanation:

explained in attachment

Download pdf
5 0
3 years ago
Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m
denis23 [38]

Answer:

4.245s

Explanation:

Given that,

Hypothetical value of speed of light in a vacuum is 18 m/s

Speed of the car, 14 m/s

Time given is 6.76 s, and we're asked to find the observed time, T

The relationship between the two times can be given as

T = t / √[1 - (v²/c²)]

The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject

t = T / √[1 - (v²/c²)]

And now, we substitute the values and insert into the equation

t = 6.76 * √[1 - (14²/18²)]

t = 6.76 * √[1 - (196/324)]

t = 6.76 * √(1 - 0.605)

t = 6.76 * √0.395

t = 6.76 * 0.628

t = 4.245 s

Therefore, the time the driver measures for the trip is 4.245s

8 0
3 years ago
A sample of helium behaves as an ideal gas as energy is
alisha [4.7K]

Answer:

0.0321 g

Explanation:

Let helium specific heat c_h = 5.193 J/g K

Assuming no energy is lost in the process, by the law of energy conservation we can state that the 20J work done is from the heat transfer to heat it up from 273K to 393K, which is a difference of ΔT = 393 - 273 = 120 K. We have the following heat transfer equation:

E_h = m_hc_h \Delta T = 20 J

where m_h is the mass of helium, which we are looking for:

m_h = \frac{20}{c_h \Delta T} = \frac{20}{5.193 * 120} \approx 0.0321 g

4 0
3 years ago
A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 c
seraphim [82]

Answer:

Flow rate 2.34 m3/s

Diameter 0.754 m

Explanation:

Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.

The area at the well head is

A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2

So the volume flow rate along the pipe is

\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s

We can use the similar logic to find the cross-section area at the refinery

A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2

The radius of the pipe at the refinery is:

A_r = \pi r^2

r^2 =A_r/\pi = 0.446/\pi = 0.141

r = \sqrt{0.141} = 0.377m

So the diameter is twice the radius = 0.38*2 = 0.754m

6 0
3 years ago
What is the minimum v (in km/s) the rocket engines must provide to allow the craft to escape from the Earth?
coldgirl [10]

Answer:

11 kilometers (7 miles) per second, or over 40,000 kilometers per hour (25,000 miles per hour)

Explanation:

3 0
3 years ago
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