Sometimes, in an intense battle, gunfire is so intense that bullets from opposite sides collide in midair. Suppose that one (with mass M = 5.12 g moving to the right at a speed V = [08]____________________ m/s directed 21.3° above the horizontal) collides and fuses with another with mass m = 3.05 g moving to the left at a speed v = 282 m/s directed 15.4° above the horizontal. a. What is the magnitude of their common velocity (m/s) immediately after the collision? b. What is the direction of their common velocity immediately after the collision? (Measure this angle in degrees from the horizontal.) c. What fraction of the original kinetic energy was lost in the collision?

Answers

Answer 1

Answer:

The magnitude of the speed is 83.0325 m\s, the direction is 62.7 degrees, and the fraction of kinetic energy lost is 0.895.

What is collision?

The collision is the phenomenon when two objects come in direct contact with each other. Then both the bodies exert forces on each other.

The mass, angle, and velocity of the first object are 5.12 g, 21.3°, and 239 m/s.

And the mass, angle, and velocity of the second object be 3.05 g, 15.4°, and 282 m/s.

The momentum (P₁) before a collision will be

$\rm P_1 = (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) \hat{x} + (m_1 u_1 sin \theta _1+ m_2 u_2 sin \theta _2) \hat{y}$

The momentum (P₂) after a collision will be

$\rm P_2 = (m_1 + m_2) u \ cos\ \theta \ \hat{x} \ + (m_1 + m_2) u \ sin \ \theta \ \hat{y}$

Applying momentum conservation, we have

$\rm (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) = (m_1 + m_2) u \ cos\ \theta \ \n\n$ ...1

$\rm (m_1 u_1 sin \theta _1+m_2 u_2 sin \theta _2) \ =(m_1 + m_2) u \ sin \ \theta$ ...2

From equations 1 and 2, we have

$\rm \theta = tan \ ^(-1) ( (m_1 u_1 cos \theta _1 +m_2 u_2cos \theta _2))/( (m_1 u_1 sin \theta _1 - m_2 u_2 sin \theta _2))\n\n\n\theta = tan \ ^(-1) (5.12*239*cos21.3+3.05*282*cos15.4)/(5.12*239*sin21.3-3.05*282*sin15.4)\n\n\n\theta = 62.7^o$

From equation 1, we have

$\rm u = ((m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) )/( (m_1 + m_2) \ cos\ \theta ) \n\n\nu = (5.12*239*cos21.3 - 3.05*282*cos15.4)/((5.12+3.05)cos62.2)\n\n\nu = 83.0325 m/s$

Then the change in kinetic energy, we have

$\rm \Delta KE = (1)/(2)m_1u_1^2+(1)/(2)m_2u_2^2-(1)/(2)(m_1+m_2)u^2\n\n\n\Delta KE = (1)/(2) * 5.12 * 239^2 + (1)/(2)*3.05*282^2 - (1)/(2)(5.12+3.05)*83.032^2\n\n\n\Delta KE = 239.34 \ J$

The fraction of kinetic energy lost will be

$\rm Energy \ lost = (239.34)/(267.5) = 0.895$

More about the collision link is given below.

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Answer 2

Answer:

Answer:

Detailed solution is given below

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Answers

Explanation:

For Part (a)

Since the apparent wavelength decreases hence galaxy moving towards the stationary observer.

Δλ/λ=v/c

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Solve for x
–30 = 5(x + 1)

Answers

Answer:

-30=5(x+1) is -7

Explanation:

distribute flip subtract 5 from both sides divide both sides by 5

-30=(5)(x)+(5)(1) (Distribute)
-30=5x+5
Flip equation
5x+5=-30
Subtract 5 from both sides
5x+5-5=-30-5
5x=-35
Divide both sides by 5
X=-7

What is the weight on Earth of an object with mass 45 kg. Hint gravity = 10 N/kg *1 point
45 N
450 N
450 kg
10N

Answers

Answer:

450N

Explanation:

weight= m*g

weight=45*10

weight=450N

By using a 2-meter stick (like the one in lab) marked in millimeters and a stopwatch that measures to 1/100h of a second, you decide to measure the speed of a motorized toy car that travels at a constant velocity. You measure out a 162.0cm interval with the 2-meter stick and time how long it takes the car to travel that distance using the stopwatch. Repeating the ex 2.95 s Calculate the average speed of the toy car What are the absolute and relative uncertainties of the distance and time measurements? Which measurement is more uncertain? Use the weakest link rule to determine the relative and absolute uncertainty in your speed estimation. Explain why it is necessary to calculate relative uncertainties. Why is absolute uncertainty not enougn ent 5 times you get the following time data: 3.11 s 3.15 s 2.84 s 2.97 s

Answers

Answer:

Explanation:

The average speed of a body is defined as the ratio between total distance and total time

v = dx / dt

v = 162.0 / 2.95

v = 54.9 m / s

The absolute errors (uncertainties) of the distance and time measurements as measured with instruments are the errors of the instruments

Δx = 0.1 cm

Δt = 0.01 s

Relative errors (uncertainties) are the absolute errors between the measured value

Er = Δx /x

Er = 0.1 / 162.0

Er = 6.2 10⁻⁴ length

Er = 0.01 / 2.95

Er = 3.4 10⁻³ time

The most uncertain measure is the time to have a greater relative error

Let's calculate the relative speed error

Δv / v = dv / dx dx + dv / dt dt

dv / dx = 1 / t

dv / dt = x (-1 / t²)

Er = Δv / v = 1 / t Δx + x / t² Δt

Er = 0.1 / 2.95 + 162.0/2.95² 0.01

Er = 0.034 + 0.19

Er = 0.22

We can observe that the relative error of time is much higher than the relative error of distance, so to reduce the speed error, time must be measured with much more precision

Absolut mistake

Er = Δv / v

Δv = Er v

Δv = 0.22 54.9

Δv = 12 cm / s

v± Δv = (5 ±1 ) 10 cm/s

When calculating the relative uncertainty, it is known which magnitude should be more precisely medical to reduce the total error of a derived magnitude

The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is constant. At a specific temperature the pressure is 102.1 kPa at sea level and 87.8 kPa at h = 1,000 m. (Round your answers to one decimal place.) (a) What is the pressure at an altitude of 4500 m? kPa (b) What is the pressure at the top of a mountain that is 6165 m high?

Answers

Final answer:

The rate of change of atmospheric pressure with respect to altitude is proportional to the current pressure. Using this information, we can calculate the pressure at different altitudes.

Explanation:

To solve this problem, we can use the fact that the rate of change of atmospheric pressure with respect to altitude is proportional to the current pressure. We can set up a proportion using the given information to find the constant of proportionality. Then, we can use this constant to find the pressure at different altitudes.

(a) Let's use the given information to find the constant of proportionality. We have P = kP, where k is the constant of proportionality. Using the values at sea level and 1000m, we can set up the proportion 102.1/87.8 = k. Solving for k, we find k ≈ 1.16.

Now, we can use this constant to find the pressure at an altitude of 4500m. We set up the proportion 102.1/x = 1.16, where x is the pressure at 4500m. Solving for x, we find x ≈ 122.0 kPa.

(b) We can use the same constant of proportionality to find the pressure at the top of a mountain that is 6165m high. We set up the proportion 102.1/x = 1.16, where x is the pressure at the top of the mountain. Solving for x, we find x ≈ 89.2 kPa.

Learn more about Rate of change of atmospheric pressure with altitude here:

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A uniform electric field, with a magnitude of 370 N/C, is directed parallel to the positive x-axis. If the electric potential at x = 2.00 m is 1 000 V, what is the change in potential energy of a particle with a charge of + 2.80 x 10-3 C as it moves from x = 1.9 m to x = 2.1 m?