PHYSICS COLLEGE

Two cars, one with mass mi = 1200 kg is traveling north, and the other a large car with mass m2= 3000 kg is traveling East. If they collide while each car is traveling with a speed v = 5.0 m/s, what is the car's final speed and direction (vector notation can be given as well). Assume the collision is perfectly inelastic.

Answers

Answer 1

Answer:

Answer:

Both cars travel at < 10 , 4 > m/s

Explanation:

Conservation of Linear Momentum

The total momentum of a system of particles of masses m1 and m2 traveling at velocities v1 and v2 (vectors) is given by

$\vec p_1=m_1\vec v_1+m_2\vec v_2$

When the particles collide, their velocities change to v1' and v2' while their masses remain unaltered. The total momentum in the final condition is

$\vec p_2=m_1\vec v'_1+m_2\vec v'_2$

We know the collision is perfectly inelastic, which means both cars stick together at a common final velocity v'. Thus

$\vec p_2=m_1\vec v'+m_2\vec v'=(m_1+m_2)\vec v'$

Both total momentums are equal:

$m_1\vec v_1+m_2\vec v_2=(m_1+m_2)\vec v'$

Solving for v'

$\displaystyle v'=(m_1\vec v_1+m_2\vec v_2)/(m_1+m_2)$

The data obtained from the question is

$m_1=1200\ kg$

$m_2=3000\ kg$

The first car travels north which means its velocity has only y-component

$\vec v_1=<0,5>$

The second car travels east, only x-component of the velocity is present

$\vec v_2=<5,0>$

Plugging in the values

$\displaystyle v'=(1200<0,5>+3000<5,0>)/(1200+3000)$

$\displaystyle v'=(<0,6000>+<15000,0>)/(1500)$

$\displaystyle v'=(<15000,6000>)/(1500)=<10,4>\ m/s$

The magnitude of the velocity is

$|v'|=√(10^2+4^2)=10.77\ m/s$

And the angle

$\displaystyle tan\alpha=(4)/(10)=0.4$

$\alpha=21.8^o$

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A car travels 13 km in a southeast direction and then 16 km 40 degrees north of east. What is the car's resultant direction?

Answers

Answer:

21.48 km 2.92° north of east

Explanation:

To find the resultant direction, we need to calculate a sum of vectors.

The first vector has module = 13 and angle = 315° (south = 270° and east = 360°, so southeast = (360+270)/2 = 315°)

The second vector has module 16 and angle = 40°

Now we need to decompose both vectors in their horizontal and vertical component:

horizontal component of first vector: 13 * cos(315) = 9.1924

vertical component of first vector: 13 * sin(315) = -9.1924

horizontal component of second vector: 16 * cos(40) = 12.2567

vertical component of second vector: 16 * sin(40) = 10.2846

Now we need to sum the horizontal components and the vertical components:

horizontal component of resultant vector: 9.1924 + 12.2567 = 21.4491

vertical component of resultant vector: -9.1924 + 10.2846 = 1.0922

Going back to the polar form, we have:

$module = √(horizontal^2 + vertical^2)$

$module = √(460.0639 + 1.1929)$

$module = 21.4769$

$angle = arc\ tangent(vertical/horizontal)$

$angle = arc\ tangent(1.0922/21.4491)$

$angle = 2.915\°$

So the resultant direction is 21.48 km 2.92° north of east.

A typical sugar cube has an edge length of 1 cm. If you had a cubical box that contained a mole of sugar cubes, what would its edge length be? (One mole = 6.02 ✕ 1023 units.)

Answers

Since volume of each cube is 1 cm^3
Then we can get the volume of 1 mole of cubes, which is $1 * 6.02 * 10^23 cm^3$
The edge $edge = v^1/3$
And the new adge that we are looking for: $new edge = (6.02*10^23)^1/3$ = $= 1.8191 * 46415888.336 = 84435142.472$
So, the final soution for the edge length of cube is 844km.

Do hope it helps!
Regards.

A skater has rotational inertia 4.2 kg-m2 with his fists held to his chest and 5.7 kg?m2 with his arms outstretched. The skater is spinning at 3.0 rev/s while holding a 2.5-kg weight in each outstretched hand; the weights are 76 cm from his rotation axis. If he pulls his hands in to his chest, so they�re essentially on his rotation axis, how fast will he be spinning? Express your answer using two significant figures. ?f=

Answers

Answer: 38.5rad/s

Explanation: The calculations can be viewed on the image attached below. Thanks

A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 21.0 N/m. The block rests on a frictionless surface. A 5.30×10?2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.97 m/s and sticking.Part AHow far does the putty-block system compress the spring?

Answers

The distance the putty-block system compress the spring is 0.15 meter.

Given the following data:

Mass = 0.454 kg

Spring constant = 21.0 N/m.

Mass of putty = $5.30* 10^(-2)\;kg$

Speed = 8.97 m/s

To determine how far (distance) the putty-block system compress the spring:

First of all, we would solver for the initialmomentum of the putty.

$P_p = mass * velocity\n\nP_p = 5.30* 10^(-2)* 8.97\n\nP_p = 47.54 * 10^(-2) \;kgm/s$

Next, we would apply the law of conservation of momentum to find the final velocity of the putty-block system:

$P_p = (M_b + M_p)V\n\n47.54* 10^(-2) = (0.454 + 5.30* 10^(-2))V\n\n47.54* 10^(-2) = 0.507V\n\nV = (0.4754)/(0.507)$

Velocity, V = 0.94 m/s

To find the compression distance, we would apply the law of conservation of energy:

$U_E = K_E\n\n(1)/(2) kx^2 = (1)/(2) mv^2\n\nkx^2 =M_(bp)v^2\n\nx^2 = (M_(bp)v^2)/(k) \n\nx^2 = ((0.454 + 5.30* 10^(-2)) * 0.94^2)/(21)\n\nx^2 = ((0.507 * 0.8836))/(21)\n\nx^2 = ((0.4480))/(21)\n\nx=√(0.0213)$

x = 0.15 meter

Answer:

Explanation:

Force constant of spring K = 21 N /m

we shall find the common velocity of putty-block system from law of conservation of momentum .

Initial momentum of putty

= 5.3 x 10⁻² x 8.97

= 47.54 x 10⁻² kg m/s

If common velocity after collision be V

47.54 x 10⁻² = ( 5.3x 10⁻² + .454) x V

V = .937 m/s

If x be compression on hitting the putty

1/2 k x² = 1/2 m V²

21 x² = ( 5.3x 10⁻² + .454) x .937²

x² = .0212

x = .1456 m

14.56 cm

In general terms, the efficiency of a system can be thought of as the output per unit input. Which of the expressions is a good mathematical representation of efficiency e of any heat engine?
Where,
Qh: the absolute value (magnitude) of the heat absorbed from the hot reservoir during one cycle or during some time specified in the problem
Qc: the absolute value (magnitude) of the heat delivered to the cold reservoir during one cycle or during some time specified in the problem
W: the amount of work done by the engine during one cycle or during some time specified in the problem

A) e=QhW
B) e=QcQh
C) e=QcW
D) e=WQh
E) e=WQc

Answers

Answer:

Efficiency e = W/Qh

Explanation:

As written above efficiency of a system is calculated as the output per unit input. For heat Engine, Efficiency is calculated by dividing the Work done by Engine by Heat absorbed from hot reservoir.

In theoretical terms The maximum efficiency of a heat engine (which no engine ever attains) is equal to the temperature difference between the hot and cold ends divided by the temperature at the hot end, each expressed in absolute temperature (Kelvin).

But in practical calculations, it is calculated as e = W/Qh , and we define the thermal efficiency, of any heat engine as the ratio of the work it does, W, to the heat input at the high temperature, Qh.

The equations for single-slit and multiple-slit interference both contain the variable θ. For the multiple-slit case, the angle is: a. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences constructive interference.
b. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences destructive interference.
c. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences constructive interference.
d. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences destructive interference.

Answers

Answer:

the answers the correct one is c

Explanation:

The diffraction pattern for a slit is

a sin θ = m λ

Where a is the width of the slit, λ the wavelength, m the order of destructive interference and θ the angle where the interference occurs.

The expression for multi-slit diffraction (diffraction grating) is

d sin θ = m λ

Where d is the distance between slits, λ the wavelength m the order of the diffraction maximums and θ the angle for these maximums.

When we compare the expressions of the answers the correct one is c

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