PHYSICS COLLEGE

Twopucksofequalmasscollideonafrictionlesssurface,asillustratedinthefigure.Immediatelyafterthe collision, the speed of the black puck is 1.5 m/s. What is the speed of the white puck immediately after the collision?

Answers

Answer 1
Answer:

Answer:

The speed of the white puck immediately after the collision is 2.6 m/s.

Explanation:

Given that,

Two pucks are equal masses.

Speed of black puck = 1.5 m/s

According to given figure,

We need to calculate the speed of the white puck immediately after the collision

Using law of conservation of momentum

mv=m_(1)v_(1)\cos\theta+m_(2)v_(2)\cos\theta

Put the value into the formula according to figure

m*3=m* v_(1)*\cos30+m*1.5*\cos60

3m=0.866m v_(1)+0.75m

v_(1)=(3-0.75)/(0.866)

v_(1)=2.6\ m/s

Hence, The speed of the white puck immediately after the collision is 2.6 m/s.

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Ted Clubber Lang. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at the elbow. During this punch, the horizontal flexor muscles of the shoulder contract and shorten at an average speed of 75 cm/s. They move through an arc length of 5 cm during the hook, while the first moves through an arc length of 100 cm. What is the average speed of the first during the hook?

Q.1- Find the distance travelled by a particle moving in a straight line with uniform acceleration, in the 10th unit of time.

Answers

Answer:

If the acceleration is constant, the movements equations are:

a(t) = A.

for the velocity we can integrate over time:

v(t) = A*t + v0

where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:

\int\limits^(10)_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)

Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.

A cart with mass 340 g moving on a frictionless linear air track at an initial speed of 1.2 m/s undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at 0.66 m/s. (a) What is the mass of the second cart? (b) What is its speed after impact?

Answers

Answer:

A) m2 = 98.71g

B) v_f2 = 1.86 m/s

Explanation:

We are given;

Mass of cart; m1 = 340g

Initial speed; v_i1 = 1.2 m/s

Final speed; v_f1 = 0.66 m/s

A)Since the collision is elastic, we can simply apply the conservation of momentum to get;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2) - - - - - (eq1)

From conservation of kinetic energy, we have;

(1/2)m1•(v_i1)² = (1/2)m1•(v_f1)² + (1/2)m2•(v_f2)² - - - - eq(2)

Let's make v_f2 the subject in eq 2;

Thus,

v_f2 = √([m1•(v_i1)² - m1•(v_f1)²]/m2)

v_f2 = √([m1((v_i1)² - (v_f1)²)]/m2)

Let's put this for v_f2 in eq1 to obtain;

m2 = {m1((v_i1) - (v_f1))}/√([m1((v_i1)² - (v_f1)²)]/m2)

Let's square both sides to give;

(m2)² = {m1•m2((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

This gives;

m2 = {m1((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

Plugging in the relevant values to get;

m2 = {340((1.2) - (0.66))²}/([(1.2)² - (0.66)²]

m2 = 98.71g

B) from equation 1, we have;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2)

Making v_f2 the subject, we have;

v_f2 = m1[(v_i1) - (v_f1)]/m2

Plugging in the relevant values to get;

v_f2 = 340[(1.2) - (0.66)]/98.71

v_f2 = 1.86 m/s

Final answer:

To determine the mass of the second cart and its speed after impact, we can use the principle of conservation of momentum. The initial momentum of the first cart is equal to its final momentum plus the momentum of the second cart. After calculating the mass of the second cart, we can use the conservation of momentum again to find its speed by equating the final velocity of the combined carts to the initial velocity of the first cart.

Explanation:

To determine the mass of the second cart, we can use the principle of conservation of momentum. The initial momentum of the first cart, with a mass of 340 g and an initial velocity of 1.2 m/s, is equal to its final momentum plus the momentum of the second cart. Using this equation, we can solve for the mass of the second cart.


After calculating the mass of the second cart, we can use the conservation of momentum again to find its speed after the impact. Since the two carts stick together after the collision, the final velocity of the combined carts is equal to the initial velocity of the first cart. Using this equation, we can solve for the speed of the second cart.

Learn more about conservation of momentum here:

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An electron moving at 3.94 103 m/s in a 1.23 T magnetic field experiences a magnetic force of 1.40 10-16 N. What angle does the velocity of the electron make with the magnetic field? There are two answers between 0° and 180°. (Enter your answers from smallest to largest.)

Answers

Answer:

10.4⁰ and 169.6⁰

Explanation:

The force experienced by the moving electron in the magnetic field is expressed as F = qvBsinθ where;

q is the charge on the electron

v is the velocity of the electron

B is the magnetic field strength

θ is the angle that the velocity of the electron make with the magnetic field.

Given parameters

F =  1.40*10⁻¹⁶ N

q = 1.6*10⁻¹⁹C

v = 3.94*10³m/s

B = 1.23T

Required

Angle that the velocity of the electron make with the magnetic field

Substituting the given parameters into the formula:

1.40*10⁻¹⁶ =  1.6*10⁻¹⁹ * 3.94*10³ * 1.23 * sinθ

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁹⁺³sinθ

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁶sinθ

sinθ = 1.40*10⁻¹⁶/7.75392 * 10⁻¹⁶

sinθ = 1.40/7.75392

sinθ = 0.1806

θ = sin⁻¹0.1806

θ₁ = 10.4⁰

Since sinθ is positive in the 1st and 2nd quadrant, θ₂ = 180-θ₁

θ₂ = 180-10.4

θ₂ = 169.6⁰

Hence, the angle that the velocity of the electron make with the magnetic field are 10.4⁰ and 169.6⁰

Do you think a baseball curves better at the top of a high mountain or down on a flat plain

Answers

The baseball curves better at a flatplain due to contacting with air.

What is a baseball curve?

A curveball is a breaking pitch with more movement than most other pitches. It is thrown slower and with more overall break than a slider and is used to throw hitters off balance.

On a flat plain, a baseball will curve down better. This is due to the curving being caused by the ball contacting air and being pushed in a specificdirection.

The spin, speed, and location of the ball's stitches in relation to the air will all influence how it changes direction when pushed against.

Consider throwing a baseball in a vacuum where there is no air. Because there is no air to push on the ball, it will not curve at all.

Thus, a flat plain area will be better for baseball curve.

For more details regarding curveball, visit:

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A baseball will curve better down on a flat plain.In view of the fact that the curving is caused by the ball contacting the air and pushing the ball in a particular direction.

If I am given a total capacitence of two capacitors, their capacitence togather is 22 F. What capacitence would the individual capacitors have if they are connected in parallel or connected in series.

Answers

Answer:

In parallel combination, the capacity of each capacitor is 11 F.

In series combination, the capacity of each capacitor is 44 F.

Explanation:

Let there are two capacitors each of capacitance C.

When they are connected in parallel:

In parallel combination, the effective capacitance is Cp.

Cp = C1 + C2 = C + C

22 = 2 C

C = 11 F

When they are connected in series:

In parallel combination, the effective capacitance is Cs.

1 / Cs = 1 / C1 + 1 / C2 = 1 / C + 1 / C = 2 / C

1 / 22 = 2 / C

C = 44 F

Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the compressor if the inlet volumetric flow rate is 0.15 m3/s. Use constant specific heats evaluated at 300 K.

Answers

Answer:

\dot W_(in) = 49.386\,kW

Explanation:

An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

\dot W_(in) + \dot m \cdot c_(p)\cdot (T_(1)-T_(2)) = 0

The power consumed by the compressor can be calculated by the following expression:

\dot W_(in) = \dot m \cdot c_(v)\cdot (T_(2)-T_(1))

Let consider that air behaves ideally. The density of air at inlet is:

P\cdot V = n\cdot R_(u)\cdot T

P\cdot V = (m)/(M)\cdot R_(u)\cdot T

\rho = (P\cdot M)/(R_(u)\cdot T)

\rho = ((104\,kPa)\cdot (28.02\,(kg)/(kmol)))/((8.315\,(kPa\cdot m^(3))/(kmol\cdot K) )\cdot (292\,K))

\rho = 1.2\,(kg)/(m^(3))

The mass flow through compressor is:

\dot m = \rho \cdot \dot V

\dot m = (1.2\,(kg)/(m^(3)))\cdot (0.15\,(m^(3))/(s) )

\dot m = 0.18\,(kg)/(s)

The work input is:

\dot W_(in) = (0.18\,(kg)/(s) )\cdot (1.005\,(kJ)/(kg\cdot K))\cdot (565\,K-292\,K)

\dot W_(in) = 49.386\,kW
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