A block of mass m slides with a speed vo on a frictionless surface and collides with another mass M which is initially at rest. The two blocks stick together and move with a speed of vo /3. In terms of m, mass M is most nearly_____.

Answers

Answer 1

Answer:

To solve this problem we will apply the concepts related to the conservation of momentum. Momentum can be defined as the product between mass and velocity. We will depart to facilitate the understanding of the demonstration, considering the initial and final momentum separately, but for conservation, they will be later matched. Thus we will obtain the value of the mass. Our values will be defined as

$m_1 = m$

$m_2 = M$

$v_(1i) =v_0$

$v_(2i) = 0$

Initial momentum will be

$P_i = m_iv_(1i)+m_2v_(2i)$

$P_i = mv_0$

After collision

$v_(1f) = v_(2f) = (v_0)/(3)$

Final momentum

$P_f = (m_1+m_2)((v_0)/(3))$

$P_f = (m+M)((v_0)/(3))$

From conservation of momentum

$P_f = P_i$

Replacing,

$(m+M)((v_0)/(3))=mv_0$

$(m+M)(1)/(3) = m$

$m+M=3m$

$M=3m-m$

$M=2m$

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A fully loaded, slow-moving freight elevator has a cab with a total mass of 1400 kg, which is required to travel upward 37 m in 3.6 min, starting and ending at rest. The elevator's counterweight has a mass of only 930 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable?

Answers

Answer:

789.8 W

Explanation:

mass of the cab = 1400 kg, the counter weight of the elevator = 930 kg

weight of the cab = 1400 × 9.81 where weight = mg and m is mass and g is acceleration due to gravity.

weight of the cab = 13734 N

counter weight of the elevator = 930 × 9.81 = 9123.3 N

the exerted force of the elevator = weight of the cab - counter weight of the elevator = 13734 - 9123.3 = 4610.7 N

Average power by the motor P = F × v = F × distance / time

where v is speed in m/s, and time is in seconds

P = 4610.7 × 37 / ( 3.6 × 60) = 789.80 W

where (3.6 × 60 ) is the time in seconds

How to Measure the mass of a coin using a triple beam balance

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Answer:

https://youtu.be/stW-C7F7QOg

A loop of wire lies flat on the horizontal surface in an area with uniform magnetic field directed vertically up. The loop of wire suddenly contracts to half of its initial diameter. As viewed from above induced electric current in the loop isa. counterclockwiseb. clockwisec. there is no current in the loop because magnetic field is uniformd. there is no current in the loop because magnetic field does not change

Answers

Complete Question

a. counterclockwise

b. clockwise

c. there is no current in the loop because magnetic field is uniform

d. there is no current in the loop because magnetic field does not change

Answer:

Option A is the correct answer

Explanation:

According to the question the loop of wire contracts to half it initial diameter and will mean that less number of electric field line will pass through the loop and this change in magnetic flux will cause current to flow in the loop of wire and from Lenz's law this current will in the opposite direction of what produced it which is the change in magnetic flux so the current will flow in a counterclockwise direction

Which of the following sets of characteristics describe what we know about the outer planets? (2 points)Lowest daily average temperature, smaller in size and few or no moons.
Many moons, smaller in size and a ring system.
Rocky surface, closest to the sun and larger in size,
Gaseous composition, larger size and many moons.

Answers

Gaseous composition, larger size and many moons describe about the outer planets.

What are outer planets?

Jupiter, Saturn, Uranus, and Neptune are the four outer planets. They are all gas giants consisting primarily of hydrogen and helium. Their interiors are liquid and contain thick gaseous outer layers. Numerous moons and planetary rings consisting of dust and other particles are present on every one of the outer planets.

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Answer: D

Explanation: Gaseous composition, larger size and many moons

A solid 0.6950 kg ball rolls without slipping down a track toward a vertical loop of radius ????=0.8950 m . What minimum translational speed ????min must the ball have when it is a height H=1.377 m above the bottom of the loop in order to complete the loop without falling off the track? Assume that the radius of the ball itself is much smaller than the loop radius ???? . Use ????=9.810 m/s2 for the acceleration due to gravity.

Answers

Answer:

The minimum transnational speed is 4.10 m/s.

Explanation:

Given that,

Mass of solid ball = 0.6950 kg

Radius = 0.8950 m

Height = 1.377 m

We need to calculate the minimum velocity of the ball at bottom of the loop to complete the track

Using formula velocity at lower point

$v_(min)=√(5gR)$

Put the value into the formula

$v_(min)=√(5*9.8*0.8950)$

$v_(min)=6.62\ m/s$

We need to calculate the velocity

Using conservation of energy

P.E at height +K.E at height = K.E at the bottom

$mgH+(1)/(2)mv^2=(1)/(2)m(√(5gR))^2$

$v^2=(√(5gR))^2-2gH$

$v^2=(6.62)^2-2*9.8*1.377$

$v^2=16.8352$

$v=√(16.8352)$

$v=4.10\ m/s$

Hence, The minimum transnational speed is 4.10 m/s.

Final answer:

The minimum translational speed the solid ball must have when it is at a height H=1.377 m above the bottom of the loop to successfully complete the loop without falling off the track is approximately 7.672 m/s. This was derived using principles of energy conservation.

Explanation:

The minimum translational speed must be sufficient enough to maintain contact with the track even at the highest point of the loop. Using the principle of energy conservation, the total energy at the height H, assuming potential energy to be zero here, should be equal to the total energy at the highest point of the loop. Here, the total energy at height H will consist of both kinetic and potential energy while at the top of the loop it consists of potential energy only. Setting these equations equal to each other: 0.5 * m * v² + m * g * H = m * g * 2R Solving the above equation for v:v = √2g (2R-H). Substituting known values henceforth gives us √2*9.81*(2*0.895-1.377) = 7.672 m/s. Hence, the ball must have a minimum translational speed of approximately 7.672 m/s at height H to complete the loop without falling.

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The universe is filled with photons left over from the Big Bang that today have an average energy of about 2 × 10−4 eV (corresponding to a temperature of 2.7 K). As derived in lecture, the number of available energy states per unit volume for photons is ????(????)????????