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Ugo [173]
3 years ago
5

A diamond with a mass of 45 g hangs motionless from a chain. what is the upward force of the chain on the diamond?

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
6 0
The upward force of the chain on the diamond would be the tension in the chain, and this tension would have to support the weight of the 45g that hangs from the chain.

mass = 45 g = 45/1000 kg = 0.045kg

Weight = mg = 0.045 * 10 ≈ 0.45N,            g ≈ 10 m/s²

<span>So the upward force is ≈ </span><span>0.45N. </span>
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Which of the following will cause an increase in gas pressure in a closed container?
Ludmilka [50]

C. Both A. and B.

Explanation:

Statement A. Reducing the volume is true because of Boyle's law, which states that for a gas at fixed temperature, the pressure p and the volume V are inversely proportional:

pV=const.

Therefore, when the volume V is reduced, the pressure p increases.

Statement B. Adding more gas is also true: in fact, if we add gas into the container, we will have more molecules of the gas hitting the wall of the container. But the pressure of a gas is exactly given by this: by the collision of the molecules against the wall of the container, so the more the molecules of gas, the greater the pressure.

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3 years ago
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Acceleration is defined as the rate of change for which of the following?
Snowcat [4.5K]
<span>B. velocity .................</span>
6 0
2 years ago
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An object is moving in a straight line along the y axis. As a function of time, its position is given by the equation y=3.0+4.8x
romanna [79]

Answer:

Explanation:

<u>Instant Velocity and Acceleration </u>

Give the position of an object as a function of time y(x), the instant velocity can be obtained by

v(x)=y'(x)

Where y'(x) is the first derivative of y respect to time x. The instant acceleration is given by

a(x)=v'(x)=y''(x)

We are given the function for y

y(x)=3.0+4.8x +6.4x^2

Note we have changed the last term to be quadratic, so the question has more sense.

The velocity is

v(x)=y'(x)=4.8+12.8x

And the acceleration is a(x)=v'(x)=12.8

5 0
3 years ago
You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C
jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0
3 years ago
Two 25.0N weights are suspended at opposite ends of a rope that passes over a frictionless pulley. What is the tension in the ro
goldenfox [79]

Answer:

tension in rope = 25.0 N

Explanation:

  • Two forces act on the suspended weight. The force coming down is the gravitational force and the upward force by the tension in the rope.
  • Since the suspended weight is not accelerating so that the net force will be zero. Therefore the tension in the rope should be 25 N.

       ∑F = F - W = 0

       so

       F = W

       so tension in rope = F  = T  = 25 N

8 0
2 years ago
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