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mel-nik [20]
3 years ago
15

A chain lying on the ground is 11 meters long and its mass is 95 kilograms. The chain is threaded through a pulley, which is fix

ed to the ground, and pulled directly up so that it forms the shape of an L. How much work is required to raise one end of the chain to a height of 7 meters?
Physics
1 answer:
Natalija [7]3 years ago
6 0

Answer: 296.1 J or 6.98 kJ

Explanation:

Given

Length of chain, l = 11 m

Mass of chain, m = 95 kg

It is worthy of note that chain has two ends. So this depends which of the two ends should end up at

the given height of 7 m.

Scenario 1. If it is the leading end:

Weight of chain raised

W = mg/l

W = 9.8×95/11

W = 84.6 N

Height raised = ½ m (which is the height of the centre of gravity of the raised portion).

Work done = 84.6 × 1/2 * 7 = 296.1 Joules

Scenario 2. If it is the trailing end:

Weight of chain raised = 9.8×95 = 931 N.

Height raised (average) = 12.5 m (5.5 m for the chain midpoint + 7 m off the ground)

Work done = 931 × 7.5 = 6.98 kJoules

Thus, the work required is either of 296.1 J or 6.98 kJ

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How much energy (in kW-h) does a 900 Watt stove use in a week if it is used for 1.5 hours each day?
ioda

Answer:

9.45 kWh

Explanation:

Energy = Power × time

E = 900 W × (1.5 h/day × 7 day)

E = 9450 Wh

E = 9.45 kWh

3 0
3 years ago
Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.3
MAXImum [283]

Answer:

Δθ = 15747.37 rad.

Explanation:

  • The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.
  • Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:

       \omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)

  • Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:

       \omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)

  • Solving for Δθ in (2):

       \theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)

  • The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:

       \theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)

  • Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:

       \omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)

  • Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:

      \theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)

  • The total angular displacement is just the sum of (3), (4) and (6):
  • Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad
  • ⇒ Δθ = 15747.37 rad.
4 0
2 years ago
The national high magnetic field laboratory holds the world record for creating the strongest magnetic field. for brief periods
max2010maxim [7]

Newton’s 2nd law states that Force is equal to the product of mass (m) and acceleration (a):

F = m a                                  ---> 1

While in magnetic forces, force can also be expressed as:

F = q v B                               ---> 2

where,

q = total charge

v = velocity = 45 cm / s = 0.45 m / s

B = the magnetic field = 85 T

First we solve for the total charge, q:

q = 3.8 × 10^-23 g (1 mol / 23 g) (6.022 × 10^23 electrons / mol) (1.602 × 10^-19 C / electron)

q = 1.594 × 10^-19 C

 

We equate equations 1 and 2 then solve for acceleration a:

m a = q v B

a = q v B / m

a = [1.594 × 10^-19 C * 0.45 m / s * 85 T] / 3.8 × 10-26 kg

a = 160,437,862.2 m/s^2

 

Therefore the maximum acceleration of Na ions is about 160 × 10^6 m/s^2.

5 0
3 years ago
Read 2 more answers
PY85
aliya0001 [1]

Answer:

460 g

Explanation:

Heat lost by the warm water = heat gained by the cold water

-mCΔT = mCΔT

-m (4.184 J/g/K) (37°C − 85°C) = (1000 g) (4.184 J/g/K) (37°C − 15°C)

-m (37°C − 85°C) = (1000 g) (37°C − 15°C)

-m (-48°C) = (1000 g) (22°C)

m = 458 g

Rounded to two significant figures, you need a mass of 460 g of water.

3 0
2 years ago
Suppose that the collector is held at a small negative voltage with respect to the grid. Will the accelerated electrons reach th
Leona [35]

Answer:

B) Yes, but only those electrons with energy greater than the potential difference established between the grid and the collector will reach the collector.

Explanation:

In the case when the collector would held at a negative voltage i.e. small with regard to grid So yes the accelerated electrons would be reach to the collecting plate as the kinetic energy would be more than the potential energy that because of negative potential

so according to the given situation, the option b is correct

And, the rest of the options are wrong

3 0
3 years ago
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