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3 years ago

10

Suppose a person is pulling a rope attached to a load then during the calculations of number of forces acting on it do i have to

take force applied by a man along with the tension or only tension (explain it with reason)

1 answer:

larisa86 [58]3 years ago

5 0

Answer:

Gravity and force can be applied with it.

Explanation:

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At what condition does a body becomes weightless at the equator?

ArbitrLikvidat [17]

Answer:

The decrease is due to the bulge at the equator (putting more distance between the rest of the planet and the surface

Explanation:

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3 years ago

1. What is technology?<br>techndegy

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Answer:

the application of scientific knowledge for practical purposes, especially in industry. Another answer:the sum of techniques,skills,methods and processes used in the production of goods or services or in the accomplishment of objectives, such as scientific investigation

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3 years ago

A Carnot engine operates between temperature levels of 600 K and 300 K. It drives a Carnot refrigerator, which provides cooling

KATRIN_1 [288]

Explanation:

Formula for maximum efficiency of a Carnot refrigerator is as follows.

$\frac{W}{Q_{H_{1}}} = \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}$ ..... (1)

And, formula for maximum efficiency of Carnot refrigerator is as follows.

$\frac{W}{Q_{C_{2}}} = \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}}$ ...... (2)

Now, equating both equations (1) and (2) as follows.

$Q_{C_{2}} \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}}$ = $Q_{H_{1}} \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}$

$\gamma = \frac{Q_{C_{2}}}{Q_{H_{1}}}$

= $\frac{T_{C_{2}}}{T_{H_{1}}} (\frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{2}} - T_{C_{2}}})$

= $\frac{250}{600} (\frac{(600 - 300)K}{300 K - 250 K})$

= 2.5

Thus, we can conclude that the ratio of heat extracted by the refrigerator ("cooling load") to the heat delivered to the engine ("heating load") is 2.5.

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3 years ago

A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.

babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8 + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

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3 years ago

Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sp

tensa zangetsu [6.8K]

Answer:

1. 14.4 m/s 2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:

Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

ω= v/r

If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

∆K = -∆U

½ *m*v² + ½* I* ω² = m*g*h

Simplifying and replacing the value of the angular velocity:

½ * v² + ½ I *(v/r)² = g*h (1)

In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

Solid Sphere I = 2/5* m *r²

Replacing in (1):

½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

Replacing by the value given for h, and solving for v:

v = √(10/7*9.8 m/s2*14.7 m) = 14. 4 m/s

Spherical shell I=2/3*m*r²

Replacing in (1):

½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

Replacing by the value given for h, and solving for v:

v = √(6/5*9.8 m/s2*14.7 m) = 13.2 m/s

Hoop I= m*r²

Replacing in (1):

½ * v² + ½ (m*r²) *(v/r)² = g*h

Replacing by the value given for h, and solving for v:

v = √(9.8 m/s2*14.7 m) = 12.0 m/s

Cylinder I = 1/2 * m* r²

Replacing in (1):

½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

Replacing by the value given for h, and solving for v:

v = 2*√(1/3*9.8 m/s2*14.7 m) = 13.9 m/s

5 0

3 years ago