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2 years ago

Suppose a 65.5 kg gymnast climbs a rope. What is the tension in the rope if she climbs at a constant speed

Physics

1 answer:

natali 33 [55]2 years ago

7 0

The tension in the rope when the gymnast climbs it at constant speed is 641.9 N.

Given:

Mass of gymnast, m = 65.5 kg

The speed 'v' of gymnast is constant

Solution:

Consider the free-body diagram of the system as shown below.

Balancing forces along the vertical axis we get:

ΣFy = 0

Thus, we get:

F = ma - (1)

where, m is mass of gymnast

a is acceleration of gymnast (a = 0m/s², as the speed is constant)

Also,

F = T - mg -(2)

where, T is tension in the rope

g is acceleration due to gravity

Equating (1) & (2), we get:

ma = T - mg

Re-arranging the equation, we get:

T = m(a+g)

Applying values in above equation we get:

T = (65.5 kg)(0 m/s²+9.8 m/s²)

T = 641.9 N

Therefore, the tension in the rope when the gymnast climbs it at constant speed is 641.9 N.

Learn more about tension here:

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When the liquid line is restricted, the supply of refrigerant to the metering device is reduced. What is the effect on suction p

Nimfa-mama [501]

Answer:

The suction pressure decreases and the superheat increases when the liquid line is restricted and the supply of refrigerant to the metering device is reduced.

Explanation:

1. The five components of refrigeration are:

Fluid refrigerant
Compressor
Condenser coil
Evaporator coil
Expansion device.

The compressor limits the vapor released by the refrigerant. This

causes a rise in pressure (in refrigerant), which then pushes the

vapor into the coils on the outside of the refrigerator.

2. Now when the cooler air meets the warm gas present in the coils, it

gets converted into liquid form.

3. Thus, when the liquid form is at high pressure, the refrigerant then

cools down as it flows through the coils placed in the fridge ( in both

freezing and normal sections).

4. The refrigerant also absorbs the warm air present in the fridge, which

causes it to evaporate and flow back through the compressor and the

cycle repeats in the same form.

Thus, when the liquid line is restricted and the supply of refrigerant to the metering device is reduced it causes a decrease in suction pressure and an increase in superheat.

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8 0

2 years ago

A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$

where

$F_{net}$ is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

$a=3.1 m/s^2$ is its acceleration

Substituting, we find the net force on the bicyclist:

$F_{net}=(60)(3.1)=186 N$

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

$F_{net}=F_a-R$

where:

$F_{net}$ is the net force

$F_a$ is the applied force (forward)

$R$ is the air drag (backward)

In this problem we have:

$F_{net}=186 N$ is the net force (found in part b)

$R=60 N$ is the magnitude of the air drag

Solving for $F_a$ , we find the force produced by the bicyclist while pedaling:

$F_a=F_{net}+R=186+60=246 N$

3 0

3 years ago

Calculate the kinetic energy of Julie, a 60 kg biker, traveling at a velocity of 8 m/s to the right

Vadim26 [7]

Answer:

1,920 Joules

Explanation:

K.E. = 1/2 mv2

so K.E. = 1/2 (60)(8x8) = 1,920 Joules

8 0

3 years ago

Three long, straight wires are carrying currents that have the same magnitude. In C the current is opposite to that in A and B.

Nadusha1986 [10]

Answer:

(b) B

Explanation:

The direction of force on a current carrying wire in a magnetic field can be found using the right hand rule, which states that-"stretch the thumb in the direction of the current, and point the fingers in the direction of magnetic field. The direction of palm will then give the direction of force on the wire

On wire B the forces due to A and C act in the same direction and so strengthen each other. they get added up because the forces act in the same direction.

on wires A and C the forces (due to B and C and A and B

respectively) act in opposite directions and therefore tend to cancel out.

5 0

4 years ago

During a compression at a constant pressure of 290 Pa, the volume of an ideal gas decreases from 0.62 m3 to 0.21 m3. The initial

Aloiza [94]

Answer:

a) -41.1 Joule

b) 108.38 Kelvin

Explanation:

Pressure = P = 290 Pa

Initial volume of gas = V₁ = 0.62 m³

Final volume of gas = V₂ = 0.21 m³

Initial temperature of gas = T₁ = 320 K

Heat loss = Q = -160 J

Work done = PΔV

⇒Work done = 290×(0.21-0.62)

⇒Work done = -118.9 J

a) Change in internal energy = Heat - Work

ΔU = -160 -(-118.9)

⇒ΔU = -41.1 J

∴ Change in internal energy is -41.1 J

b) V₁/V₂ = T₁/T₂

⇒T₂ = T₁V₂/V₁

⇒T₂ = 320×0.21/0.62

⇒T₂ = 108.38 K

∴ Final temperature of the gas is 108.38 Kelvin

5 0

4 years ago