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Scilla [17]
3 years ago
11

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 60 cm long and has a mass of 3.8 kg, with the center of

mass at 40% of the arm length from the shoulder.
a. What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight out to his side, parallel to the floor?
b. What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight, but 45∘ below horizontal?
Physics
1 answer:
Serggg [28]3 years ago
3 0

Answer:

(a) τ = 26.58 Nm

(b) τ = 18.79 Nm

Explanation:

(a)

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m

τ₁ = (29.4 N)(0.6 m)

τ₁ = 17.64 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m

τ₂ = (37.24 N)(0.24 m)

τ₂ = 8.94 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 17.64 Nm + 8.94 Nm

<u>τ = 26.58 Nm</u>

<u></u>

(b)

Now, the arm is at 45° below horizontal line.

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

42.42 cm = 0.4242 m

τ₁ = (29.4 N)(0.4242 m)

τ₁ = 12.47 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m

τ₂ = (37.24 N)(0.1696 m)

τ₂ = 6.32 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 12.47 Nm + 6.32 Nm

<u>τ = 18.79 Nm</u>

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Alborosie

(amount of heat)Q = ? , (Mass) m= 4 g , ΔT = T f - T i = 180 c° - 20 °c = 160 °c ,

Ce = 0.093 cal/g. °c

Q = m C ΔT

Q = 4 g × 0.093 cal/g.c° × ( 180 °c- 20 °c )

Q= 4×0.093 × 160

Q = 59.52 cal

I hope I helped you^_^

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3 years ago
A student is given three wires that are made from different materials, but each wire has the same physical dimensions. For a giv
Ymorist [56]

Answer:

Use the ammeter to measure the current that flows through each wire, because a larger current that flows through the wire corresponds to a smaller resistivity

Explanation:

Since they are connected to a constant voltage power source, the potential difference does not change. The potential difference is proportional to the product of the current and the resistance and, the resistance opposes the flow of electric current. It is clear to see that a large current that flows through the current means there is a lesser resistance to the flow of current at constant potential difference across the circuit.

6 0
3 years ago
Which form of energy is due to the movement of electrons?
Kruka [31]
Electrical Energy : b
3 0
3 years ago
If a builder of mass 75kg climbs a vertical ladder of 25m how much energy has she gained ?
erica [24]
So,

GPE (graviational potential energy) = mass x g x height

GPE is depends on where zero height is defined.  In this situation, we define h = 0 as the initial height.

GPE = 75 \ kg*9.8 \ \frac{m}{s^2}*25 \ m

GPE = 18,375 \frac{kg*m^2}{s^2}

GPE = 18,375 \ joules(J) \ or \ 18.375 \ kilojoules(kJ)

The builder has gained 18.375 kJ of PE.
4 0
3 years ago
at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the g
Readme [11.4K]

The mass of water that must be raised is 5.25\cdot 10^7 kg

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

P_{out} = 0.70 P_{in}

where P_{in} is the power in input.

The power in input can be written as

P_{in} = \frac{W}{t}

where

W is the work done in lifting the water

t = 3 h = 10,800 s is the time elapsed

The work done in lifting the water is given by

W=mgh

where

m is the mass of water

g=9.8 m/s^2 is the acceleration of gravity

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

P_{out} = 0.70 \frac{mgh}{t}

Where

P_{out} = 150 MW = 150\cdot 10^6 W

And solving for m, we find:

m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

3 0
3 years ago
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