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Zina [86]
3 years ago
5

What average mechanical power (in W) must a 64.5 kg mountain climber generate to climb to the summit of a hill of height 285 m i

n 49.0 min? Note: Due to inefficiencies in converting chemical energy to mechanical energy, the amount calculated here is only a fraction of the power that must be produced by the climber's body.
Physics
1 answer:
kakasveta [241]3 years ago
7 0

Answer:61.275 W

Explanation:

Given

Mass of climber m=64.5 kg

Height to climb h=285 m

Time taken t=49 min \approx 2940 s

Now Using  Work-energy Theorem work done is given by

W=m g h

W=64.5\times 9.8 \times 285

W=180148.5 J

W=180.148 kJ

Now  Average Mechanical Power is given by

P=\frac{W}{\Delta t}

P=\frac{180.148}{2940}=0.0612 kW

P=61.275 W

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The coefficient of linear expansion for some common materials are listed in the table to the right. You would like to design an
Stella [2.4K]

Answer:

A)    T_{f} = 77.6  ⁰C ,  B) T_{f} = 341.6 C

Explanation:

The thermal expansion is given by

           ΔL = α L₀ ΔT

The thermal expansion coefficient for polycarbonate varies between 65 and 70 10⁻⁶ ⁰C⁻¹ and for cast iron 12 10⁻⁶ C⁻¹

Let's look for the final temperature that says same shift change

          ΔT = ΔL / α L₀

          T_{f} - T₀ = ΔL /α L₀

          T_{f}= T₀ + ΔL / α L₀

Let's calculate for each material

A) Polycarbonate use as thermal expansion coefficient 70 10⁻⁶ C⁻¹

         T_{f} = 23.0 + 0.0268 / (70 10⁻⁶ 7.01)

         T_{f} = 23.0 + 54.6

        T_{f} = 77.6  ⁰C

B) cast iron with thermal expansion coefficient 12 10⁻⁶ C⁻¹

      T_{f} = 23.0 + 0.0268 / (12 10⁻⁶ 7.01)

      T_{f} = 23.0 + 3.1859 10²

      T_{f} = 341.6 C

8 0
4 years ago
What force pulls all objects with mass toward<br> one another?
Vilka [71]
Gravity

Hope this helps:)
5 0
3 years ago
I need to know if i have this correct
kherson [118]

Answer:

YES! you do have the right answer

Explanation:

4 0
3 years ago
While water skiing behind her father’s boat, Letty is pulled at constant speed by a force of 164 N from the tow rope that makes
kap26 [50]

Answer:

0.265

Explanation:

Draw a free body diagram.  There are four forces:

Normal force Fn pushing up.

Weight force mg pulling down.

Tension force T at an angle θ.

Friction force Fn μ pushing left.

Sum the forces in the y direction:

∑F = ma

Fn + T sin θ − mg = 0

Fn = mg − T sin θ

Sum the forces in the x direction:

∑F = ma

T cos θ − Fn μ = 0

Fn μ = T cos θ

μ = T cos θ / Fn

μ = T cos θ / (mg − T sin θ)

Given T = 164 N, θ = 10.0°, m = 65.0 kg, and g = 9.8 m/s²:

μ = (164 N cos 10.0°) / (65.0 kg × 9.8 m/s² − 164 N sin 10.0°)

μ = 0.265

7 0
3 years ago
A 20-pound object is dropped from a 50-foot bridge onto the ground below it. A 50-pound object is dropped from a 100-foot cliff
Anvisha [2.4K]

Answer:

here 20 lb object will hit the ground first

Explanation:

Acceleration of the object while it is dropped in air is given as

F = mg

a = \frac{F}{m}

a = g

so acceleration of the object is independent of the mass of the object

now time to reach the ground is given as

h = \frac{1}{2}gt^2

t = \sqrt{\frac{2h}{g}}

for 20 pound object

t_1 = \sqrt{\frac{2(50)}{32}}

t_1 = 1.77 s

for 50 pound object

t_2 = \sqrt{\frac{2(100)}{32}}

t_2 = 2.5 s

5 0
3 years ago
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