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Andrews [41]
2 years ago
13

A jogger jogs around a circular track with a diameter of 275 m in 14.0 minutes. what was the jogger's average speed in m/s?

Physics
1 answer:
Pie2 years ago
5 0

The jogger's average speed 1.03 m/s

<h3>The Speed and the Velocity of a Particle in a Circle</h3>

The speed of a particle is a circle will always be constant while the velocity will not. That is, velocity varies.

Given that a jogger jogs around a circular track with a diameter of 275 m in 14.0 minutes. First convert the minutes to seconds

Given parameters are;

  • Diameter = 275 m
  • Radius r = D/2 = 137.5 m
  • Time t = 14 minutes = 14 x 60s = 840 s

Speed V = 2πr ÷ t

V = ( 2 × π × 137.5 )  ÷ 840

V = 863.9 / 840

V = 1.028 m/s

Therefore, the jogger's average speed 1.03 m/s approximately

Learn more about Circular Motion here: brainly.com/question/20905151

#SPJ1

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Explanation:

W = PE

W = mgh

1500 J = (20 kg) (9.8 m/s²) h

h = 7.65 m

Round as needed.

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A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the a
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Answer:

C) one-half as great

Explanation:

We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

\Delta x=v_0t+\frac{gt^2}{2}

In this case, the sphere starts from rest, so v_0=0. Replacing the given values and solving for g':

g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}

The acceleration due to gravity near Earth's surface is g=9.8\frac{m}{s^2}. So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

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3 years ago
The average kinetic energy of the molecules of an ideal gas is directly proportional to the
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The average kinetic energy<span> of a </span>gas<span> particle is </span>directly proportional<span> to the </span>temperature<span>.</span>
3 0
3 years ago
An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,
insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

Where

t_{hf} = Temperature at the inside of the pipe = 300 °C

t_{f} = Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

k_A = 15 W/m·K

k_B = 0.038 W/m·K

h_{hf} = 75 W/m²·K

h_{cf} = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

Q = \frac{t_A-t_B}{R_T} Where R_T for the air film on the pipe outer surface is given by

R_T= \frac{1}{\alpha A}

where A =area of the outside of the pipe

= \frac{1}{10*2\pi*0.07*1 } = 0.227 K/W

Therefore

131.32 W = \frac{t_A-20}{0.227} which gives

t_A = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

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Answer: C

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