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Vesnalui [34]
3 years ago
7

The record time for a Tour de France cyclist to ascend the 1100-mm-high Alpe d'Huez is 37.5 minmin. The rider and his bike had a

mass of 65 kg. At the average speed of the ride, the metabolic power to ride on level ground would be 700 W. To this, we must add the power to make the climb. What was his total metabolic power?
Physics
1 answer:
katen-ka-za [31]3 years ago
7 0

Answer:

1945.6 W

Explanation:

We are given that

Height,h=1100 m

Time,t=37.5 min=37.5\times 60=2250 s

1 min=60 s

Mass,m=65 kg

Power,P=700 W

We have to find the his total metabolic power.

Power,=P'=\frac{W}{t}=\frac{mgh}{t}=\frac{65\times 9.8\times 1100}{2250}=311.4 W

Where g=9.8m/s^2

Efficiency =25%

Therefore,Power,P'=\frac{311.4}{0.25}=1245.6 W

Total metabolic power=P+P'=700+1245.6=1945.6 W

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A ball is thrown straight up with a speed of 30 m/s, and air resistance is negligible. How long does it take the ball to reach t
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Answer:

3 seconds

Explanation:

Applying,

Applying,

v = u±gt................ Equation 1

Where v = final velocity, u = initial velocity, t = time, g = acceleration due to gravity.

From the question,

Given: v = 0 m/s ( at the maximum height), u = 30 m/s

Constant: g = -10 m/s

Substitute these values into equation 1

0 = 30-10t

10t = 30

t = 30/10

t = 3 seconds

6 0
2 years ago
At the surface, atmospheric pressure is 1.013 × 10^5 Pa. People can normally snorkel down to a depth of roughly one meter. What
natulia [17]

Answer:

1.01 × 10⁵ Pa  

Explanation:

At the surface, atmospheric pressure is 1.013 × 10⁵ Pa.

We need to find the total pressure on the air in the lungs of a person to a depth of 1 meter.

Pressure at a depth is given by :

P=\rho gh

Where

\rho is the density of air, \rho=1.225\ kg/m^3

So,

P=1.225\times 9.8\times 1\\\\=12\ Pa

Total pressure, P = Atmospheric pressure + 12 Pa

= 1.013 × 10⁵ Pa + 12 Pa

= 1.01 × 10⁵ Pa

Hence, the total pressure is 1.01 × 10⁵ Pa.

5 0
2 years ago
There is a 50 g sample of Ra-229. It has a half-life of 4 minutes.
Sloan [31]

Via half-life equation we have:


A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }


Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes.  By plugging those values in we get:

A_{final}=50(\frac{1}{2})^\frac{12}{4}=50(\frac{1}{2})^{3}=50(\frac{1}{8})=6.25g


There is 6.25 grams left of Ra-229 after 12 minutes.

4 0
3 years ago
Which of the following statement(s) is
alexgriva [62]

Answer:

Explanation:

A rectifier is an electrical device that converts alternating current (AC) to direct current (DC), a process known as rectification. Rectifiers have many uses including as components of power supplies and as amplitude modulation detectors (envelope detectors) of radio signals. Rectifiers are most commonly made using solid state diodes but other type of components can be used when very high voltages or currents are involved. When only a single diode is used to rectify AC (by blocking the negative or positive portion of the waveform), the difference between the term diode and the term rectifier is simply one of usage. The term rectifier describes a diode that is being used to convert AC to DC. Most rectifier circuits contain a number of diodes in a specific arrangement to more efficiently convert AC power to DC power than is possible with only a single diode.

5 0
2 years ago
61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A
Mademuasel [1]

Answer:

Part a)

percentage = 21.3%

Part b)

percentage = 2.13 \times 10^{-5}%

Explanation:

As we know that total power used in the room is given as

P = P_1 + P_2 + P_3 + P_4

here we have

P_1 = (110)(3) = 330 W

P_2 = 100 W

P_3 = 60 W

P_4 = 3 W

P = 330 + 100 + 60 + 3

P = 493 W

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

110\times i = 493

i = 4.48 A

now resistance of transmission line

R = \frac{\rho L}{A}

R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}

R = 5.23 \ohm

now power loss in line is given as

P = i^2 R

P = (4.48)^2(5.23)

P = 105 W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{105}{493} \times 100

percentage = 21.3%

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

110 \times 10^3 (i ) = 493

i = 4.48\times 10^{-3} A

now power loss in line is given as

P = i^2 R

P = (4.48 \times 10^{-3})^2(5.23)

P = 1.05 \times 10^{-4} W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{1.05 \times 10^{-4}}{493} \times 100

percentage = 2.13 \times 10^{-5}%

6 0
3 years ago
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