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umka21 [38]
3 years ago
9

While running, a person dissipates about 0.60 J of mechanical energy per step per kilogram of body mass. If a 52-kg person devel

ops a power of 80 W during a race, how fast is the person running
Physics
1 answer:
Yuliya22 [10]3 years ago
3 0

Answer:

The person is running at a speed of 2.564 m/s

Explanation:

Given;

mechanical energy dissipated per kilogram per step, E/kg/S = 0.6 J/kg/S

mass of the person, m = 52 kg

power developed by the person, P = 80 W

mechanical energy of the person per step, E = 0.6 J/kg x 52 kg

                                                        E_{step} = 31.2 J

mechanical energy for the total step, E_{total} = 31.2 J x S

P = E / t

P_{avg} = \frac{E_{total}}{t} \\\\P_{avg} = \frac{E_{step}*S}{t}\\\\\frac{P_{avg}}{E_{step}} = \frac{S}{t} \\\\\frac{S}{t} = \frac{80}{31.2} \\\\\frac{S}{t} = 2.564 \ m/s

Therefore, the person is running at a speed of 2.564 m/s

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A futuristic train rushes past you at an incredible, relativistic speed such that it appears to only be 80% as long as you know
Fed [463]

Answer:

v = 0.6c = 1.8 x 10⁸ m/s

Explanation:

From Einstein's theory of relativity, we know that the length of an object contracts while traveling at a speed relative to speed of light. The contraction is according to following equation:

L = L_{0}\sqrt{1 - \frac{v^2}{c^2}}

where,

L = Relative Length

L₀ = Rest Length

According to given:

L = 0.8 L₀

c = speed of light = 3 x 10⁸ m/s

v = relative speed of train = ?

Therefore,

0.8L_{0} = L_{0}\sqrt{1 - \frac{v^2}{c^2} } \\\\0.8 = \sqrt{1 - \frac{v^2}{c^2} }

squaring on both sides:

0.64 = 1-\frac{v^2}{c^2}\\\\\frac{v^2}{c^2} = 1-0.64 \\v^2= 0.36c^2\\

taking square root on both sides:

<u>v = 0.6c = 1.8 x 10⁸ m/s</u>

3 0
3 years ago
A motorcyclist drives from a to b with the uniform speed of 30 km/h-1 and returns back with the speed of 20 km/h-1.find the aver
maxonik [38]
30+20 =50
For average 50/2=25
4 0
3 years ago
Read 2 more answers
The blades of a fan running at low speed turn at 26.2 rad/s. When the fan is switched to high speed, the rotation rate increases
Lady bird [3.3K]

Answer: 1.79\ rad/s^2

Explanation:

Given

Initial angular speed is \omega_1=26.2\ rad/s

Final angular speed is \omega_2=36.5\ rad/s

Time period t=5.75\ s

Magnitude of the fan's acceleration is given by

\Rightarrow \alpha=\dfrac{\omega_2-\omega_1}{t}

Insert the values

\Rightarrow \alpha=\dfrac{36.5-26.2}{5.75}\\\\\Rightarrow \alpha=\dfrac{10.3}{5.75}\\\\\Rightarrow \alpha=1.79\ rad/s^2

Thus, fan angular acceleration is 1.79\ rad/s^2

8 0
3 years ago
Read 2 more answers
A long wire carrying a 5.0 A current perpendicular to the (xy)-plane intersects the x-axis at x = -2.00 cm. A second parallel wi
zimovet [89]

Answer:

a. 05cm from x axis

b. 8cm from x axis

Explanation:

If the net magnetic field is zero and the currents are in the same direction then the thanks point is between the currents i1 and i2 as show in the attachment below

a. Given that i1= 5A and i2=3A

Let assume the null point is xcm from current i1, then the null point will be (4-x)cm from current i2 since the total length is 4cm.

Now the magnetic field of the current i1 from the null point= to magnetic field of current i2 from the null point

B1=B2

μi1/2πx=μi2/2π(4-x)

i1/x=i2/(4-x)

5/x=3/(4-x)

20-5x=3x

8x=20

8x=2.5cm

since from the left of x axis is 2cm, then the null point is 2.5-2 which 0.5cm from the origin x axis.

The null point is 0.5cm from the origin x axis

b. If both current are flowing in opposite direction, the null point lies outside of the current.

Then with same analysis let assume the first current i1 is xcm from the null point and since the total length is 4cm the second current i2 will be (x-4)cm from the null point.

Also the magnetic field of the current i1 from the null point = to magnetic field of current i2 from the null point

B1=B2

μi1/2πx=μi2/2π(x-4)

i1/x=i2/(x-4)

5/x=3/(x-4)

5x-20=3x

2x=20

x=10cm.

This shows that the distance of the null point from current i1 is 10cm and the current i1 is 2cm from the x axis, then the null point is 10-2=8cm from the origin x axis.

The null point is 8cm from the x axis.

Check the attachment to see the diagram of the current and the null points

6 0
3 years ago
The specific heat of a solid Y is 11.5 cal/g°C. A sample of this
fgiga [73]

The sample of the solid in grams is 5.5g

HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

  • The quantity of heat absorbed or released by a substance can be calculated using the following formula:

Q = m × c × ∆T

Where;

Q = quantity of heat absorbed/released (cal)

m = mass of the substance (g)

c = specific heat (cal/g°C)

∆T = change in temperature (°C)

  • According to the information provided in this question:

Q = 7.90Kcal = 7900cal

m = ?

c = 11.5 cal/g°C

T1 = 135K = 135K − 273.15 = -138.1°C

T2 = 260K = 260K − 273.15 = -13.15°C

∆T = -13.15 - (-138.1) = 124.95°C

  • Hence, the mass of the substance can be calculated as follows:

m = Q ÷ c∆T

m = 7900 ÷ (11.5 × 124.95)

m = 7900 ÷ 1436.93

m = 5.5grams.

Therefore, the sample of the solid in grams is 5.5g.

Learn more: brainly.com/question/21643161?referrer=searchResults

5 0
2 years ago
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