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masya89 [10]
3 years ago
12

A man of mass 80 kg runs up a flight of stairs 20 m high in 10 s. (a) how much power is used to lift the man? (b) If the man’s b

ody is 25% efficient, how much power does he expend? (c) This man consumes approximately 1.05 × 107 J (2500 food calories) of energy per day while maintaining a constant weight. What is the average power he produces over a day? (d) Compare this with his power production when he runs up the stairs.
Physics
1 answer:
ad-work [718]3 years ago
7 0

Answer:

a: 1568w

Explanation:

P = F*d/t = mgd/t = 1568w

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A car moves with a speed of 72 km/h for 15 minutes and then with a speed of 80 km/h for the next 12 minutes. The total distance
sukhopar [10]

Answer:

The total distance covered by the car is 3,810.08 m

Explanation:

Given;

initial speed of the car, u = 72 km/hr = 20 m/s

initial time, t₁ = 15 minutes = 900 s

final speed of the car, v = 80 km/hr = 22.22 m/s

final time, t₂ = 12 minutes = 720 s

The acceleration of the car is given as;

a = \frac{v-u}{t_2 -t_1} \\\\a = \frac{22.22-20}{900-720}\\\\a = 0.0123 \ m/s^2

The total distance covered by the car is given as;

v² = u² + 2as

where;

s is the total distance covered by the car

22.22² = 20² + 2(0.0123)s

22.22²  - 20² = 2(0.0123)s

93.728 = 0.0246s

s = 93.728 / 0.0246

s = 3,810.08 m

Therefore, the total distance covered by the car is 3,810.08 m

8 0
3 years ago
Read 2 more answers
(Brainliest) How are radio waves used in cell phone wireless communication technology?
poizon [28]

Radio waves are used in cell phone wireless communication by this: Radio waves are released from the cell phone and travel to satellites in the stratosphere. After they reach the satellite, they are redirected to the recipient of the call/text.

Hope this helps! Can I have brainliest please?

5 0
3 years ago
A teacher wants to demonstrate that the radioactive source emits alpha beta and gamma radiation. Describe a method the teacher c
Mila [183]

By using an electric field, it is feasible to differentiate between these different forms of radiation.

<h3>What is a radioactive source?</h3>

A source that emits radiation like gamma, beta, and alpha rays is said to be radioactive. Using an electric field, we can discriminate between these different forms of radiation.

The field does not deflate the gamma rays, but it does deflate the alpha and beta rays, with the alpha being deflated to the field's negative portion and the beta to its positive part.

Hence, by using an electric field, it is feasible to differentiate between these different forms of radiation.

To learn more about the radioactive source refer;

brainly.com/question/12741761

#SPJ1

8 0
1 year ago
Sarah and Maisie are analysing data from their school sports day. Looking at the 1500 m results for Stephen, Maisie believes tha
Dahasolnce [82]

Answer:

Sarah is right

Explanation:

This is an exercise that differentiates between scalars and vectors.

A scalar is a number, instead a vector is a number that represents the module in addition to direction and sense.

In this case, the distance (scalar) traveled is a number, which is why it is worth 1500m, but the displacement is a vector and since the point where it leaves is the same point where the vector's modulus arrives is zero, so the DISPLACEMENT VECTOR is zero

consequently Sarah is right

4 0
3 years ago
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
3 years ago
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