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natulia [17]
2 years ago
14

How many slices of bread did each climber have to eat to compensate for the increase of the gravitational potential energy of th

e system climbers-Earth? (One piece of bread releases about 1.0×106J of energy in the body.) Assume that the height of Mt. Everest is 8850 m, the weight of a person is about 79 kg , and efficiency in converting the body's chemical energy into mechanical work is 10.5 %.
Physics
1 answer:
N76 [4]2 years ago
3 0

Answer:

So No of slices to be consumed by each person = n = 65

Explanation:

Energy released by one slice = E1

E1=10^6\ J

h = 8850 m ; m = 79 kg ,η= 10.5%

We know that potential energy given as

u = m g h

u = 79 x 9.81 x 8850

u=6.8\times 10^6\ J

we know from the defination of efficiency that,  η= E(out)/E(in)

Now amount of PE has to be compensated, In our case, E(out) =u

0.105=\dfrac{E(out)}{E(in)}

0.105=\dfrac{6.8\times 10^6}{E(in)}

E(in)=64.76\times 10^6\ J

Let n be the number of bread slices to be consumed.

n = E(in)/E1

n=\dfrac{64.76\times 10^6}{10^6}

n=64.76

So No of slices to be consumed by each person = n = 65

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Vesnalui [34]

Answer:

The magnitude of the resultant acceleration is 2.2 m/s^2

Explanation:

Mass (m) of the sailboat =  2000 kg

Force acting on the sailboat due to ocean  tide is F_1 = 3000N

Eastwards means takes place along the positive x direction

ThenF_{1x} = 3000N and F_{1y}= 0

Wind Force acting on the Sailboat isF_2  = 6000N directed towards the northwest that means at an angle  45 degree above the negative x axis

Then  

F_{2x} = -(6000N) cos 45 degree = -4242.6 N

F_{2y}  = (6000N) cos 45 degree = 4242.6 N

Hence  , the net force acting on the sailboat in x direction is  

F_x = F_{1x}+ F_{2x}

=  - 3000 N + 4242.6 N

=  - 3000 N +4242.6 N

= 1242.6N

Net Force acting on the sailboat in y direction is  

F_y = F_{1y}+ F_{2y}

= 0+ 4242.6N

= 4242.6N

The magnitude of the resultant force =

Using pythagorean theorm of 1243 N and 4243 N

\sqrt{(1242.6)^2 + (4242.6)^2

\sqrt{(1544054.76) + (17999654.8)}

\sqrt{(19543709.5)^2}

4420.8 N

F = ma

a = \frac{F}{m}

a =\frac{4420.8}{ 2000}

=2.2 m/s^2

4 0
2 years ago
Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3>s. (a) How fast will it shoot out of a hole 4
kati45 [8]

Answer:

velocity  = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

0.750 = \frac{\pi }{4} × (4.50×10^{-2})²  × velocity

solve it we get

velocity  = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

0.750 = \frac{\pi }{4} × 3 ×  (4.50×10^{-2})²  × velocity

velocity = 52.4 m/s

5 0
2 years ago
Suppose that sunlight is incident upon both a pair of reading glasses and a pair of sunglasses. Which pair would you expect to b
Ainat [17]

Answer: the pair of sunglasses

Explanation:

A good pair of sunglasses are composed of abosorbent lenses that filter the sunlight that affects the eyes retina, especially ultraviolet (UV). So, these sunglasses are used to reduce the amount of light or radiant energy transmitted.

On the other hand, normal reading glasses (in which the lens glass has not been treated to filter ultraviolet sunlight) will let UV rays pass through.

Therefore, if both glasses are exposed to sunlight, the sunglasses are expected to be warmer by absorbing that radiant energy and preventing it from reaching the eyes.

4 0
2 years ago
Which of the following quantities provide enough information to calculate the tension in a string of mass per unit length μ that
Bad White [126]

Answer:

A. the wave speed v and Wavelength

Explanation:

Given that

Mass density per unit length=μ

Frequency = f

The velocity V given as

\mu=\dfrac{T}{V^2}\ kg/m

V=\sqrt{\dfrac{T}{\mu}}

T=Tension

V=Velocity

V= f λ

λ=Wavelength

Therefore to find the tension ,only wavelength and speed is required.

The answer is A.

8 0
2 years ago
(a) A small object with mass 3.75 kg moves counterclockwise with constant speed 1.55 rad/s in a circle of radius 2.55 m centered
Eddi Din [679]

Answer:3.95 m/s

Explanation:

Given

mass of object m=3.75 kg

\omega =1.55 rad/s

radius of circle =2.55 m

initial Position r=2.55 \hat{i}

angular displacement \theta _0=8.95 rad

8.95 radian can be written as

1.42 (2\pi )

i.e. Particle is at first quadrant with \theta =0.4242\pi \times \frac{180}{\pi }

\theta =76.36^{\circ}

(c)velocity is v=\omgea \times r

v=1.55\times 2.55=3.95 m/s

8 0
2 years ago
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