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AVprozaik [17]
3 years ago
5

Calculate the efficiency of a machine if the input work is 200J and the output work is 150J.​

Physics
1 answer:
lidiya [134]3 years ago
8 0

Answer:

75% is the efficency of this maxhine

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"at sea level, atmospheric pressure is about _______ mm mercury."
RUDIKE [14]
Standard sea level pressure by definition equals 760 mm (29.92 inches) of mercury
6 0
4 years ago
Read 2 more answers
The relationship between energy (E), power (P), and time (t) is
NemiM [27]

Answer:

The avarage power of the body is 96.898 watts.

Explanation:

We must notice that given definition of power implies a constant consumption of energy, so that we should assume that energy consumption is constant. A Calorie is equal to 4186 joules. If we know that E = 2000\,Cal and t = 1\,day, the power of body, measured in watts, is:

P = \frac{(2000\,Cal)\cdot \left(4186\,\frac{J}{Cal} \right)}{(1\,day)\cdot \left(24\,\frac{h}{day} \right)\cdot \left(3600\,\frac{s}{h} \right)}

P = 96.898\,W

The avarage power of the body is 96.898 watts.

3 0
4 years ago
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and
Svetlanka [38]

Answer:

a = 1.152s

b = 0.817 m

c = 7.29m/s

Explanation: let the following

From the first equation of linear motion

V = u+at..........1

parameters be represented as :

t = Time taken

v = Final velocity

a = Acceleration due to gravity = 9.8m/s²

u = Initial velocity = 4 m/s

s = Displacement

V = 0

Substitute the values into equation 1

0 = 4-9.8(t)

-4 = -9.8t

t = 4/9.8

t = 0.408s

From : s = ut+1/2at^2.........2

S = 4×0.408+0.5(-9.8)×0.408^2

S= 1.632-4.9(0.166)

S = 1.632-0.815

S = 0.817m

Her highest height above the board is 0.817 m

Total height she would fall is 0.817+1.90 = 2.717 m

From equation 2

s = ut+1/2at^2

2.717 m = 0t+0.5(9.8)t^2

2.717 m = 0+4.9t^2

2.717 m = 4.9t^2

2.717/4.9 = t^2

0.554 =t^2

t =√0.554

t = 0.744s

Hence, her feet were in the air for 0.744+0.408seconds

= 1.152s

Also recall from equation 1

V= u+at

V = 0+9.8(0.744)

V = 7.29m/s

Hence, the velocity when she hits the water is 7.29m/s

Finally,

a = 1.152s

b = 0.817 m

c = 7.29m/s

4 0
3 years ago
The term that describes the direction closest to the point of origin is:
andreyandreev [35.5K]
The term that describes the direction closest to the point of origin is Proximal. Dorsal is the directional term for the movement toward the back of the body. Cephalic is the term that describes the movement towards the top of the body. Ventral on the other hand describes the movement toward the front of the body.  
3 0
3 years ago
How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s
poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

W = \Delta KE

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

Here,

m = mass

v_{f,i} = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2

W_1 = 150(m) J

Second case) When the particle goes from 20m/s to 30m/s

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2

W_1 = 250(m) J

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

5 0
4 years ago
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