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ziro4ka [17]
3 years ago
8

What is the average power output of an athlete who can life 9.0 * 10^2 kg 2.5 m in 2.0 s?

Physics
1 answer:
eduard3 years ago
3 0

Answer:

Average power output of athlete = 11025 Watts

Explanation:

Work done is defined as the product of force applied and the displacement perpendicular to the force.

Work = Force\times Displacement

Power is defined as work done per unit time.

Power = \frac{Work done}{Time}

Here the person lifts 900 kg.

Height = 2.5 m

Time interval = 2 seconds

Force = weight \times gravity

          = 900 \times 9.8

          = 8820 N

Work done = Force\times Displacement

                   = 8820 \times 2.5

                   = 22050 J

Power = \frac {22050}{2}

           = 11025 Watt

                 

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An object is placed 4.0 cm to the left of a convex lens with a focal length of +8.0 cm . Where is the image of the object?
Serjik [45]

The image of the object is 8cm to the left of the lens (D)

<h3></h3>

What is the image of an object?

The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.

It is calculated thus:

1÷v = 1÷f - 1÷u

<h3>How to calculate the image of an object</h3>

From the formula

1÷v = 1÷f - 1÷u

<h3>Where </h3>

V = image distance fromthe object

U = object

f = focal length

Substitute the values

1÷v = 1÷8 - 1÷ 4

1÷v = - 1÷8

Make v the subject of formula

v = -8cm

Therefore, the image of the object is 8cm to the left of the lens (D)

Learn more on focal length here:

brainly.com/question/25779311

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6 0
2 years ago
A parallel-plate capacitor has 2.10 cm × 2.10 cm electrodes with surface charge densities ±1.00×10-6 C/m2. A proton traveling pa
Darya [45]

Answer:

x=0.53x10^{-3} m

Explanation:

Using Gauss law the field is uniform so

E=ζ/ε

Charge densities ⇒ζ=1.x10x^{-6} \frac{C}{m^{2}}

ε=8.85x10^{-12} \frac{C^{2}}{n*m^{2}}

E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}

Force of charge is

F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N

F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s

So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2}  \\x_{f}=0.53x^{-3}m

5 0
2 years ago
use the instantaneous center of zero velocity to determine the angular velocity !ab of link ab and the velocity vb of collar b f
allsm [11]

Instantaneous center:

It is the center about a body moves in planer motion. The velocity of Instantaneous center is zero and Instantaneous center can be lie out side or inside the body. About this center every particle of a body rotates.

From the diagram

Where these two lines will cut then it will the I-Center.Point A and B is moving perpendicular to the point I.

If we take three link link1,link2 and link3 then I center of these three link will be in one straight line It means that they will be co-linear.

Therefore, when the mass is at its equilibrium position (which corresponds to x=0), the velocity of the mass will be maximum.

To know more about velocity, refer: brainly.com/question/12413963

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7 0
1 year ago
Consider the system consisting of two blocks connected by a rope and a pulley. The coefficient of static friction between the ra
RSB [31]

Answer:

1.2 kg

Explanation:

Let UP ramp be the positive direction

                                                  F = ma

   T     -   Wt ||      -         Ff            = m(0)

  mg   -  Μgsinθ -      μΜgcosθ    = 0

m(9.8) - 13sin35 - 0.36(13)cos35 = 0        

                                                 m = 13(sin35 + 0.36cos35) / 9.8

                                                 m = 1.15205... ≈ 1.2 kg

5 0
2 years ago
two cars are travelling side by side down a straight road in opposite directions. both cars are travelling at a speed of 22 m/s.
Nimfa-mama [501]

Answer: 44m/s

Explanation:

Speed of Car A travelling left = 22m/s

Speed of Car B travelling right = 22m/s

Now recall that relative speed of objects moving in opposite directions is equal to the sum of each speed

Hence, Relative speed = (Speed of Car A + Speed of Car B)

= (22m/s + 22m/s)

= 44m/s

3 0
3 years ago
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