A completely submerged object always displaces its own

A weightlifter raises a 200-kg barbell through a height of 2 m in 2.2 s. the average power he develops during the lift is

Lunna [17]

Power is calculated as work per unit time, and work in turn is calculated as force multiplied by distance. In this case, the force required is equivalent to the weight of the barbell multiplied by acceleration due to gravity.
P = W/t = Fd/t = mgd/t = (200 kg)(9.81 m/s^2)(2 m)/2.2 s = 1783.64 Watts.

4 0

4 years ago

Read 2 more answers

The fictional rocket ship Adventure is measured to be 65 m long by the ship's captain inside the rocket.When the rocket moves pa

kipiarov [429]

Answer:

1.04 s

Explanation:

The computation is shown below:

As we know that

t = t' × 1 ÷ (√(1 - (v/c)^2)

here

v = 0.5c

t = 1.20 -s

So,

1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)

1.20 = t' × 1 ÷ (√(1 - (0.5)^2)

1.20 = t' ÷ √0.75

1.20 = t' ÷ 0.866

t' = 0.866 × 1.20

= 1.04 s

The above formula should be applied

8 0

3 years ago

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 15.6

VladimirAG [237]

To solve this problem it is necessary to take into account the kinematic equations of motion and the change that exists in the volume flow.

By definition the change in speed is given by

$v_f^2-v_i^2 = 2ax$

Where,

x= distance

$v_f =$ final velocity

$v_i =$ initial velocity

a = acceleration

On the other hand we know that the flow of a fluid is given by

$\dot{V} = Av$

Where,

A = Area

v = Velocity

PART A )

Applying this equation to the previously given values we have to

$v_f^2-v_i^2 = 2ax$

$v_f^2-0 = 2*(9.8)(15.6)$

$v_f^2=305.76$

$v_f = 17.48$

Therefore the velocity of the water leaving the hole is 17.48m/s

PART B )

In the case of the hole we take the area of a circle, therefore replacing in the flow equation we have to,

$\dot{V} = \pi r^2 v$

$r = \sqrt{\frac{\dot{V}}{\pi v}}$

$r = \sqrt{\frac{3*10^{-3}*\frac{1}{60}}{\pi (17.48)}$

$r = \sqrt{9.10*10^{-7}}$

$r = 0.54*10^{-4}$

The diameter is 2 times the radius, then is $1.91*10^{-3}$ m or 1.91mm

<em>Note: The rate flow was converted from minutes to seconds.</em>

8 0

3 years ago

You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As so

grigory [225]

Answer:

Explanation:

Time duration during which acceleration exists in bicycle =

23 / 12 = 1.91 s

Time duration during which acceleration exists in car

= 47 / 8 = 5.875 s

Distance covered by bicycle during acceleration ( t = 1.91 s )

= 1/2 x 12 x (1.91)²

= 21.88 mi

Distance covered by car during this time ( t = 1.91 s )

= 1/2 x 8 x (1.91)²

7.64 mi ,

velocity of car after 1.91 s

= 8 x 1.91 = 15.28 mi/h

Let after time 1.91 , time taken by them to meet each other be t

Total distance covered by cycle = total distance covered by car

21.88 + 23 t = 7.64 + 15.28t + 4 t²

21.88 = 7.64 - 7.72t +4 t²

4 t² -7.72 t -14.24 = 0

t = 2.83 s

Total time taken

= 2.83 + 1.91

= 4.74 s

So after 4.74 s they will meet each other.

b ) Maximum distance occurs when velocity of both of them becomes equal .

Velocity after 1.91 s of bicycle

12 x 1.91 = 23 mi/h

Velocity after 1.91 s of car

8 x 1.91 = 15.28 mi/h . Let after time t , the velocity of car becomes 23

15.28 + 8t = 23

t = .965 s

So after time .965 s , car has velocity equal to that of bicycle.

The bicycle will travel a distance of

= 21.88 + .965 x 23 = 44.075 mi

car will travel a distance of

7.64 + 15.28 x .965 + .5 x 8 x .965²

= 7.64 + 14.75 + 3.72

= 26.11 mi

Distance between car and bicycle

= 44.075 - 26.11 = 17.965 mi

= 17.965 x 1760

= 31618.4 ft.

5 0

3 years ago

When light passes from air into water at an angle of 60° from the normal, what is the angle of refraction? (40.6°)

matrenka [14]

Answer:

40.6°

Explanation:

5 0

3 years ago