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bulgar [2K]
3 years ago
10

a 40 kilogramstudent runs up a staircase to a floor that is 5.0 meters higher than her starting point in 7.0 seconds. the studen

t's power output is
Physics
1 answer:
Wittaler [7]3 years ago
3 0

When she reaches the top, she has more potential energy than she had at the bottom.

         Potential energy = (mass) x (gravity) x (height)

                                     = (40 kg) x (9.8 m/s²) x (5 m)

                                     =    1,960 more joules .

         Power = (energy) / (time)

                     = (1,960 joules) / (7 seconds)

                     =         280 watts

                    (about 0.375 horsepower) 
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A body is projected from the ground at an angle of 30° with the horizontal at an initial speed of 128 ft/s. Ignoring air frictio
Elanso [62]

Answer

given,

v = 128 ft/s

angle made with horizontal = 30°

now,

horizontal component of velocity

vx = v cos θ = 128 x cos 30° = 110.85 ft/s

vertical component of velocity

vy = v sin θ = 128 x sin 30° = 64 m/s

time taken to strike the ground

using equation of motion

v = u + at

0 =-64 -32 x t

t = 2 s

total time of flight is equal to

T = 2 t = 2 x 2 = 4 s

b) maximum height

using equation of motion

 v² = u² + 2 a h

 0 = 64² - 2 x 32 x h

 64 h = 64²

  h = 64 ft

c) range

R = v_x × time of flight

R = 110.85 × 4

R = 443.4 ft

4 0
3 years ago
Which type of lens is shown in the picture below?<br> plane<br> refractional<br> concave
TiliK225 [7]
It is a concave lens

Have a nice day
8 0
3 years ago
4. How do ordinary objects behave in microgravity?
Nikolay [14]

Answer:

Bubbles paused

Explanation:

the air bubble doesn't rise because it is no lighter than the water around it—there's no buoyancy. The droplet doesn't fall from the leaf because there's no force to pull it off. It's stuck there by molecular adhesion.

for instance, onto the International Space Station, gravity becomes negligible, and the laws of physics act differently than here on Earth

On Earth, the buoyancy of the air bubbles causes them to rise to the top together, creating a segregation between air and water. However, in microgravity, nothing forces the air bubbles to interact and thus rise together, Green said.

8 0
3 years ago
A mass of 1000 kg drops from a height of 10 m on a platform of negligible mass. It is desired to design a spring and dashpot on
Juliette [100K]

Answer:

k = 5\times 10^{4}\ N/m

b = 0.707\times 10^{3}

t = 7.1\times 10^{- 5}\ s

Solution:

As per the question:

Mass of the block, m = 1000 kg

Height, h = 10 m

Equilibrium position, x = 0.2 m

Now,

The velocity when the mass falls from a height of 10 m is given by the third eqn of motion:

v^{2} = u^{2} + 2gh

where

u = initial velocity = 0

g = 10m/s^{2}

Thus

v = \sqrt{2\times 10\times 10} = 10\sqrt{2}\ m/s

Force on the mass is given by:

F = mg = 1000\times 10 = 10000 N = 10\ kN

Also, we know that the spring force is given by:

F = - kx

Thus

k = \frac{F}{x} = \frac{10000}{0.2} = 5\times 10^{4}\ N/m

Now, to find the damping constant b, we know that:

F = - bv

b = \frac{F}{v} = \frac{10000}{10\sqrt{2}} = 0.707\times 10^{3}

Now,

Time required for the platform to get settled to 1 mm or 0.001 m is given by:

t = \frac{0.001}{v} = \frac{0.001}{10\sqrt{2}} = 7.1\times 10^{- 5}\ s

4 0
3 years ago
A 2.74 g coin, which has zero potential energy at the surface, is dropped into a 12.2 m well. After the coin comes to a stop in
nata0808 [166]

Answer:

- 0.328J

Explanation:

POTENTIAL ENERGY = mgh

                                   = 0.00274 × 9.81×12.2  here 2.74g = 0.00274kg

                                   = 0.32792868 J

                                   = 0.328J

AS IT IS BELOW THE SURFACE HENCE MUST BE NEGATIVE

hence potential energy = - 0.328J

8 0
3 years ago
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