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Mademuasel [1]
3 years ago
8

Consider a system of a cliff diver and the earth. the gravitational potential energy of the system decreases by 28,000 j as the

diver drops to the water from a height of 32.0 m. determine her weight in newtons.
Physics
1 answer:
olga2289 [7]3 years ago
5 0

The decrease in gravitational potential energy of the system is given by

\Delta U = (mg) \Delta h

where

m is the mass, g is the gravitational acceleration, and \Delta h is the variation of height of the system.


(mg) also corresponds to the weight of the diver, therefore if we rearrange the equation and we use \Delta U=28000 J and \Delta h=32.0 m, we can find her weight:

mg=\frac{\Delta U}{\Delta h}=\frac{28000 J}{32 m}=875 N

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What mass of a material with density Ï is required to make a hollow spherical shell having inner radius r1 and outer radius r2?
inysia [295]
<span>Yes, there are! r1 and r2 are numbers. The volume of the hollow shell is 4 π 3 ( r 3 1 − r 3 2 ) 4π3(r13−r23). Now multiply by ρ to get the mass.</span>
3 0
3 years ago
If you wanted to
VLD [36.1K]

Answer:

Option C

Explanation:

According to the formula

  • \\ \boxed{\sf R=\rho\dfrac{\ell}{A}}

So

If we use wide wire we increase the area of cross section so resistance decreases

5 0
2 years ago
A spool whose inner core has a radius of 1.00 cm and whose end caps have a radius of 1.50 cm has a string tightly wound around t
White raven [17]

Answer:

v₁ = 37.5 cm / s

Explanation:

For this exercise we can use that angular and linear velocity are related

        v = w r

in the case of the spool the angular velocity for the whole system is constant,

They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,

         w = v₀ /r₀

for the outside of the spool r₁ = 1.5 cm

         w = v₁ / r₁1

since the angular velocity is the same we set the two expressions equal

        \frac{v_o}{r_o} = \frac{v_1}{r_1}

        v1 = \frac{r_1}{r_o} \ \ v_o

let's calculate

       v₁ = \frac{1.50}{1.00} \ \ 25.0

       v₁ = 37.5 cm / s

4 0
3 years ago
Please help me solve all of them ( a, b, c and d ) thankiew !! <br> I’m also kind of in a rush
Sergio039 [100]

Answer:

a-

V= IR

9V = I ×( 12+6)

I = 9/ 18 A = 0.5 A

b

V=IR

240 = 6 A ×( 20 + R)

40 = 20 + R

R = 20 ohm

c

resultant resistance of the 2 parallel resistances= Ro

1/Ro = 1/ 5 + 1/ 20

1/Ro =( 20+5)/100

= 1/Ro = 1/4

Ro= 4 ohm

V=IR

V = 2A × ( 1+ 4 OHM)

V = 10V

d

equivalent resistance = Ro

1/Ro = 1/(2+8) + 1/(5+5)

1/Ro = 1/10 +1/10

2/10 = 1/ Ro

Ro= 10/2 = 5 ohm

V = IR

12V = I × 5Ohm

I=2.4 A

6 0
2 years ago
Why were the rings of Uranus not observed directly from telescopes on the ground on Earth? How were they discovered?
leonid [27]

Answer:Explained below.

Explanation:

Uranus rings is made up of jet black, coal-like particles in small bands, making them difficult  to perceive from Earth.This indicates that they are probably composed of a mixture of the ice and a dark material. The nature of  material is dismal, but it might be some organic compounds greatly darkened by the charged particle irradiation from the Uranian magnetosphere. Rings were discovered by using a infrared telescope throughout  the occultation of a star as Uranus passed in front of it. The light from the star dimmed many times before it was obstructed by the disk of Uranus and subsequently, showing the presence of various distinct rings.

6 0
3 years ago
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