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kobusy [5.1K]
3 years ago
13

A flowerpot that has a mass of 1.5 kg is sitting on a windowsill 30 meters from the ground. Is the energy of the flowerpot poten

tial of kinetic?
How much energy does the flowerpot have?

​
Physics
1 answer:
zzz [600]3 years ago
8 0

Answer:

It is Potential and corresponding to

90

J

.

Explanation:

Potential Energy is energy stored in the system and ready (potentially) to be changed into, say, Kinetic Energy through movement.

The general expression for Gravitational Potential Energy is:

U=mgh

where:

mg is the weight; h is the height relative to a certain reference level (the ground in this case) choosen to have zero Potential Energy.

With your data we get:

U=3⋅30=90J

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Satellite A has an orbital radius 3.00 times greater than that of satellite B. Satellite B's orbital period around Earth is 120
igomit [66]

Answer:

To find the circumference (orbit) of an object, you use Pi x Diameter. 

As you have the circumference of B, you divide it by Pi to get the Diameter. 

So 120 divided by 3.141592654 = 38.2 minutes for the Diameter. 

As' radius and Diameter will be 3x greater than B. 

38.2 x 3 = 114.6 

To get back to the orbital period, times 114.6 by Pi, and you will get 360 minutes

HOPE THIS HELPS AND PLS MARK AS BRAINLIEST

THNXX :)

7 0
3 years ago
Sophia is planning on going down an 8-m high water slide. Her weight is 50 N. What is Sophia's Gravitational Potential Energy at
pogonyaev
Gpe = 50 x 8 = 400 Joules
6 0
3 years ago
A flat coil having 40 turns, each one of cross-sectional area 12.0 cm^2, is oriented with its plane perpendicular to a uniform m
iragen [17]

what means emf ? what it means

4 0
3 years ago
On sunday a strong gale blows near beth’s house. the best estimate for the wind speed is ____ miles/hour.
Rufina [12.5K]

Answer:

B. 47-54 miles/hour

Explanation:

Gale is a strong wind which is depicted by red warning flag. According to U.S. National Weather Service gale is a sustained surface wind. It is also used to refer winds from tropical coastal areas.

On the basis of force of wind gale is divided in four groups:

  1. Near gale   - 32-38 mph
  2. Gale            - 39-46 mph
  3. Strong gale - 47-54 mph
  4. Storm           - 55-63 mph
8 0
4 years ago
A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate
exis [7]

Answer:

The velocity of water at the bottom, v_{b} = 28.63 m/s

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, P_{gauge} = 2.90 atm

Solution:

Now,

Atmospheric pressue, P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa

At the top, the absolute pressure, P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa

Now, the pressure at the bottom will be equal to the atmopheric pressure, P_{b} = 1 atm = 1.01\times 10^{5} Pa

The velocity at the top, v_{top} = 0 m/s, l;et the bottom velocity, be v_{b}.

Now, by Bernoulli's eqn:

P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}

where

h_{t} -  h_{b} = 12.8 m

Density of sea water, \rho = 1030 kg/m^{3}

\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} =  v_{b}

\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}

v_{b} = 28.63 m/s

5 0
4 years ago
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