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kondor19780726 [428]

3 years ago

8

A satellite of mass 1000kg is in a circular orbit around a planet. The centripetal acceleration of the satellite in its orbit is

5m/s^2. What is the gravitational force exerted on the satellite by the planet?
A. 0N
B. 200N
C. 5000N
D. 10000N

1 answer:

ludmilkaskok [199]3 years ago

8 0

Answer: 5000N

Explanation:

The basic principle of a circular orbit is that Fg = m × ac, so as we have the mass and the centripetal acceleration (also called normal acceleration) we just have to operate. Fg = 1000kg × 5m/s² = 5000N

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Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti

Alex

Answer:

A) The speed of the water must be 8.30 m/s.

B) Total kinetic energy created by this maneuver is 70.12 Joules.

Explanation:

A) Mass of squid with water = 6.50 kg

Mass of water in squid cavuty = 1.55 kg

Mass of squid = $m_1=6.50 kg- 1.55 kg=4.95 kg$

Velocity achieved by squid = $v_1=2.60 m/s$

Momentum gained by squid = $P=m_1v_1$

Mass of water = $m_2=1.55 kg$

Velocity by which water was released by squid = $v_2$

Momentum gained by water but in opposite direction = $P'=m_2v_2$

P = P'

$m_1v_1=m_2v_2$

$v_2=\frac{m_1v_1}{m_2}=\frac{4.95 kg\times 2.60 m/s}{1.55 kg}=8.30 m/s$

B) Kinetic energy does the squid create by this maneuver:

Kinetic energy of squid = K.E = $\frac{1}{2}m_1v_1^{2}$

Kinetic energy of water = K.E' = $\frac{1}{2}m_2v_2^{2}$

Total kinetic energy created by this maneuver:

$K.E+K.E'=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}$

$=\frac{1}{2}\times 4.95 kg\times (2.60 m/s)^2+\frac{1}{2}\times 1.55 kg\times (8.30 m/s)^2=70.12 Joules$

4 0

3 years ago

What happens when a resultant electric field exist in conductor

klio [65]

Answer:

When an electric field exists in a conductor a current will flow.

This implies a voltage difference between two points on the conductor.

Electrostatics pertains to static charge distributions.

That means that an object such as a charged spherical conductor will be at the same potential (voltage) on both its outer and inner surfaces.

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4 years ago

Jack has two boxes. One is 148g and one is 78g. If jack pushes both boxes with the same amount of force, which will accelerate f

Step2247 [10]

The 78g box, since it has less weight, would accelerate faster. If you had a frictionless surface, and you conducted this experiment, both boxes, without any outside forces, would accelerate at the same rate forever. However, in this problem we must assume the surface is not frictionless. Friction is determined by weight; the more weight, the more friction. Since the 78g box has less weight, it has less friction, making it easier to push with less force.

8 0

4 years ago

Encontrar la distancia y desplazamiento de las dos trayectorias si se mueve el móvil Desde A hasta B

Lerok [7]

Answer:

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Explanation:

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3 years ago

A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv

cestrela7 [59]

<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":

$P=\sigma A T^{4}$ (1)

Where:

$P=300J/min=5J/s=5W$ is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing $1W=\frac{1Joule}{second}=1\frac{J}{s}$

$\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}}$ is the Stefan-Boltzmann's constant.

$A=5cm^{2}=0.0005m^{2}$ is the Surface area of the body

$T=727\°C=1000.15K$ is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close. So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

$P=\sigma A \epsilon T^{4}$ (2)

Where $\epsilon$ is the body's emissivity

(the value we want to find)

Isolating $\epsilon$ from (2):

$\epsilon=\frac{P}{\sigma A T^{4}}$ (3)

Solving:

$\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}}$ (4)

Finally:

$\epsilon=0.17$ (5) This is the body's emissivity

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3 years ago