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3 years ago

Compared to the amount of heat jupiter receives from the sun, jupiter radiates _____ heat.

Physics

1 answer:

uysha [10]3 years ago

3 0

C
http://education.seattlepi.com/planet-gives-off-2-times-much-heat-receives-sun-5420.html

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A 1600 kg helicopter has 250,000 J of gravitational energy. What is its

aleksandrvk [35]

Answer:

15.9 m

Explanation:

Use the equation for gravitational potential energy (Ug):

Ug= mass*g*height

250,000 J= 1600 kg*9.8 m/s^2*height

height= 15.9 m

Hope this helps!!! :)

6 0

3 years ago

Dana has a sports medal suspended by a long ribbon from her rearview mirror. as she accelerates onto the highway, she notices th

Dominik [7]

Refer to the diagram shown below.

m = the mass of the medal
T = the tension in the string
g = 9.8 m/s²

Part (a)
The inertial force, F, tends to move the medal away from the windshield when the vehicle accelerates to the right.

Answer: The medal leans away from the windshield.

Part (b)
For force balance,
T cos(15°) = mg
0.9659 T = 9.8m
T = 10.1457m N

Also,
F = Tsin(15°) = 10.1457m*0.2588 = 2.6257m N
The inertial force is equal to the accelerating force, therefore
the acceleration, a, is given by
(2.6257m N) = (m kg)*(a m/s²)
a = 2.6257 m/s²

Answer: 2.626 m/s²

5 0

4 years ago

Use the work-energy theorem to determine the force required to stop a 1000 kg car moving at a speed of 20.0 m/s if there is a di

Vlad1618 [11]

Answer:

4.44 kN in the opposite direction of acceleration.

Explanation:

Given that, the initial speed of the car is, $u=20m/s$

And the mass of the car is, $m=1000 kg$

The total distance covered by the car before stop, $s=45m$

And the final speed of the car is, $u=0m/s$

Now initial kinetic energy is,

$KE_{i}=\frac{1}{2}mu^{2}$

Substitute the value of u and m in the above equation, we get

$KE_{i}=\frac{1}{2}(1000kg)\times (20)^{2}\\KE_{i}=20000J$

Now final kinetic energy is,

$KE_{f}=\frac{1}{2}mv^{2}$

Substitute the value of v and m in the above equation, we get

$KE_{f}=\frac{1}{2}(1000kg)\times (0)^{2}\\KE_{i}=0J$

Now applying work energy theorem.

Work done= change in kinetic energy

Therefore,

$F.S=KE_{f}-KE_{i}\\F\times 45=(0-200000)J\\F=\frac{-200000J}{45}\\ F=-4444.44N\\F=-4.44kN$

Here, the force is negative because the force and acceleration in the opposite direction.

6 0

3 years ago

There are initially No nuclei in a radioactive source. After a time interval of three half-lives, what is the number of nuclei t

ladessa [460]

The number of nuclei that have decayed after three half-lives is 12.5% of the original mass.

<h3>What is half live of radioactive isotope?</h3>

The half life of a radioactive isotope is the time taken for the element to decay to half of its original mass.

<h3>Number of nuclei decayed after 3 half lives</h3>

N = N₀/(2³)

N = N₀/8

N = 0.125N₀

N = 12.5% of N₀

where;

N is n number of nuclei that have decayed after 3 half lives
N₀ is the original mass

Thus, the number of nuclei that have decayed after three half-lives is 12.5% of the original mass.

Learn more about half life here: brainly.com/question/10525285

#SPJ1

6 0

2 years ago

Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150

Phoenix [80]

Answer:

The temperature of the core raises by $2.8^{o}C$ every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius

Thus reactor core whose mass is $1.60\times 10^{5}kg$ will require

$0.3349\times 1.60\times 10^{5}kJ\\\\=0.53584\times 10^{5}kJ$

energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

$\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s$

5 0

3 years ago