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Debora [2.8K]
4 years ago
7

Kid A has jumped off the sled! (RIP)

Physics
1 answer:
balandron [24]4 years ago
5 0

Answer:

46.45 m/s

Explanation:

Total momentum before jump = Total momentum after jump

11.7 * ( 36.7 + 45.2 ) = ( 36.7 * (-31.1) ) + ( 45.2 * v )

v*45.2 - 1141.37 = 958.23

v = 2099.6/45.2 = 46.45

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Consider a planet with a surface temperature of 50 K an escape speed of 30 km/s. What type of planet must this be?
Ilya [14]

Answer:

option A. Jovian

Explanation:

If we consider the temperature of the planet which is 50 K, shows that the planet is quite distant from the central point of the solar system which itself shows one of the characteristics of Jovian planets.

If escape velocities of the planets are to be considered than for terrestrial planets like that of our Earth, the escape velocity must be similar to that of the Earth which is 11.2 Km/s, quite a smaller value as the gravitational pull of Earth is stronger than that of the Jovian planets with much higher values of escape velocities as the mentioned one here is 30 km/s which is again indicative of the planet being a Jovian planet.

4 0
3 years ago
Which of these processes are chemical reactions?
Rudiy27
I’m not very good at this subject but I’d say it’s A Boiling an egg
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4 years ago
¿Una persona de 60 corriendo a 4 m/s, tiene más energía cinética que un proyectil de 10 gramos a 300 m/s?
Svet_ta [14]

Explanation:

The question says that "Does a 60 kg person running at 4 m/s have more kinetic energy than a 10 gram projectile at 300 m/s ? "

Speed of a person is 4 m/s

Mass of a person is 60 kg

Kinetic energy of a person is : K=\dfrac{1}{2}mv^2

So,

K=\dfrac{1}{2}\times 60\times 4^2\\\\K=480\ J

Mass of a projectile is 10 grams or 0.01 kg

Speed of a projectile is 300 m/s

Kinetic energy of a projectile is :

K=\dfrac{1}{2}mv^2

K=\dfrac{1}{2}\times 0.01\times (300)^2\\\\K=450\ J

So, it is clear that the kinetic energy of a person is more than that of the kinetic energy of a projectile.

8 0
3 years ago
Explain your problem of<br>Federalism<br>​
telo118 [61]

That is more of a History of English question.

6 0
3 years ago
The voltage applied across a given parallel-plate capacitor is doubled. How is the energy stored in the capacitor affected?
ikadub [295]

Answer:

The energy stored in the capacitor quadruples its original value.

Explanation:

The energy stored in a capacitor is given by the equation

U=\frac{1}{2}CV^2

where

C is the capacitance

V is the voltage across the plates

The capacitance, C, depends only on the properties of the capacitor, so it does not change when the voltage applied is changed.

Instead, in this problem the voltage applied is doubled:

V' = 2V

So the new energy stored is

U'=\frac{1}{2}C(2V)^2=4(\frac{1}{2}CV^2)=4U

so, the energy stored has quadrupled.

8 0
3 years ago
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