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worty [1.4K]
3 years ago
13

A rocket has a mass of 156,789 kg and is traveling at 45.6 m/s. How much kinetic energy does the rocket

Physics
1 answer:
Flura [38]3 years ago
7 0

Explanation:

K.E =1/2 mv^2

=1/2(156789)(45.6)^2

=78,394.5 × 2,079.36

=163,010,387.52 kg m/s

This should be your answer.

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The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -50 °C and 900 Pa, respect
Sidana [21]

Answer:

T = 273 + (-50) = 273 – 50 = 223 K

R = 188.82 J / kg K for CO2

Density (Martian Atmosphere) = P / RT = 900 / 188.92 x 223 = 900 / 42129.16 = 0.0213 kg / m^{3}

T = 273 +18 = 291 K, R = 287 J / kg k (for air) P = 101.6 k Pa = 101600 Pa

Density (Earth Atmosphere) = P / RT = 101600 / 287 x 291 = 1.216 kg / m^{3}

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4 years ago
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How can you degauss a magnet temporarily?
EleoNora [17]

Answer:

Demagnetization processes include heating past the Curie point, applying a strong magnetic field, applying alternating current, or hammering the metal.

Explanation:

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3 years ago
a parent swings a 18.5 kg child in a circle of radius 1.05m, making 5 revolutions in 13.4s. what centripetal acceleration does t
Zolol [24]

Answer: 0.146 m/s^{2}

Explanation:

The <u>centripetal acceleration</u> a_{c} of an object moving in a uniform circular path is given by the following equation:

a_{c}=\frac{V^{2}}{r}  (1)

Where:

V is the tangential velocity

r=1.05 m is the radius of the circle

On the other hand, the tangential velocity  is expressed as:

V=\omega r (2)

Where \omega is the angular velocity, which can be found knowing the child makes 5 revolutions in 13.4s:

\omega=\frac{5 rev}{13.4 s}=0.37 rev/s (3)

Substituting (3) in (2):

V=(0.37 rev/s)(1.05 m) (4)

V=0.39 m/s (5)

Substituting (5) in (1):

a_{c}=\frac{(0.39 m/s)^{2}}{1.05 m}  (6)

Finally:

a_{c}=0.146 m/s^{2}  

6 0
3 years ago
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A charge q of 1.3 × 10-16 coulombs moves from point A to a lower potential at point B in an electric field of 3.2 × 102 newtons/
LUCKY_DIMON [66]

Explanation :

It is given that,

Charge, q=1.3\times 10^{-16}\ C

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Distance, d=1.1\times 10^{-2}\ m

The work done is stored in the form of potential energy.

W=F.d

\because F=qE

So, W=qE\ d

W=1.3\times 10^{-16}\ C\times 3.2\times 10^2\ N/C\times 1.1\times 10^{-2}\ m

W=4.576\times 10^{-16}\ J

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a moving ball rolls into a stationary ball. the total momentum of both balls after the collision will be
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The momentum of the rolling ball will have less momentum than before the collision and the stationary ball will have more momentum after the collusion.
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