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3 years ago

13

This planet gets its blue-green color because its atmosphere is full of methane gas and hydrogen.

2 answers:

Effectus [21]3 years ago

5 0

The correct is -

Uranus

Iteru [2.4K]3 years ago

4 0

The correct answer is Uranus

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Newton’s first law describes the tendency calles

beks73 [17]

The first law is about force or push and pull

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3 years ago

Read 2 more answers

Which process requires more energy: completely vaporizing 1 kg of saturated liquid water at 1 atm pressure or completely vaporiz

Lilit [14]

The correct answer would be the first option. The process that would need more energy would be vaporizing 1 kg of saturated liquid water at a pressure of 1 atmosphere. This can be seen from the latent heat of vaporization of each system. For the saturated water at 1 atm, the latent heat is equal to 40.7 kJ per mole while, at 8 atm, the latent heat is equal to 36.4 kJ per mole. The latent heat of vaporization is the amount of heat needed in order to vaporize a specific amount of substance without any change in the temperature. As we can observe, more energy is needed by the liquid water at 1 atm.

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3 years ago

A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the

olasank [31]

Explanation:

The given data is as follows.

Spring constant (k) = 78 N/m, $\theta = 30^{o}$

Mass of block (m) = 0.50 kg

According to the formula of energy conservation,

mgh sin $\theta = \frac{1}{2}kx^{2}$

h = $\frac{1}{2} \times \frac{kx^{2}}{mg Sin \theta}$

= $\frac{78 \times 0.04}{2 \times 0.5 \times 9.8 \times 0.5}$

= 0.64 m

Thus, we can conclude that the distance traveled by the block is 0.64 m.

4 0

2 years ago

20.0 m [N] - 15 m [S20degreesE]

denis-greek [22]

Answer:

thank for making me give up on life

Explanation:

I thought the stuff I had was hard wth is even that

3 0

3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago