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damaskus [11]
3 years ago
13

Which statement best describes what happens during a solar eclipse?

Physics
2 answers:
Bond [772]3 years ago
5 0

Answer:

C

Explanation:

the moon is in between the sun and the earth when a solar eclipse is happening

Varvara68 [4.7K]3 years ago
4 0

Answer: A.

Explanation:

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Does the coefficient of kinetic friction depend on the weight of the block
ivolga24 [154]
No it does not because this weight is being balanced by a normal source
8 0
4 years ago
Identify the forces acting on the object of interest. From the list below, select the forces that act on the piano.
bekas [8.4K]

The correct answer is the following.

The forces that act on the piano are: 2) gravitational force acting on the piano (piano's weight). 5) force of the floor on the piano (normal force). 7) force of Chadwick on the piano.

As we see in the picture that I have attached is Chadwick pushing the piano in a horizontal plane. So Chadwick is applying a force that produces an acceleration. It is his force on the piano plus the acceleration of the weight of the piano, it's a gravitational force. This is pure physics applied to an object.

5 0
3 years ago
A bin is given a push across a horizontal surface. The bin has a mass m, the push gives it an initial speed of 1.60 m/s, and the
emmasim [6.3K]

Answer:

The bin moves 0.87 m before it stops.

Explanation:

If we analyze the situation and apply the law of conservation of energy to this case, we get:

Energy Dissipated through Friction = Change in Kinetic Energy of Bin (Loss)

F d = (0.5)(m)(Vi² - Vf²)

where,

F = Frictional Force = μR    

but, R = Normal Reaction = Weight of Bin = mg

Therefore, F = μmg

Hence, the equation becomes:

μmg d = (0.5)(m)(Vi² - Vf²)

μg d = (0.5)(Vi² - Vf²)

d = (0.5)(Vi² - Vf²)/μg

where,

Vf = Final Velocity = 0 m/s (Since, bin finally stops)

Vi = Initial Velocity = 1.6 m/s

μ = coefficient of kinetic friction = 0.15

g = 9.8 m/s²

d = distance moved by bin before coming to stop = ?

Therefore,

d = (0.5)[(1.6 m/s)² - (0 m/s)²]/(0.15)(9.8 m/s²)

<u>d = 0.87 m</u>

5 0
3 years ago
How does light energy work
OLEGan [10]

Answer:

Light consists of photons, which are produced when an object's atoms heat up. Light travels in waves and is the only form of energy visible to the human eye.

Explanation:

6 0
3 years ago
Read 2 more answers
Hockey puck B rests on a smooth ice surface and is struck by a second puck A, which has the same mass. Puck A is initially trave
Marina86 [1]

Answer:

a.v_{b2}=6.8544 \ m/s\\\\b. v_{a2}=12.891 \ m/s\\\\c. \theta _b=62\textdegree

Explanation:

Puck A's initial speed is v_a_1=14.6m/s and move in a direction \theta_b=28.0\textdegree after the collision.

#P_1=P_2 since there's no external force on the system(P=mv).

#The collision equation can be written as;

m_av_a_1+m_bv_b_1=m_av_a_2+m_bv_b_2

The kinetic energies before and after the collision are expressed as:

K_a_1+K_{b1}=K_a_2+K_{b2}, \ K=0.5mv^2

0.5m_av_a_1+0.5m_b(0)=0.5m_av_a_2+0.5m_bv_b_2\\\\14.6^2=v_{a2}^2+v_{b2}^2\\\\v_{b2}^2=213.16-v_{a2}^2

Let +x along A's initial direction and +y along A's final direction makes the angle 62\textdegree

\dot v_{a1}=14.6i\\\\\dot v_{a2}=(v_{a2} \ cos \ 28\textdegree)i+(v_a_2\ sin \ 28\textdegree)j\\\\\dot v_{b2}=v_{b2x}i+v_{b2y}j

#Substitute in v_{b2}^2=213.16-v_{a2}^2:

\dot v_{b2}=(14.6i)-\dot v{a2}\\\\\dot v_{b2}.\dot v_{b2}=v_{b2}^2\\\\\#Right \ side\\\\(14.6i-\dot v{a2}).(14.6i-\dot v{a2})=(14.6i)^2+\dot v_{a2}^2.\dot v_{a2}^2-2(14.6i).\dot v_{a2}\\\\=213.16+v_{a2}^2-2(14.6i).(v_{a2}\ cos 28\  \textdegree i+v_{a2} \ sin \ 28\textdegree j)\\\\v_{b2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\ \  v_{b2}^2=213.16-v_{a2}^2\\\\213.16-v_{a2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\\\\2v_a2}^2=29.2\ cos\ 28 \textdegree v_{a2}\\\\

v_{a2}=12.891\ m/s

Hence, the the speed of puck A after the collision is 12.891  m/s

#. The velocity of A after the collision is;

\dot v_{a2}=12.891 \ cos \ 28 \textdegree i+12.891 \ sin \ 28\textdegree j\\\\=11.382i+6.052j

Substitute \dot v_{a2} into \dot v_{b2}=(14.6i)-\dot v{a2}:

\dot v_{b2}=14.6i-(11.382i+6.052j)\\\\=3.218i-6.052j

This is the velocity of puck B after the collision, it's speed is:

v_{b2}=\sqrt{v_{b2x}^2+v_{b2y}^2}\\\\=\sqrt{3.218^2+(-6.052)^2}\\\\v_{b2}=6.8544 \ m/s

The velocity of puck B after the collision is 6.8544 m/s

c. The direction of puck B after the collision is:

\theta _b=tan^{-1}\frac{v_{b2y}}{v_{b2x}}\\\\=tan^{-1} \frac{-6.052}{3.218}\\\\\approx 62\textdegree

Hence, the direction of B's velocity after the collision is 62°

4 0
4 years ago
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