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Pani-rosa [81]
3 years ago
6

The Earth orbits the Sun at a speed of 30 km/s. At that speed it completes one path around the Sun every year. Of course, as tha

t happens we are heading toward a different place in space constantly, always accelerating toward the Sun to stay on the circular path. Suppose at a paritcular instant we are headed in the direction of a distant star and we measure the speed of the light from that star as it arrives on Earth. In the vacuum of space we would measurea. 299,792,458 m/sb. 299,792,428 m/sc. 299,792,488 m/sd. 0 m/s
Physics
1 answer:
romanna [79]3 years ago
3 0

Answer:

a. 299,792,458 m/s

Explanation:

Since the speed of light in a vacuum is invariant and has the value of 299,792,458 m/s, we would measure this value of 299,792,458 m/s for the speed of light from the star as it arrives on Earth.

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How do I find the tension? its supposed to be in the mid 40s​
babunello [35]

Answer:

a = 2.3 m/s²

T = 45 N

Explanation:

Draw a free body diagram for each mass.

For the mass on the incline, there are four forces:

Weight force mg pulling down.

Normal force N perpendicular to the incline.

Friction force Nμ pushing down the incline.

Tension force T pulling up the incline.

For the hanging mass, there are two forces:

Weight force Mg pulling down.

Tension force T pulling up.

Sum of the forces on the hanging mass in the -y direction:

∑F = ma

Mg − T = Ma

T = Mg − Ma

Sum of the forces on the sliding mass in the perpendicular direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces on the sliding mass in the parallel direction:

∑F = ma

T − mg sin θ − Nμ = ma

Substitute:

Mg − Ma − mg sin θ − mgμ cos θ = ma

Mg − mg (sin θ + μ cos θ) = ma + Ma

Mg − mg (sin θ + μ cos θ) = (m + M) a

a = [ Mg − mg (sin θ + μ cos θ) ] / (m + M)

Plug in values:

a = [ 6.0×9.8 − 5.0×9.8 (sin 30° + 0.20 cos 30°) ] / (5.0 + 6.0)

a = 2.3 m/s²

Now find tension:

T = Mg − Ma

T = 6.0×9.8 − 6.0×2.3

T = 45 N

3 0
4 years ago
What is 52,427 in scientific notation?
Shkiper50 [21]

Answer: Why is 52,427 written as 5.2427 x 104 in scientific notation

PLEASE BRAINLIEST

7 0
4 years ago
A delivery truck travels with a constant velocity up an 8 slope. A 60 kg box sits on the floor of the truck and, because of sta
Blababa [14]

Explanation:

Given that,

The slope of the ramp, \theta=8^{\circ}

Mass of the box, m = 60 kg

(a) Distance covered by the truck up the slope, d = 300 m

Initially the truck moves with a constant velocity. We know that the net work done on the box is equal to 0 as per work energy theorem as :

W=\dfrac{1}{2}m(v^2-u^2)=0

u and v are the initial and the final velocity of the truck

(b) The work done on the box by the force of gravity is given by :

W=Fd\ cos\theta

Here, \theta=90+8=98^{\circ}

W=mgd\ cos\theta

W=60\times 9.8\times 300\ cos(98)

W = -24550.13 J

(c) What is the work done on the box by the normal force is equal to 0 as the angle between the force and the displacement is 90 degrees.

(d) The work done by friction is given by :

W_f=-W

W_f=24550.13\ J

Hence, this is the required solution.

6 0
3 years ago
Mike is watching a Disney movie with his son Jason Jason likes the movie so much that he becomes excited and stands in front of
shusha [124]
You could put it in two terms.

Rebellion- rebelling, or acting against, someone's orders or actions.

Disobedience- the opposite of obidience; not following the rules; breaking the law.
5 0
3 years ago
Humberto builds two circuits using identical components.Circuit 1: A series circuit with three lightbulbs Circuit 2: A parallel
PolarNik [594]
<span>Humberto builds two circuits using identical components,
and then adds components to each circuit.

Circuit 1:
A series circuit with three lightbulbs. 
Then add three more lightbulbs in series.

Circuit 2:
A parallel circuit with three lightbulbs
Then add two more lightbulbs on new branches
in parallel with each original bulb.

After adding the new lightbulbs in Circuit 1:
-- the voltage across each of the original bulbs is less,
-- the current through the whole series circuit is less,
-- the original three bulbs shine dimmer than before, and
-- the total power delivered from the battery is less.
-- The battery lasts longer.

After adding the new lightbulbs in Circuit 2:  
</span>-- the voltage across each of the original bulbs is doesn't change, 
-- the current through each original bulb doesn't change,
-- the original three bulbs shine just as bright as before, 
-- the total currrent drawn by the circuit, and the total current
delivered by the battery, increases, and
-- the total power delivered from the battery increases.
-- The battery runs down sooner.
5 0
3 years ago
Read 2 more answers
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