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4 years ago

What object in the universe that reflect light from stars

Physics

2 answers:

alexdok [17]4 years ago

7 0

The moon reflects light from stars and then the light is moon light.

SSSSS [86.1K]4 years ago

3 0

Every object that is not a star is visible only when a star's light shines
on it. Then, some of the star's light may be reflected from the object,
to our eyes.

Objects that do not generate light, and are visible only when illuminated
by a star's light, include planets, moons, comets, asteroids, meteoroids,
and artificial satellites without batteries and headlights.

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A skateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angle of 58 above the horizontal. The end of the ram

sweet-ann [11.9K]

Answer:

a) Maximum height reached above ground = 2.8 m

b) When he reaches maximum height he is 2 m far from end of the ramp.

Explanation:

a) We have equation of motion v²=u²+2as

Considering vertical motion of skateboarder.

When he reaches maximum height,

u = 6.6sin58 = 5.6 m/s

a = -9.81 m/s²

v = 0 m/s

Substituting

0²=5.6² + 2 x -9.81 x s

s = 1.60 m

Height above ground = 1.2 + 1.6 = 2.8 m

b) We have equation of motion v= u+at

Considering vertical motion of skateboarder.

When he reaches maximum height,

u = 6.6sin58 = 5.6 m/s

a = -9.81 m/s²

v = 0 m/s

Substituting

0= 5.6 - 9.81 x t

t = 0.57s

Now considering horizontal motion of skateboarder.

We have equation of motion s =ut + 0.5 at²

u = 6.6cos58 = 3.50 m/s

a = 0 m/s²

t = 0.57

Substituting

s =3.5 x 0.57 + 0.5 x 0 x 0.57²

s = 2 m

When he reaches maximum height he is 2 m far from end of the ramp.

8 0

4 years ago

The weight of a luggage is 69.3 N on the moon. Find its weight on the Earth.

kiruha [24]

Answer:

415.8N

Explanation:

Given:

weight of a luggage = 69.3 N

But

✓(weight on the Earth) is equivalent to 6 times weight if the moon.

✓Then, weight on the Earth) =( 69.3N × 6)

✓weight on the Earth) = 415.8N

7 0

3 years ago

1. A SUV along with 5 passengers has a mass of 3500 kg. It has a driving force of 2500 N directed along west on a perfectly hori

Dovator [93]

Answer:

The net acceleration of the SUV is 0.429 meters per square second due west.

Explanation:

Statement is incomplete. Description is presented below:

<em>A SUV along with 5 passengers has a mass of 3500 kg. It has a driving force of 2500 N directed along west on a perfectly horizontal road. The surface of the road exerts a resistance force of 500 N due east. At the same time a high wind is blowing a force of 500 N due east in the opposite direction of the car's drive force. Does the car has any acceleration? If yes, then what are the magnitude and direction of the car's acceleration?</em>

According to Newton's Laws of Motion, the SUV will accelerate if and only if net acceleration is different of zero. Let suppose as positive the direction of driving force ( $F$ ), measured in newtons:

$\Sigma F = F - R -f = F_{net}$ (1)

Where:

$R$ - Resistance force, measured in newtons.

$f$ - Wind force, measured in newtons.

$F_{net}$ - SUV net force, measured in newtons.

If we know that $F = 2500\,N$ , $R = 500\,N$ and $f = 500\,N$ , then net force experimented by the SUV is:

$F_{net} = 2500\,N-500\,N-500\,N$

$F_{net} = 1500\,N$

The car has acceleration.

By definition of force for systems with constant mass, we calculate the acceleration of the vehicle below:

$a_{net} = \frac{F_{net}}{m}$ (2)

Where $m$ is the mass of the SUV, measured in kilograms.

If we know that $F_{net} = 1500\,N$ and $m = 3500\,kg$ , then the net acceleration of the car is:

$a_{net} = \frac{1500\,N}{3500\,kg}$

$a_{net} = 0.429\,\frac{m}{s^{2}}$

The net acceleration of the SUV is 0.429 meters per square second due west.

8 0

3 years ago

Points A and B lie above an infinite plane of negative charge that produces a uniform electric field of strength E=10 N/C. What

olasank [31]

There is no illustration of the problem provided but I'll attempt to provide an answer.

The relationship between the electric potential difference between two points and the average strength of the electric field between those two points is given by:

║E║ = ΔV/d

║E║ is the magnitude of the average electric field, ΔV is the potential difference between A and B, and d is the distance between A and B.

We are given the following values:

║E║= 10N/C

d = 3m

Plug these values in and solve for ΔV

10 = ΔV/3

ΔV = 30V

4 0

4 years ago

A nail driven into a board increases in temperature.

denis23 [38]

Answer:

ΔT = 40.91 °C

Explanation:

First we find the kinetic energy of one hit to the nail:

K.E = (1/2)mv²

where,

K.E = Kinetic energy = ?

m = mass of hammer = 1.6 kg

v = speed of hammer = 7.7 m/s

Therefore,

K.E = (1/2)(1.6 kg)(7.7 m/s)²

K.E = 47.432 J

Now, for 10 hits:

K.E = (10)(47.432 J)

K.E = 474.32 J

Now, we calculate the heat energy transferred (Q) to the nail. As, it is the 59% of K.E. Therefore,

Q = (0.59)K.E

Q = (0.59)(474.32 J)

Q = 279.84 J

The change in energy of nail is given as:

Q = mCΔT

where,

m = mass of nail = 7.6 g = 0.0076 kg

C = specific heat capacity of aluminum = 900 J/kg.°C

ΔT = Increase in temperature = ?

Therefore,

279.84 J = (0.0076 kg)(900 J/kg.°C)ΔT

ΔT = (279.84 J)/(6.84 J/°C)

5 0

3 years ago