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Nataliya [291]
3 years ago
9

Because not all stars are the same distance from earth, it can be difficult to determine?

Physics
2 answers:
Inessa [10]3 years ago
8 0
I think answer is A. The Brightness of a Star
ehidna [41]3 years ago
8 0
The answer is b because some star are really bright which make us think that they are close but they are probably really far
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Spaceship 1 and Spaceship 2 have equal masses of 300kg. They collide. Spaceship 1's final speed is 3 m/s, and Spaceship 2's fina
fiasKO [112]

Answer:

B. 1500 kg*m/s

Explanation:

Momentum p = m* v

In any type of collision, the total momentum is preserved!

The total momentum before and the total momentum after the collision is the same. We know the mass and speed after the collision so we can calculate the total momentum.

p1 + p2 =

m1*v1 + m2*v2

m1 = me = 300 kg

v1 = 3 m/s

v2 = 2 m/s

Substitute the given numbers:

300*3 + 300+2

900 + 600

1500 kg*m/s, which is answer B.

3 0
3 years ago
8. Il An 8.00 kg package in a mail-sorting room slides 2.00 m down a
Vitek1552 [10]

Answer:

See below

Explanation:

Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N

  no work is done by this force

Force friction = coeff friction * force normal = .4 * 47.1823 = 7.55 N

   work of friction = 7.55 * 2 m = 15.1 j

Force Downplane = mg sin 53 = 62.61  N

    work = 62.61 * 2 = 125.22 j

Net Force downplane =   force downplane - force friction = 55.06 N

net Work = force * distance = 55.06 N * 2 M = 110.12 j

3 0
2 years ago
A bird lands on a bare copper wire carrying a current of 51
nirvana33 [79]
The resistance of the piece of wire is
R= \frac{\rho L}{A}
where
\rho = 1.68 \cdot 10^{-8}\Omega m is the resistivity of the copper
L=5.1 cm=0.051 m is the length of the piece of wire
A=0.13 cm^2 = 0.13 \cdot 10^{-4} m^2 is the cross sectional area of the wire
By substituting these values, we find the value of R:
R= \frac{\rho L}{A}=6.6 \cdot 10^{-5} \Omega

Then, by using Ohm's law, we find the potential difference between the two points of the wire:
V=IR=(51 A)(6.6 \cdot 10^{-5} \Omega )=3.4 \cdot 10^{-3} V
7 0
3 years ago
A model plane has a mass of 0.75 kg and is flying 12 m above the ground
Grace [21]

Answer:

Option C. 210 J.

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 0.75 Kg

Height (h) = 12 m

Velocity (v) = 18 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Total Mechanical energy (ME) =?

Next, we shall determine the potential energy of the plane. This can be obtained as follow:

Mass (m) = 0.75 Kg

Height (h) = 12 m

Acceleration due to gravity (g) = 9.8 m/s²

Potential energy (PE) =?

PE = mgh

PE = 0.75 × 9.8 × 12

PE = 88.2 J

Next, we shall determine the kinetic energy of the plane. This can be obtained as follow:

Mass (m) = 0.75 Kg

Velocity (v) = 18 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.75 × 18²

KE = ½ × 0.75 × 324

KE = 121.5 J

Finally, we shall determine the total mechanical energy of the plane. This can be obtained as follow:

Potential energy (PE) = 88.2 J

Kinetic energy (KE) = 121.5 J

Total Mechanical energy (ME) =?

ME = PE + KE

ME = 88.2 + 121.5

ME = 209.7 J

ME ≈ 210 J

Therefore, the total mechanical energy of the plane is 210 J.

8 0
2 years ago
The temperature of 1 m^3 of water is decreased by 10°C. If this thermal energy is used to lift the water vertically against grav
Rama09 [41]

Answer:

h = 4271.43 m

Explanation:

given,

Volume of the water = 1 m³

temperature decrease by = 10°C

heat removed from water

Q = m c ΔT                            

Q = ρ V c ΔT                            

   = 1000 × 1 × 4186 × 10

   = 4.186 × 10⁷ J

energy is used to do work to move the water against its weight

Q = force  × displacement

4.186 × 10⁷ J =  m g × h                    

4.186 × 10⁷ J =  1000 × 1 × 9.8 × h                

h = 4271.43 m                                

hence, the change in height of is equal to h = 4271.43 m

8 0
3 years ago
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