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krok68 [10]
3 years ago
14

What is the momentum of an 18-kg object moving at 0.5 m/s?

Physics
1 answer:
Otrada [13]3 years ago
6 0

sorry Idk the answer ..???

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A body moving at .500c with respect to an observer
lakkis [162]

Answer:

0.8c and -0.14c

Explanation:

The first fragment will have a speed of +0.5c respect of a frame of reference moving at +0.5c

Lest name v the velocity of the frame of reference, and u' the velocity of the object respect of this moving frame of reference.

The Lorentz transform for velocity is:

u = (u' + v) / (1 + (u' * v) / c^2)

u = (0.5c + 0.5c) / (1 + (0.5c * 0.5c) / c^2) = 0.8c

The other fragment has a velocity of u' = -0.6c respect of the moving frame of reference.

u = (-0.6v + 0.5c) / (1 + (0.5c * 0.5c) / c^2) = -0.14c

6 0
3 years ago
Identify the elements that come before iron (Fe) in the periodic table Fe, Co, Cu, K, Ni, Mn
harkovskaia [24]
FE is iron CO is cobalt CU is copper K is potassium NI is nickle MN is magnemese
5 0
3 years ago
Read 2 more answers
Two men are trying to carry a wooden pole. If one of them is weaker than other, how can they carry the pole hence making small l
Romashka-Z-Leto [24]

Answer: Answer down below.

Explanation:

6 0
2 years ago
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A cube has a drag coefficient of 0.8. what would be the terminal velocity of a sugar cube 1 cm on a side in air (ρ = 1.2 kg/m3)?
kramer
To answer this question, we should know the formula for the terminal velocity. The formula is written below:

v = √(2mg/ρAC)
where
m is the mass
g is 9.81 m/s²
ρ is density
A is area
C is the drag coefficient

Let's determine the mass, m, to be density*volume.
Volume = s³ = (1 cm*1 m/100 cm)³ = 10⁻⁶ m³
m = (1.6×10³ kg/m³)(10⁻⁶ m³) = 1.6×10⁻³ kg
A = (1 cm * 1 m/100 cm)² = 10⁻⁴ m²

v = √(2*1.6×10⁻³ kg*9.81 m/s²/1.6×10³ kg/m³*10⁻⁴ m²*0.8)
<em>v = 0.495 m/s</em>
7 0
3 years ago
Army is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s west
Aleks04 [339]

Explanation:

Given that,

Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.

Taking eastward as positive direction, we have:

v_B=+5\ m/sis the velocity of Bill with respect to Amy (which is stationary)

v_c=15\ m/s is the velocity of Carlos with respect to Amy.

Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is

v_B=+5\ m/s

Therefore, Carlos velocity in Bill's reference frame will be

v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s

So, the magnitude is 20 m/s and the direction is westward (negative sign).

7 0
3 years ago
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