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egoroff_w [7]
3 years ago
8

If earth had no atmosphere , would a falling object ever reach terminal velocity?

Physics
1 answer:
Bas_tet [7]3 years ago
6 0
The right answer is no.
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The acceleration due to gravity on Jupiter is 2.5 times what it is here on earth. An object weighing 347.9 N here on earth will
inessss [21]

Answer:  weight on Jupiter = 869.75 N

              mass  on Earth = mass on Jupiter = 35.5 Kg

Explanation:

W = mg

W = weight

m = mass

g = gravitational acceleration [ on the Earth, g₁ = 9,8 N/kg ]

On the Earth,

G₁ = m x g₁  = 347,9 N

On the Jupiter,

G₂ = mg₂  

mass on the Earth = mass on the Jupiter !  

m = G₁ : g = 347.9 N : 9,8 N/kg = 35.5 kg

G2 : G1 = 2.5

G₂ = 2,5 G₁ = 2,5 x 347.9 N =  869,75 N

5 0
3 years ago
Diferencie energia cinética de energia potencial gravitacional.
Vikentia [17]

Answer:

Featured snippet from the web

The atoms and molecules in it are in constant motion. The kinetic energy of such a body is the measure of its temperature. Potential energy is classified depending on the applicable restoring force. Gravitational potential energy – potential energy of an object which is associated with gravitational force

4 0
3 years ago
A bus starts to move from rest. If its velocity becomes 90 km/hr after 8 seconds, calculate its acceleration.
a_sh-v [17]

Answer:

The acceleration is 11.25 km/hr

Explanation:

Divide the velocity over time to find the acceleration. The bus is accelerating 11.25 km/hr per second in the range of 8 seconds

7 0
3 years ago
A 167-? resistor and a 114-? resistor are both connected in parallel across a power supply. If the potential drop across the 167
Troyanec [42]
19-? Is the exact p.d across the 114-?resistor.
Current will different
But p.d will same in parallel circuit .
7 0
3 years ago
An arrow is shot at an angle of 35° and a velocity of 50 m/s. How long does it take to return to its original starting height?
Ostrovityanka [42]

Answer:

4.02 s

Explanation:

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

Angle of projection (θ) = 35°

Initial velocity (u) = 50 m/s

Acceleration due to gravity (g) = 10 m/s²

Time of flight (T) =?

The time of flight of the arrow can be obtained as follow:

T = 2uSineθ / g

T = 2 × 35 × Sine 35 / 10

T = 70 × 0.5736 / 10

T = 7 × 0.5736

T = 4.02 s

Therefore, the time taken for the arrow to return is 4.02 s

4 0
3 years ago
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