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julsineya [31]
3 years ago
13

On which planet would you weigh less?

Physics
2 answers:
shtirl [24]3 years ago
7 0

Answer:

i think f witch is mars

Explanation:

....

Reil [10]3 years ago
4 0
Mars. its less massive than the earth causing it to pull you less
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The waves with the lowest energy and lowest frequencies of the electromagnetic spectrum are the
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5 0
3 years ago
A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.550 m/s. The to
Arisa [49]

Answer: 0.5 m/s

Explanation:

Given

Speed of the sled, v = 0.55 m/s

Total mass, m = 96.5 kg

Mass of the rock, m1 = 0.3 kg

Speed of the rock, v1 = 17.5 m/s

To solve this, we would use the law of conservation of momentum

Momentum before throwing the rock: m*V = 96.5 kg * 0.550 m/s = 53.08 Ns

When the man throws the rock forward

rock:

m1 = 0.300 kg

V1 = 17.5 m/s, in the same direction of the sled with the man

m2 = 96.5 kg - 0.300 kg = 96.2 kg

v2 = ?

Law of conservation of momentum states that the momentum is equal before and after the throw.

momentum before throw = momentum after throw

53.08 = 0.300 * 17.5 + 96.2 * v2

53.08 = 5.25 + 96.2 * v2

v2 = [53.08 - 5.25 ] / 96.2

v2 = 47.83 / 96.2

v2 = 0.497 ~= 0.50 m/s

3 0
3 years ago
What can you infer from the fact that although CFC's have been banned, refrigerators and aerosol cans are still being sold?
oksian1 [2.3K]
It would be "Researchers found environmental safe replacements for CFC's"
3 0
3 years ago
Read 2 more answers
In the parallel spring system, the springs are positioned so that the 37 N weight stretches each spring equally. The spring cons
Reptile [31]

Answer:

x = 5.29 m

Explanation:

given,

weight of stretch = 37 N

left-hand spring constant (k₁)= 2.7 N/cm

right hand spring constant(k₂)= 4.3 N/ cm

spring are connected in parallel

F = F₁ + F₂

F = k₁x + k₂x

F = (k₁+ k₂)x

37= (4.3+ 2.7)x

7 x = 37

x = 5.29 m

3 0
3 years ago
Objects with masses of 120 kg and a 420 kg are separated by 0.380 m. (a) Find the net gravitational force exerted by these objec
SOVA2 [1]

Answer:

F_{net} = 6.879\times 10^{- 7}\ N

Solution:

As per the question:

Mass of first object, m = 120 kg

Mass of second object, m' = 420 kg

Mass of the third object, M = 69.0 kg

Distance between the m and m', d = 0.380 m

Now,

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:

F = \frac{GMm}{\frac({d}{2}^{2})}

F = \frac{6.67\times 10^{-11}\times 120\times 0.69}{\frac({0.380}{2}^{2})} = 1.529\times 10^{- 7}\ N

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':

F = \frac{GMm}{\frac({d}{2}^{2})}

F' = \frac{6.67\times 10^{-11}\times 420\times 0.69}{\frac({0.380}{2}^{2})} = 5.35\times 10^{- 7}\ N

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':

F_{net} = F + F'

F_{net} = 1.529\times 10^{- 7} + 5.35\times 10^{- 7} = 6.879\times 10^{- 7}\ N

6 0
3 years ago
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