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2 years ago

Compare the gravitational force on a 4 kg

Physics

1 answer:

Eddi Din [679]2 years ago

6 0

Answer:

The gravitational force by the Earth on the object, and by the object on the Earth is

F = GMm/R²

= 6.674×10−11 m3⋅kg−1⋅s−2 × 6 × 10^24 kg × 44.5 kg/(6.4 × 10^6 m)²

= 435 N

<u>Please note that the ration between the gravitation force 435 and the mass 44.5</u>

should be gravitational acceleration

435/44.5 = 9.78

I attribute the discrepancy between 9.78 and the usual 9.81 to rounding off in the

Earth's weight and radius.

The mass of the Moon is M / 81.3.

The radius of the Moon is R × 0.27.

The gravitational force on the moon would be

G(M/81.3)m/(R×0.27)² = 0.17×GMm/R²

The gravitational force on the moon is smaller by the factor of about 0.17.

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A block of aluminum at a temperature of T1 = 32 degrees C has a mass of m1 = 12 kg. It is brought into contact with another bloc

Mariulka [41]

Answer: $T=21.33 ^{\circ}$

Explanation:

Given

mass of First Block $m_1=12 kg$

Temperature $T_1=32^{\circ}$

mass of second block $m_2=0.5 m_1=6 kg$

Temperature $T_2=9^{\circ}$

Heat capacity of aluminium c=899 J/kg-K

Final Temperature acquired by both blocks at steady state

Heat loss first block =Heat gain by second block

$m_1\times c\times (32-T)=m_2\times c\times (T-9)$

$12\times 899\times (32-T)=6\times 899\times (T-9)$

$2\times 32=3T$

$T=\frac{64}{3}=21.33 ^{\circ}$

5 0

3 years ago

A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,

egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

6 0

3 years ago

A 8.34 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is

Mnenie [13.5K]

Answer:

21870.3156 N

Explanation:

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 1.6 m/s²