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nexus9112 [7]
3 years ago
5

50 POINTS!!! PLEASE!!!!!! Pls help me with this been stuck on it in like FOREVER!! PLSS.

Physics
2 answers:
Alenkinab [10]3 years ago
5 0

Answer:

it will move in 356.5g in diagonal direction(resultant or between the both direction)

Explanation:

285+428/2=356.5g

Hope u like it

PLZ MARK ME AS BRAINLIEST

Akimi4 [234]3 years ago
3 0
Backward. The most force is the 428 g block.
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A ball with a volume of 3 cm3 and a density of 9.0 g/cm3 increases its velocity from 2 m/s to 6 m/s over a 12
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Answer:

M = ρ V = 9 gm/cm^ 3 * cm^3 = 27 gm

a = (V2 - V1) / t = (6 - 2) m/s / 12 s = 1/3 m/s^2     the acceleration

F = M a = 27 gm * 1/3 m/s^2 = 9 dynes      net force applied

4 0
2 years ago
A balloon and a mass are attached to a rod that is pivoted at P.
Brrunno [24]

Answer:

The correct option is;

B Move both the balloon and mass 10 cm to the right

Explanation:

Given that the system is in equilibrium, we have;

Force of balloon = F_b↑

Force of mass = F_m ↓

The direction of the balloon is having an upward motion which gives a clockwise moment or motion to the rod while the direction of the force of the mass weight is downwards, giving the rod an anticlockwise moment

for the rod to rotate clockwise, the moment of the balloon should be larger than that of the rod

At the present equilibrium we have;

F_b × 30 =  F_m × 20

Therefore;

F_m = 1.5×F_b

Moving both balloon and mass 10 cm to the right gives;

The moment of the balloon = F_b × (30 - 10)  = F_b × 20 = 20×F_b,

The moment of the mass =  F_m × (20 - 10) =  F_m × 10

When we substitute  F_m = 1.5×F_b in the moment equation for the mass, we have;

The moment of the mass = F_m × 10 = 1.5×F_b ×10 = 15×F_b

Therefore, the balloon now has a larger momentum than that of the mass and the rod will rotate clockwise.

4 0
3 years ago
A satellite orbits the Earth (mass = 5.98 x 1024 kg) once every = 43200 s. At what radius does the satellite orbit?
kap26 [50]

Answer:

26621 km

Explanation:

We are given;

Mass: m = 5.98 x 10^(24) kg

Period; T = 43200 s

Formula for The velocity(v) of the satellite is:

v = 2πR/T

Where R is the radius

Formula for centripetal acceleration is;

a_c = v²/R

Thus; a_c = (2πR/T)²/R = 4π²R/T²

Formula for gravitational acceleration is:

a_g = Gm/R²

Where G is gravitational constant = 6.674 × 10^(-11) m³/kg.s²

Now the centripetal acceleration of the satellite is caused by its gravitational acceleration. Thus;

Centripetal acceleration = gravitational acceleration.

Thus;

4π²R/T² = Gm/R²

Making R the subject gives;

R = ∛(GmT²/4π²)

Plugging in the relevant values;

R = ∛((6.674 × 10^(-11) × 5.98 x 10^(24) × 43200²)/(4 × π²))

R = 26.621 × 10^(6) m

Converting to km, we have;

R = 26621 km

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