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tia_tia [17]
4 years ago
13

An object has an acceleration of 6.0 m/s/s. if the net force acting upon this object were doubled, then its new acceleration wou

ld be _____ m/s/s.
Physics
1 answer:
Flauer [41]4 years ago
5 0
F has direct relation with a
then doubling F cause acc. to get double i:e 6×2=12
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United States citizens have many ways to participate in national life. The most common form of participation is?
Travka [436]

Answer:

You would have to give better explanation on subject.

Explanation:

3 0
3 years ago
A 40 kg person sits on top of a 400 kg rock. What is the person’s weight? 390 N
algol13

The weight of the person is given by:

W = mg

W = weight, m = mass, g = gravitational acceleration

Given values:

m = 40kg, g = 9.81m/s²

Plug in and solve for W:

W = 40(9.81)

W = 390N

7 0
3 years ago
A cylinder with rotational inertia I1 = 3.0 kg · m2 rotates clockwise about a vertical axis through its center with angular spee
erastova [34]

Answer:

ω = 1.83 rad/s clockwise

Explanation:

We are given:

I1 = 3.0kg.m2

ω1 = -5.4rad/s (clockwise being negative)

I2 = 1.3kg.m2

ω2 = 6.4rad/s  (counterclockwise being positive)

By conservation of the momentum:

I1 * ω1 + I2 * ω2 = (I1 + I2) * ω

Solving for ω:

\omega = \frac{I1 * \omega1 + I2*\omega2}{I1+I2}=-1.83rad/s

Since it is negative, the direction is clockwise.

8 0
3 years ago
a baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby weighs 12 Kg. how much energy does
Tcecarenko [31]

Answer:

The carriage has the energy, W = 2469.6 J

Explanation:

Given data,

The height of the hill, h = 21 m

The carriage with the baby weighs, m = 12 kg

The energy possessed by the body due to its position is the potential energy,

                                      <em>W = P.E = mgh joules</em>

Substituting the values,

                                       W = 12 x 9.8 x 21

                                            = 2469.6 J

Hence, the carriage has the energy, W = 2469.6 J

8 0
3 years ago
A solid 200-g block of lead and a solid 200-g block of copper are completely submerged in an aquarium filled with water. Each bl
finlep [7]

Answer:

B. The buoyant force on the copper block is greater than the buoyant force on the lead block.

Explanation:

Given;

mass of lead block, m₁ = 200 g = 0.2 kg

mass of copper block, m₂ = 200 g = 0.2 kg

density of water, ρ = 1 g/cm³

density of lead block, ρ₁ = 11.34 g/cm³

density of copper block, ρ₂ = 8.96 g/cm³

The buoyant force on each block is calculated as;

F_B = mg(\frac{density \ of \ fluid}{density \ of \ object} )

The buoyant force of lead block;

F_{lead} = 0.2*9.8(\frac{1}{11.34} )\\\\F_{lead} = 0.173  \ N

The buoyant force of copper block

F_{copper} = 0.2*9.8(\frac{1}{8.96})\\\\F_{copper} = 0.219  \ N

Therefore, the buoyant force on the copper block is greater than the buoyant force on the lead block

3 0
3 years ago
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