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Vlad1618 [11]
3 years ago
8

A net force applied to a 15.0 kg box produced an acceleration of 4.2 m/s2. If the same net force was applied to a 10 kg box, wha

t would be the acceleration to the nearest tenth of a m/s2?
Physics
1 answer:
igor_vitrenko [27]3 years ago
8 0

Answer:

6.3\ m/s^2

Explanation:

Given that,

If the mass of a body is 15 kg and produced an 4.2 m/s², we need to find the acceleration if the mass is 10 kg and the same force is applied.

Force is given by :

F = ma

Since force is same

m_1a_1=m_2a_2\\\\a_2=\dfrac{m_1a_1}{m_2}\\\\a_2=\dfrac{15\times 4.2}{10}\\\\a=6.3\ m/s^2

So, if the mass is 10 kg, acceleration is 6.3\ m/s^2

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If the pressure above a solution containing a gas solute is reduced, the limit of the gas's solubility will decrease.
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What would happen if the pilot did not keep the airplane "trimmed"
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Answer:

In explanation

Explanation:

Pilots who dont use trim often like the feeling of holding constant back pressure because The heavier control forces makes it more difficult to over-control the airplane inside the turn, so it gives the sense of a more stable flight

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3 years ago
Which form of the law of conservation of energy describes the motion of the block as it slides on the floor from the bottom of t
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Answer:

Explanation:

Let assume begins movement at zero point, that is, height is equal to zero. The block has an initial linear kinetic energy and no gravitational potential energy and end with no linear kinetic energy, some gravitational potential energy and work losses due to slide friction. In mathematical terms, this system can be model as follows:

K_{1} = U_{2} + W_{loss, 1 \longrightarrow 2}

Where K, U, W are linear kinetic energy, gravitational potential energy and work, respectively.

8 0
3 years ago
A certain spring has a spring constant k1 = 660 N/m as the spring is stretched from x = 0 to x1 = 35 cm. The spring constant the
pantera1 [17]

(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².

(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.

(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.

<h3>Work done in the spring</h3>

The work done in stretching the spring is calculated as follows;

W = ¹/₂kx²

W(1 to 2) = ¹/₂K₂Δx²

W(1 to 2)  =  ¹/₂(250)(0.65 - 0.35)²

W(1 to 2)  = 11.25 J

W(0  to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃

W(0  to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)

W(0  to 3) = 64.28 J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

6 0
2 years ago
Trey races his bicycle for 192m. A wheel of his bicycle turns 48 times as the bicycle travels this distance. What is the diamete
chubhunter [2.5K]

Answer:

1.27 m

Explanation:

Distance = 192 m

number of rotations = 48

Distance traveled in one rotation = 2 x π x r

Where, r be the radius of wheel.

so, distance traveled in 48 rotations = 48 x 2 x 3.14 x r

It is equal to the distance traveled.

192 = 48 x 2 x 3.14 x r

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diameter of wheel = 2 x radius of wheel = 2 x 0.637 = 1.27 m

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3 years ago
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