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2 years ago

Which factors describe the motion of an object?

Physics

1 answer:

makkiz [27]2 years ago

7 0

Answer:

Position, speed, direction, and acceleration.

Explanation:

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A carpenter is driving a 15.0-g steel nail into a board. His 1.00-kg hammer is moving at 8.50 m/s when it strikes the nail. Half

bekas [8.4K]

Answer: The increase in temperature of the nail after the three blows is 8.0636 Kelvins. The correct option is (d).

Explanation:

Kinetic energy of the hammer ,K.E.=

$\frac{1}{2}mv^2=\frac{1}{2}1.00 kg\times (8.50 m/s)^2=36.125 J$

Half of the kinetic energy of the hammer is transformed into heat in the nail.

Energy transferred to the nail in one blow =

$\frac{1}{2}K.E.=\frac{1}{2}\times 36.125 J=18.0625 J$

Total energy transferred after 3 blows,Q = $3\times 18.0625 J=54.1875 J$

Mass of the nail = 15 g = 0.015 kg

Change in temperature = $\Delta T$

Specif heat of the steel = c = 448 J/kg K

$Q=mc\Delta T$

$54.1875 J=0.015 kg\times 448 J/kg K\times \Delta T$

$\Delta T=8.0636 K\approx 8.1 K$

The increase in temperature of the nail after the three blows is 8.1 Kelvins.Hence, correct option is (d).

4 0

2 years ago

The greater the mass is in an object, the higher resistance to a change in movement the object will have.

Bond [772]

True. Since Inertia states this law, we are bound to believe greater mass = greater resistance in change of motion

3 0

2 years ago

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which of these types of radiation do you suppose we should fear the most if administered internally? •gamma rays •neutrons •alph

4vir4ik [10]

Alpha particles, because they are the heaviest ones (helium nuclei) and will travel around the body.

4 0

2 years ago

A line of charge starts at x = +x0 and extends to positive infinity. The linear charge density is λ = λ0x0/x. Determine the elec

kari74 [83]

Explanation:

it is given that, the linear charge density of a charge, $\lambda=\dfrac{\lambda_ox_o}{x}$

Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

$dE=\dfrac{k\ dq}{x^2}$ ..........(1)

The linear charge density is given by :

$\lambda=\dfrac{dq}{dx}$

$dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx$

Integrating equation (1) from x = x₀ to x = infinity

$E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx$

$E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}$

$E=\dfrac{k\lambda_o}{2x_o}$

Hence, this is the required solution.

5 0

2 years ago

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Graphs help you see ______.

Anastaziya [24]

D- if i was right please mark me as a brainliest answer and a thank you

4 0

3 years ago

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