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3 years ago

It takes a 7N force to cause a 0.028 kg object to begin moving. What is the acceleration of the object as it begins to move?

Physics

2 answers:

galben [10]3 years ago

6 0

Answer:

250 m/s2

Hope u find this helpful

Explanation:

Force = mass x acceleration

7N = (0.028kg) (a)

7N/0.028kg = a

a = 250m/s2

lbvjy [14]3 years ago

3 0

A is the answer to the question of what jersey is the name of the the team that is he still playing with the soccer team game today and he has a soccer ball with the wwwwwwww

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Radon-220 is a radioactive gas. It escapes from underground rocks and causes a large part of the natural background radiation in

Sergeu [11.5K]

Answer: 134 neutrons are there in the nucleus of Radon-220

Explanation:

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

Thus, number of protons = atomic number = 86

Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.

Mass number = Number of protons + Number of neutrons

220 = 86 + Number of neutrons

Number of neutrons = 134

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3 years ago

The surface air pressure on the planet mars is about 0.007 atm. if you were standing on mars, the surface air pressure would be

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3 years ago

In mechanics in what do we apply v=u+at

Trava [24]

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2 years ago

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$