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3 years ago

Acceleration problem Show work plz

Physics

1 answer:

Dennis_Churaev [7]3 years ago

3 0

Answer:

The answer to your question is: vo = 25 m/s

Explanation:

data

a = -7.5 m/s²

d = 42 m

vf = 0 m/s

vo = ?

Formula

vf² = vo² - 2ad

Substitution

0² = vo² - 2(7.5)(42)

We clear vo from the equation

vo² = 2(7.5)(42)

vo² = 630 simplifying

vo = 25 m/s result

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Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.

Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge $q_{1}= 3.8\times10^{-6}\ C$

Second charge $q_{2}=3.2\times10^{-6}\ C$

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

$E_{1}=E_{2}$

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}$

$\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}$

$\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}$

$x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}$

Put the value into the formula

$x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}$

$x=1.69\ m$

Hence, The distance is 1.69 m.

5 0

3 years ago

A point charge with charge q1 = 3.40 μC is held stationary at the origin. A second point charge with charge q2 = -4.90 μC moves

expeople1 [14]

Answer:

-0.79 J

Explanation:

We are given that

$q_1=3.4\mu C=3.4\times 10^{-6} C$

$1\mu C=10^{-6} C$

$q_2=-4.9\mu C=-4.9\times 10^{-6} C$

$x_1=0.125,y_1=0$

$x_2=0.280,y_2=0.235$

We have to find the work done by the electric force on the moving point charge.

$r_1=\sqrt{x^2_1+y^2_1}=\sqrt{(0.125)^2+0}=0.125$

$r_2=\sqrt{(0.280)^2+(0.235)^2}=0.366$

Work done, $W=kq_1q_2(\frac{1}{r_1}-\frac{1}{r_2})$

Where $k=9\times 10^9$

Using the formula

$W=9\times 10^9\times 3.4\times 10^{-6}\times(-4.9\times 10^{-6})(\frac{1}{0.125}-\frac{1}{0.366})$

$W=-0.79 J$

5 0

3 years ago

Describe what happens to the system inside of a refrigerator or freezer in terms of heat transfer, work, and conservation of ene

Over [174]

Answer: A. Work is done on the system and heat is transferred from the system for a net decrease in internal energy.

Explanation:

A refrigerator is a device which dispenses heat from the close system to a warmer area or in the surrounding. By dispensing the heat the internal temperature of the refrigerator drops. The system of refrigerator violates the second law of thermodynamics. As it performs the work to cool the region instead of heating the region. The work is done on the system and the internal energy decreases and the heat energy is liberated to the surrounding area. A refrigerator is an open system.

6 0

2 years ago

Why in drinking hot water, a thin-bottomed glass is taken?

taurus [48]

Answer:

beacause it's contracts

Explanation:

when using a large bottomed glass the hot water cools that's why is good to use thin bottomed glass

6 0

2 years ago

Diesel fuel is used in the engine of trucks that carry dirt fruit and other cargo. Fuel is burned in engines to make the motor m

Marina86 [1]

Answer: Fuel is burned in engines to make the motor move.

( Chemical to Mechanical )

Explanation:

during combustion of the diesel ( when the fuel is burnt in the engine of the vehicle, the diesel ( chemical energy ) is transformed or converted to Mechanical energy. This mechanical energy is what the truck uses in moving. Without the combustion of the fuel the vehicle won’t move and the combustion of diesel is achieved through compression unlike that of fuel.

4 0

3 years ago