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3 years ago

Quick answer please

Physics

2 answers:

timama [110]3 years ago

6 0

Answer:

tq1=EoA/D

-2q1=-2EoAV/d

<u>Option C.</u>

Alekssandra [29.7K]3 years ago

6 0

Answer:

tq1=EoA/D

-2q1=-2EoAV/D

Explanation:

option C is correct

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What is the most common fuel used in nuclear power plants

strojnjashka [21]

Answer:

uranium

Explanation:

it is radioactive

8 0

3 years ago

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Cylinder A has a mass of 2kg and cylinder B has a mass of 10kg. Determinethe velocity of A after it has displaced 2m from its or

Marta_Voda [28]

The velocity of A is 5.16m/s²

<u>Explanation:</u>

Given-

mass of cylinder A, mₐ = 2kg

mass of cylinder B, mb = 10kg

Distance, s = 2m

Velocity of A, v = ?

Let acceleration due to gravity, g = 10m/s²

We know,

$a = \frac{mb * g - ma * g}{ma + mb} \\\\a = \frac{10 * 10 - 2 * 10}{ 2 + 10} \\\\a = \frac{80}{12} \\\\a = 6.67m/s^2$

We know,

$v = \sqrt{2as}$

$v = \sqrt{2 X 6.67 X 2} \\\\v = \sqrt{26.68} \\\\v = 5.16m/s^2$

Therefore, the velocity of A is 5.16m/s²

5 0

3 years ago

50 grams of ice cubes at -15°C are used to chill a water at 30°C with mass mH20 = 200 g. Assume that the water is kept in a foam

Arada [10]

Answer : The final temperature is, $25.0^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ice = $2.09J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ice = 50 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature = ?

$T_1$ = initial temperature of ice = $-15^oC$

$T_2$ = initial temperature of water = $30^oC$

Now put all the given values in the above formula, we get:

$50g\times 2.09J/g^oC\times (T_f-(-15))^oC=-200g\times 4.184J/g^oC\times (T_f-30)^oC$

$T_f=25.0^oC$

Therefore, the final temperature is, $25.0^oC$

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4 years ago

A step-up transformer has 50 turns on its primary coil and 2000 turns on its secondary

ryzh [129]

Answer:

Vs= 6000 v

Ip=9.6 A

Explanation:

Ns/Np = Vs/Vp

2000/50 = Vs/150

40 x 150 = Vs

Vs= 6000 v

Vs/Vp=Ip/Is

6000/150=Ip /0.24

Ip=9.6 A

7 0

3 years ago

The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12. How many grains are there in the ball

Fudgin [204]

Given :

The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12.

To Find :

How many grains are there in the ball?

Solution :

Volume of ball of the ballpoint is :

$V = \dfrac{4 \pi r^3}{3}\\\\V = \dfrac{4\times 3.14 \times 0.5^3}{3}\ mm^3\\\\V = 0.523\ mm^3$

Now, grain size of 12 has about 520000 grains/mm³.

Therefore, number of grains are :

$n = 520000\times 0.523\ grains\\\\n = 271960\ grains$

8 0

3 years ago