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3 years ago

A metal ball has a mass of 2kg and a volume of 6 m3. What is its density

Physics

2 answers:

Lena [83]3 years ago

8 0

Answer: The answer is 333.3333 repeating

Explanation:

Divide the mass by the volume.

Whitepunk [10]3 years ago

7 0

Answer:

Density of the metal ball, $\rho=0.334\ kg/m^3$

Explanation:

It is given that,

Mass of the metal ball, m = 2 kg

Volume of the metal ball, $V=6\ m^3$

We need to find the density of the metal ball. The total mass divided by total volume of an substance is called its density. Its formula is given by :

$\rho=\dfrac{m}{V}$

$\rho=\dfrac{2\ kg}{6\ m^3}$

$\rho=0.334\ kg/m^3$

So, the density of the metal ball is $0.334\ kg/m^3$ . Hence, this is the required solution.

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A camcorder has a power rating of 15 watts. If the output voltage from its battery is 5 volts, what current does it use?

MrMuchimi

Formula

W = E * I

Givens

E = 5 volts

W = 15 watts

I = ?

Solution

W = E * I

15 = 5 * I

15/5 = I

I = 3 amps. Answer

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3 years ago

What rapid changes to earths surface are caused by shifting plates

PtichkaEL [24]

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3 years ago

Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

7 0

3 years ago

Identifying Mechanical Energy

arsen [322]

Bruh ima be honest with you idek

3 0

3 years ago

Read 2 more answers

A motorcycle moving at a constant velcoity suddenly accelerates at a rate of 4.0 m/s/s to a speed of 35 m/s in 5.0 s. What was t

oee [108]

Answer:

15 m/s

Explanation:

v = u+ at

35 = u + 20

35-20 = u

u= 15 m/s

7 0

3 years ago