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3 years ago

A block of lead with dimensions 2.0 dm x 8.0 cm x 35 mm, has a mass of 6.356 kg. calculate the density of lead in g/cm

Physics

1 answer:

insens350 [35]3 years ago

8 0

First, let us calculate for the volume of the block of lead using the formula:

V = l * w * h

But we have to convert all units in terms of cm:

l = 2.0 dm = 20 cm

w = 8 cm

h = 3.5 cm

Therefore the volume is:

V = (20 cm) * (8 cm) * (3.5 cm)
V = 560 cm^3

Next we convert the mass in terms of g:

m = 6.356 kg = 6356 g

Density is mass over volume, so:

density = 6356 g / 560 cm^3

density = 11.35 g / cm^3 (ANSWER)

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Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha

Inessa05 [86]

Answer:

$+1.11\mu C$

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

$Q_1$ is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to $Q_2$ and $Q_3$ at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of $Q_1$ .

Let the electric field intensity due to $Q_2$ be + $E_2$ and that due to $Q_3$ be - $E_3$ since the charge is negative. Hence at the origin;

$+E_2-E_3=0..................(1)$

From equation (1) above, we obtain the following;

$E_2=E_3.................(2)$

From Coulomb's law the following relationship holds;

$+E_2=\frac{kQ_2}{r_2^2}\\$

$-E_3=\frac{kQ_3}{r_3^2}$

where $r_2$ is the distance of $Q_2$ from the origin, $r_3$ is the distance of $Q_3$ from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

$\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)$

k can cancel out from both side of equation (3), so that we finally obtain the following;

$\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)$

Given;

$Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m$

Substituting these values into equation (4); we obtain the following;

$\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\$

$Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C$

6 0

3 years ago

Are killer whales warm blooded.

defon

Answer:

Yes, they are warm blooded.

because their body temperature is close to that of human about 36.4º to 38ºC (97.5º to 100.4ºF)

Explanation:

7 0

3 years ago

A rectangular copper strip 1.5cm wide and 0.10cn thick carries a current of 5.0A. Find the Hall voltage for a 1.2T magnetic fiel

Scrat [10]

Answer:

4.4345× 10^-7V

Explanation:

The computation of the half voltage for a 1.2T magnetic field applied is shown below

The volume of one mole of copper is

v = m ÷p

= 63.5 ÷ 8.92

= 7.12cm

Now the density of free electrons in copper is

n = Na ÷ V

= 6.02 × 10^23 ÷ 7.12

= 8.456× 10^28/m^3

Now the half voltage is

= IB ÷ nqt

= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)

= 4.4345× 10^-7V

7 0

3 years ago

In which graph is acceleration the slope?

olga2289 [7]

Answer:

Velocity -time graph 

Explanation:

hope it helps ✌️✌️

4 0

2 years ago

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3) Calculate the kinetic energy of a 7 kg mass traveling at a velocity of 4 m/sec.

murzikaleks [220]

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Let's solve ~

Given terms :

Mass (m) = 7 kg

velocity (v)= 4 m/s

The formula to find kinetic Energy is ~

$\boxed{ \boxed{ \sf{ \frac{1}{2} m{v}^{2} }}}$

Now, apply the formula according to given situation

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \: \dfrac{1}{2} \times 7 \times ( {4)}^{2}$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \: \dfrac{1}{2} \times 7 \times 16$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:7 \times 8$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:56 \: \: joules$

Therefore, the kinetic Energy of the car is 56 joules

4 0

2 years ago

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