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3 years ago

Lead has a density of 11.3 g/cm3 and a mass of 282.5 g. What is its volume

1 answer:

4 0

The volume is 25cm^3, 282.5/11.3 g/cm = 25

Please in the future use ^ to show exponents, they make life way easier for brainly users to help solve your problem. :)

You might be interested in

Shawn and his bike have a total mass of

Trava [24]

Answer:

Shawn's kinetic energy is 61.45 J.

Explanation:

Given;

Mass of the system is, $m=52.3\ kg$

Displacement of the bike is, $d=1.6\ km=1.6\times 1000=1600\ m$

Time taken is, $t=17.4\ min=17.4\times 60=1044\ s$

Let the constant velocity be $v$ .

For constant velocity, magnitude of velocity is given as distance by time.

Therefore, $v=\frac{d}{t}=\frac{1600}{1044}=1.533\ m/s$

Now, kinetic energy of a body is given as:

$KE=\frac{1}{2}mv^2$

Here, $m=52.3\ kg,\ v=1.533\ m/s$

$\therefore KE=\frac{1}{2}\times 52.3\times (1.533)^2\\KE=0.5\times 52.3\times 2.35\\KE=61.45\ J$

Therefore, Shawn's kinetic energy is 61.45 J.

3 0

3 years ago

Differentiate between angular displacement and linear displacement.

Sauron [17]

Answer:

The angular displacement is not a length (not measured in meters or feet), so an angular displacement is different than a linear displacement. ... As the object rotates through the angular displacement phi, the point on the edge of the disk moves distance sa along a circular path.

7 0

2 years ago

Charge X has twice as much charge as particle Y. The two charges are placed near each other. Compared to the force on particle X

Art [367]

The force on charge Y is the same as the force on charge X

Explanation:

We can answer this problem by applying Newton's third law of motion, which states that:

"When an object A exerts a force on object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"

In this problem, we can identify object A as charge X and object B as charge Y. The magnitude of the electrostatic force between them is given by

$F=k\frac{q_x q_y}{r^2}$ (1)

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_x, q_y$ are the two charges

r is the separation between the two charges

According to Newton's third law, therefore, the magnitude of the force exerted by charge X on charge Y is the same as the force exerted by charge Y on charge X (and it is given by eq.(1)), however their directions are opposite.

Learn more about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

6 0

3 years ago

For a particular pipe in a pipe-organ, it has been determined that the frequencies 296 Hz and 370 Hz are two adjacent natural fr

user100 [1]

Answer:

fundamental frequency of pipe will be equal to 74 Hz

Explanation:

We have given for a particular organ pipe two adjacent frequency are 296 Hz and 370 Hz

Speed of the sound in air is 343 m/sec

We have to find the fundamental frequency for the pipe

Fundamental frequency will be equal to difference of the two adjacent frequency

So fundamental frequency = 370 - 296 = 74 Hz

So fundamental frequency of pipe will be equal to 74 Hz

8 0

3 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

3 years ago