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3 years ago

I need help!!! 60 POINTS!!!!

Physics

2 answers:

galina1969 [7]3 years ago

8 0

The answer is "A and B are different elements, while C is an isotope of B".

Scrat [10]3 years ago

3 0

Correct one is b
Good luck

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If an object has a mass of 1417g and it is moved 47 meters in 90 seconds, how much power was used?

astraxan [27]

Mass of the object is given as

$m = 1417 g = 1.417 kg$

now the speed of object is given as

$v = \frac{d}{t}$

here we know that

$d = 47 m$

$t = 90 s$

now we will have

$v = \frac{47}{90} = 0.52 m/s$

now we will have kinetic energy of the object as

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

$KE = \frac{1}{2}(1.417)(0.52)^2$

$KE = 0.19 J$

now the power is defined as rate of energy

so here we can find power as

$P = \frac{KE}{t}$

$P = \frac{0.19}{90} = 2.14\times 10^{-3} W$

so above is the power used for the object

3 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

I will make brainliest! PLS HELPPP!!

inysia [295]

Answer:

1.A compound is when two substances are put together and there is a reaction creating a new substance Ex. H20. A mixture is when two substances are put together, there is no reaction and they can be separated easier than compounds. Ex. salad dressing.

2.Atomic bonds

3.Compound

4.Element

5.Mixture

Explanation:

7 0

2 years ago

Calculate the magnitude of the linear momentum for the following cases. (a) a proton with mass 1.67 10-27 kg, moving with a spee

abruzzese [7]

Answer:

Explanation:

Mass of proton :