Ask question

Ask question

All categories

Marina CMI [18]

3 years ago

11

A particle with a charge of +7.4 C is separated from another charged particle with a charge of

1 answer:

mel-nik [20]3 years ago

6 0

Explanation:

9 × 10^-3 × 13.59 = 0.122 N

You might be interested in

A gas is collected from a radioactive material; upon inspection, the gas is identified as helium. the presence of the helium ind

Flura [38]

The presence of helium gas indicates the radioactive sample is most likely decaying by α-decay, or alpha decay. α-decay is the type of radioactive decay in which an atomic nucleus emits α particles. α particles are Helium nuclei. So the correct answer would be α-decay.

7 0

3 years ago

American football uses a field that is 100.0yd long, whereas the a soccer field is 100.0m long. Which field is longer and by how

Vitek1552 [10]

The soccer field is 9.36% longer.

8 0

3 years ago

Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and

dexar [7]

Kepler’s
third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496
× 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work
with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel
them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/(
86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The
closest answer is 1.99
× 10^30

(it may vary
a little with rounding – the difference is less than 1%)

8 0

3 years ago

Read 2 more answers

What is your opinion on abortion?

zloy xaker [14]

Ratio

L nobody cares and bozo

5 0

3 years ago

Read 2 more answers

A yet-to-be-built spacecraft starts from Earth moving at constant speed to the yet-tobe-discovered planet Retah, which is 20 lig

Gre4nikov [31]

Answer:

The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame

Explanation:

From the question we are told that

The distance between earth and Retah is $d = 20 \ light \ hours = 20 * 3600 * c = 72000c \ m$

Here c is the peed of light with value $c = 3.0*10^8 m/s$

The time taken to reach Retah from earth is $t = 25 \ hours = 25 * 3600 =90000 \ sec$

The velocity of the spacecraft is mathematically evaluated as

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

substituting values

$v_s = \frac{72000 * 3.0*10^{8} }{90000}$

$v_s = 2.40*10^{8} \ m/s$

The time elapsed in the spacecraft’s frame is mathematically evaluated as

$T = t * \sqrt{ 1 - \frac{v^2}{c^2} }$

substituting value

$T = 90000 * \sqrt{ 1 - \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }$

$T = 54000 \ s$

=> $T = 15 \ hours$

So The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame

7 0

3 years ago