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yawa3891 [41]
2 years ago
11

Choices are 10.7 16.9 15.2 17.5

Physics
1 answer:
madreJ [45]2 years ago
6 0

Answer:

D

Explanation:

correct me if I'm wrong on your answer and it's the right answer from my son

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In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point
jarptica [38.1K]
The difference in electric potential energy between the two points is
\Delta U = q \Delta V
where q is the magnitude of the charge and \Delta V is the electric potential difference.

But for energy conservation, the difference in electric potential energy \Delta U between the two points is equal to the work done to move the charge between A and B:
W=\Delta U
so we have
W=q \Delta V

and by substituting the numbers of the problem, we find the value of \Delta V:
\Delta V =  \frac{W}{q}= \frac{0.90 J}{0.45 C}=2 V
3 0
3 years ago
What is the term used for sediments building up at the mouth of the river? (Hint: think New Orleans)
meriva
I believe it’s called Alluvium! It’s where the river mouth is build up of gravel,sand,silt, and clay!!
4 0
3 years ago
7. Two bikes travelling in the same direction move at a speed of 30 km/hr. The bikes are separated by a distance of 5 km. What w
Fantom [35]

Answer:

Explanation:

Call the bike on the right A

Call the bike on the left B

The car begins it's time when it passes A

4 minutes later, it passes B.

But B has moved in 4 minutes and that is the key to the problem.

How far has B moved.

t = 4 minutes = 4/60 hours = 1/15 of an hour.

d = ?

rate = 30 km / hr

d = r * t

d = 30 km/hr * 1/15 hours = 2 km

The distance between the bikes is 5 km.

So the car has traveled 5 - 2 = 3 km

d = 3 km

r = ?

t = 4 minutes = 1/15 hour

r = d/t = 3/(1/15)= 3 / 0.066666666  = 45 km/hr.

6 0
3 years ago
I need help with these questions.
strojnjashka [21]

I'll go ahead and answer the ones here without an answer. For reference, the half-life formula is <em>final amount = original amount(1/2)^(time/half-life)</em>

<em />

4) 12.5g

x = 100(1/2)^(63/21)

5) 50g

3.125 = x(1/2)^(0.1/0.025)

6) 500g

x = 4000(1/2)^(525/175)

7) 0.24g

0.06 = x(1/2)^(11430/5730)

8) 125g

x = 1000(1/2)^(17100/5700)

Hope this helps! :)

4 0
3 years ago
The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume tha
belka [17]

Given that,

Energy H=2.7\times10^{31}\ W

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

H=Ae\sigma T^4

A=\dfrac{H}{e\sigma T^4}

4\pi R^2=\dfrac{H}{e\sigma T^4}

R^2=\dfrac{H}{e\sigma T^4\times4\pi}

Put the value into the formula

R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}

R=5.0\times10^{10}\ m

(b). Given that,

Radiates energy H=2.1\times10^{23}\ W

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

R^2=\dfrac{H}{e\sigma T^4\times4\pi}

Put the value into the formula

R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}

R=5.42\times10^{6}\ m

Hence, (a). The radius of the star is 5.0\times10^{10}\ m

(b). The radius of the star is 5.42\times10^{6}\ m

8 0
3 years ago
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