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3 years ago

7

What is the final temperature if it requires 5000 J of heat to warm 2.38892 x10-2 kg of water that starts at 5oC? Remember Cp fo

r water is 4186 J/kgC

1 answer:

Tanya [424]3 years ago

3 0

Explanation:

Hi ythere sorry do not know

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A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0

3 years ago

Explanation for question i & ii. Thank you.

suter [353]

Do u know Chinese ?if yes I can explain to u easily

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2 years ago

5. Calculate the acceleration of a 2 kg block across a table if you push with a force of 20 N and the frictional force is 4 N.

frozen [14]

Answer:

Explanation:hey do you go to GOC too?

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2 years ago

When describing electromagnetic radiation, there is a(n) _____________ relationship between wavelength and frequency and the gre

disa [49]

Its actually C. I did the question on USA test prep and it said the correct answer was C.

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3 years ago

Read 2 more answers

A 40.0 kg wagon is towed up a hill inclined at 18.5 degrees with respect to the horizontal. The tow rope is parallel to the incl

sp2606 [1]

Answer:

7.9m/s

Explanation:

We are given that

Mass of wagon=40 kg

$\theta=18.5^{\circ}$

Tension= $140 N$

Initial velocity of wagon= $u=0$

Displacement=s=80 m

Net force applied on wagon= $F=T-mgsin\theta=140-40(9.8)sin18.5=15.7 N$

By using $g=9.8m/s^2$

$a=\frac{F}{a}=\frac{15.7}{40}=0.39m/s^2$

We know that

$v^2-u^2=2as$

Using the formula

$v^2=2\times 0.39\times 80$

$v=\sqrt{2\times 0.39\times 80}=7.9m/s$

5 0

3 years ago