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makvit [3.9K]
3 years ago
7

The furnace keeps houseAat 25◦C, while thefurnace in houseBkeeps it at 20◦C. Which house requires heat to be supplied by its fur

nace at a faster rate?
1.House B requires heat at a slightly faster rate than A.
2.House A requires heat at about twice the rate of B.
3.House A requires heat at about four times the rate of B.
4.Houses A and B require heat at the same rate.
5.House A requires heat at a slightly faster rate than B.
6.House B requires heat at about twice the rate of A.
7.House B requires heat at about four times the rate of A.
8.None of the above answers is correct.
Physics
1 answer:
EleoNora [17]3 years ago
5 0

Answer:

House A requires heat at a slightest faster rate than B

Explanation:

House A requires heat at a slightest faster rate than B due to the slight high temperature the furnace A is.

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By the second law of Newton we get the relation

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Four seconds after being launched, what is the height of a ball that starts from a height of 12 m with an initial upward velocit
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Answer:

15.24 m/s in the downward direction

Explanation:

Given that the initial upward velocity of the ball is 24 m/s.

Assuming that the upward direction is positive.

As gravitational force acts in the downward direction and the direction of acceleration is the same as the direction of force, so the acceleration due to gravity will be negative.

Now, from the equation of motion, when an object is launched with initial velocity u, the final velocity, v, of an object after time t is v=u+at.

Given that u=24 m/s, t=4 seconds, g=-9.81 m/s^2.

So, the final velocity is

v= 24 + (-9.81)\times 4 \\\\\Rightarrow v= 24-9.81\times 4

\Rightarrow v=-15.24 m/s

Here, the negative sign means the final velocity is in the downward direction.

Hence, the velocity after 4 seconds is 15.24 m/s in the downward direction.

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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-
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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}

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\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.

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