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Sholpan [36]
3 years ago
11

An ice cube is placed in a microwave oven.

Physics
1 answer:
Juli2301 [7.4K]3 years ago
5 0

Answer:

It takes 266 seconds to melt the ice.

Explanation:

Given data

  • Power of the microwave oven (P): 125 Watt
  • Heat supplied to the ice (Q): 33,200 Joule
  • Time for the melting (t): to be determined

In order to determine the time required to melt the ice, we can use the following expression.

P = Q/t

t = Q / P = 33,200 J/ 125 W = 266 s

It takes 266 seconds to melt the ice.

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A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p
levacccp [35]

Answer:

Q = 8,345 * v

Explanation:

So, we are looking for an expression of the amount of water that has been drained from the tub. The expression is in terms of v that represent the number of gallons of water drained since the plug was pulled. Since we are interested in the pounds of water that has been drained from the tub we need to take into account that for every gallon of water drained, 8.345 pounds have left the tub. Therefore, the expression for the weight of water Q that has been drained from the tub in terms of v is simply :

Q = 8,345 * v

Where v is the amount of gallons that has been drained from the tub.

Have a nice day. let me know if I can help with anything else

8 0
3 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
TRUE OR FALSE
Lynna [10]

Answer:

True

Explanation:

If you balance calorie intake it i very good FOR YA BODY

6 0
3 years ago
Read 2 more answers
You lay a mirror flat on the floor with one edge against a wall and aim a laser at the mirror. The ray reflects from the mirror
joja [24]

Answer:

the angle of incidence θ is 45.56 º

Explanation:

Given data

strikes the mirror before wall x = 30.7 cm

reflected ray strikes the wall y =  30.1 cm

to find out

the angle of incidence θ

solution

let us consider ray is strike at angle  θ so after strike on surface ray strike to wall at angle 90 - θ

we will apply here right angle triangle rule that is

tan( 90 - θ) = y /x

tan( 90 - θ)  = 30.1 / 30.7

90 - θ = tan^-1 (30.1/30.7)

90 - θ = 44.4345

θ = 45.56 º

the angle of incidence θ is 45.56 º

4 0
3 years ago
Rank the following in terms of increasing inertia:
Naddik [55]
C. A 1200kg car is going 15m/s
7 0
3 years ago
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