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3 years ago

When a hailstone is at a height of 2.00km it’s mass is 2.50g what is it’s potential energy?

Physics

1 answer:

scZoUnD [109]3 years ago

6 0

Answer:

EP = 49.05Joules (J)

Explanation:

The equation for Potential energy (EP) is

EP = m g h

We are given the values below (do convert them into SI units)

m = 0.0025kg

h = 2000m

g = 9.81m/ $s^{2}$

Substitute the values into the equation and solve for EP

EP = 0.0025 * 2000 * 9.81

EP = 49.05Joules (J)

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Find the magnitude of the resultant force and the angle it makes with the positive x-axis. (Let a = 200 N and b = 400 N. Round y

omeli [17]

Answer: R = 346.4N and angle 30° to the horizontal negative axis

Explanation:

To find the resultant force, we need to sum up the forces on the vertical and horizontal axis.

For the horizontal axis;

Rx = -b + acos60

Rx = -400N +200cos60

Rx = -400N +100N

Rx = -300N

For the vertical axis;

Ry = asin60 = 200sin60

Ry = 173.2N

The resultant force R can be given as;

R = √(Rx^2 +Ry^2)

R = √((-300)^2 + 173.2^2)

R = 346.4N

Angle z can be written as

Tanz = Ry/Rx

z = taninverse (Ry/Rx)

z = taninverse (173.2/300)

z = 30°

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3 years ago

An 1,840 W toaster, a 1,460 W electric frying pan, and a 50 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (T

shutvik [7]

Answer:

Explanation:

Given

Power drawn by toaster $P_1=1840\ W$

Power drawn by Electric fan $P_2=1460\ W$

Power drawn by lamp $P_3=50\ W$

Voltage applied $V=120\ V$

If appliances are applied in parallel then Voltage applied is same

Power is given by $P=\frac{V^2}{R}$

Resistance of toaster $R_1=\frac{120^2}{1840}=7.82\ \Omega$

Resistance of electric Fan $R_2=\frac{120^2}{1460}=9.86\ \Omega$

Resistance of toaster $R_3=\frac{120^2}{50}=288\ \Omega$

Current is given by

$I=\frac{V}{R}$

$I_1=\frac{120}{7.82}=15.34\ A$

$I_2=\frac{120}{9.86}=12.17\ A$

$I_3=\frac{120}{288}=0.416\ A$

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Answer:

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