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Marizza181 [45]
3 years ago
8

A lamp is rated for 240v,2.5A.what is the cost of using the lamp for 3hrs if 1kWh of electricity cost #12​

Physics
1 answer:
marshall27 [118]3 years ago
6 0

Answer:

≈ 22¢

Explanation:

240 / 1000 = 0.240 kV

0.240 kV(2.5 A)(3 hr) = 1.8 kW•hr

1.8 kW•hr($0.12/kW•hr) = $0.216

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A power lifter performs a dead lift, raising a barbell with a mass of 305 kg to a height of 0.42 m above the ground, giving the
Ann [662]

Answer:

Explanation:

Before it hits the ground:

The initial potential energy = the final potential energy + the kinetic energy

mgH = mgh + 1/2 mv²

gH = gh + 1/2 v²

v = √(2g (H - h))

v = √(2 * 9.81 m/s² * (0.42 m - 0.21 m))

v ≈ 2.0 m/s

When it hits the ground:

Initial potential energy = final kinetic energy

mgH = 1/2 mv²

v = √(2gH)

v = √(2 * 9.81 m/s² * 0.42 m)

v ≈ 2.9 m/s

Using a kinematic equation to check our answer:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2(9.8 m/s²)(0.42 m)

v ≈ 2.9 m/s

3 0
3 years ago
Which of the following is group 1? a. alkali metals b. alkaline earth metals c. halogens d. transition metals
Elanso [62]
The answer is A. Alkali metals
4 0
3 years ago
Read 2 more answers
A large rock of mass me materializes stationary at the orbit of Mercury and falls into the sun. Itf the Sun has a mass ms and ra
son4ous [18]

Answer:

The answer is v = \sqrt{2G\frac{M_s}{R^2}(R-r_s)}.

Explanation:

From the law of gravity,

F = G \frac{Mm}{r^2}

considering F as a conservative force, F = - \nabla U,

the general expression for gravitational potential energy is

U = -G \frac{Mm}{r},

where G is the gravitational constant, M and m are the mass of the attracting bodies, and r is the distance between their centers. The negative sign is because the force approaches zero for large distances, and we choose the zero of gravitational potential energy at an infinite distance away.

However, as the mass of the Sun is much greater than the mass of the rock, the gravitational acceleration is defined as

g = -G \frac{M}{r^2},

(the negative sign indicates that the force is an attractive force), and the potential energy between the rock and the Sun is

U = g M_e R,

which is actually the total energy of the system, because the rock materializes stationary at this point (there is no radial kinetic energy).

When the rock hits the surface of the Sun, almost all potential energy is converted to kinetic energy, but not all because the Sun is not a puntual mass. So the potential energy converted to kinetic energy is

U_p = g M_e(R- r_s),

then, the kinetik energy when the rock hits the surface is

U_k =\frac{1}{2}M_e v^2 = g M_e(R- r_s),

so

v = \sqrt{2g(R-r_s)}

where g is the gravitational acceleration generated by the Sun at R,

g = G \frac{M_s}{R^2}.

8 0
2 years ago
An object has a kinetic energy of 14 J and a mass of 17 kg , how fast is the object moving?
lisabon 2012 [21]
v ^{2} = Joules ÷ (0.5×Kilograms)

14J ÷ 8.5 = 1.64705882

Remember, 1.64705882 = v², so we need to find the square root.

The square root of 1.64705882 is 1.283377894464448

Hope this helps! 
6 0
3 years ago
Read 2 more answers
An object moving with a speed or 67m/s and has kinetic energy of 500 J what is the mass of the object
saveliy_v [14]
To solve this equation, simply plug the values into the equation for calculating kinetic energy.

KE = 1/2mv^2
500 = 1/2(m)(67^2)
500 =2244.5m
m = 500/2244.5 = 0.222 kg.
8 0
2 years ago
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