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2 years ago

Which statement is TRUE?

Physics

1 answer:

tatyana61 [14]2 years ago

8 0

<span>
In a series circuit, if one bulb goes out the other bulbs will stay lit. This is the correct one</span>

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A student attempted to measure the specific latent heat of vaporisation of water.

tensa zangetsu [6.8K]

Answer:

The latent heat of vaporization of water is 2.4 kJ/g

Explanation:

The given readings are;

The first (mass) balance reading (of the water) in grams, m₁ = 581 g

The second (mass) balance reading (of the water) in grams, m₂ = 526 g

The first joulemeter reading in kilojoules (kJ), Q₁ = 195 kJ

The second joulemeter reading in kilojoules (kJ), Q₂ = 327 kJ

The latent heat of vaporization = The heat required to evaporate a given mass water at constant temperature

Based on the measurements, we have;

The latent heat of vaporization = ΔQ/Δm

∴ The latent heat of vaporization of water = (327 kJ - 195 kJ)/(581 g - 526 g) = 2.4 kJ/g

The latent heat of vaporization of water = 2.4 kJ/g

6 0

2 years ago

A pole AB of length 10.0m and weight 600N has its center of gravity 4.0m from the end A, and lies on horizontal ground .Calculat

postnew [5]

Answer:

The force required to begin to lift the pole from the end 'A' is 240 N

Explanation:

The given parameters for the pole AB are;

The length of the pole, l = 10.0 m

The weight of the pole, W = 600 N ↓

The distance of the center of gravity of the pole from the side 'A' = 4.0 m

Let ' $F_A$ ' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive

For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have

$F_A$ × 10.0 m - W × 4.0 m = 0

∴ $F_A$ × 10.0 m = W × 4.0 m = 600 N × 4.0 m

$F_A$ × 10.0 m = 600 N × 4.0 m

∴ $F_A$ = 600 N × 4.0 m/(10.0 m) = 240 N

The force required to begin to lift the pole from the end 'A', $F_A$ = 240 N.

8 0

2 years ago

Convert 100 millimeters to meters

drek231 [11]

100 millimeters equals 0.1 meters.

3 0

3 years ago

Delilah does 170 Joules of work in 30 seconds. Adam does 260 Joules of work in 20 seconds. Who was more powerful?

sp2606 [1]

Power = $\frac{Work}{Time}$

Delilah: 170J/30s = 5.66 W

Adam: 260J/20s = 13 W

6 0

2 years ago

Find the electric field at a point midway between two charges of 30.0×10 power -9 and 60.0×10 power -9 separated by a distance o

KATRIN_1 [288]

Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹ N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹ N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea = +1.8 * 10⁴ N/C

Electric field due to charge B

Eb = -(9.0 * 10⁹ N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb = -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴ + -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

7 0

2 years ago