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Alex787 [66]
3 years ago
5

. The p.d. at the terminals of a battery is 25V when no load is connected and 24V when a load taking 10A is connected. Determine

the internal resistance of the battery.
Physics
1 answer:
makvit [3.9K]3 years ago
6 0

Answer:

the internal resistance of the cell is 0.1 ohm.

Explanation:

Given;

p.d at the terminals of a battery at no load, E₁ = 25 V

p.d at the terminals of a battery at a load, E₂ = 24 V

current through the circuit, I = 10 A

The potential drop across the circuit, V = E₁ - E₂

                                                              = 25 V -  24 V

                                                               = 1 V

The internal resistance of the cell is calculated as follows;

r = V/I

r = 1 / 10

r = 0.1 ohm

Therefore, the internal resistance of the cell is 0.1 ohm.

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You are riding in an elevator that is accelerating upward. Suppose you stand on a scale. The reading on the scale is __________.
tatyana61 [14]

Given the following choices;

A) less than your true weight, mg.

B) equal to your true weight, mg.

C) more than your true weight, mg.

D) could be more or less than your true weight, mg, depending on the value of the speed.

The answer is; C

This is due to G-force. These are the perception of the weight of an object that is accelerating against gravity. We experience 1 g force on the surface of the earth because the ground exerts an upward exertion against gravity preventing as from falling to the center of the earth.  

8 0
2 years ago
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Velocities of two bodies A and B are given in vectors notation as va =i+2j-3k and Vb=3i+2j-k what will be the relative velocity
ArbitrLikvidat [17]

Answer:

V_{B/A}=2i+2k

Explanation:

The relative velocity can be calculated by means of the difference between vector B minus vector A.

V_{A}=i+2j-3k\\V_{B}=3i+2j-k\\V_{B}-V_{A}=(3-1)i + (2-2)j+(-1-(-3))k\\V_{B/A}=2i+2k

5 0
3 years ago
An aluminum-alloy rod has a length of 10.0 cm at 20°C and a length of 10.015 cm at the boiling point of water (1000C). (a) What
nikitadnepr [17]

Answer:

a.  9.99625 cm b. 68 °C

Explanation:

(a) What is the length of the rod at the freezing point of water (0 0C)?

Before we find the length of the rod, we need to find the coefficient of linear expansion, α = (L - L₀)/[L₀(T - T₀)] where L₀ = length of rod at temperature T₀ = 10.0 cm, T₀ = 20 °C, L = length of rod at temperature T = 10.015 cm and T = 100 °C

Substituting the values of the variables into the equation, we have

α = (L - L₀)/[L₀(T - T₀)]

α = (10.015 cm - 10.0 cm)/[10.0 cm(100 °C - 20 °C)]

α = 0.015 cm/[10.0 cm × 80 °C]

α = 0.015 cm/[800.0 cm °C]

α = 0.00001875 /°C

We now find the length L₁ at T₁ = 0 °C from

L₁ = L₀(1 + α(T₁ - T₀))

So, substituting the values of the variables into the equation, we have

L₁ = L₀(1 + α(T₁ - T₀))

L₁ = 10.0 cm[1 +  0.00001875 /°C(0° C - 20 °C)]

L₁ = 10.0 cm[1 +  0.00001875 /°C × -20° C]

L₁ = 10.0 cm[1 - 0.000375]

L₁ = 10.0 cm[0.999625]

L₁ = 9.99625 cm

(b) What is the temperature if the length of the rod is 10.009 cm?

With length L₃ = 10.009 cm at temperature T₃, using

L₃ = L₀(1 + α(T₃ - T₀))

making T₃ subject of the formula, we have

L₃/L₀ = 1 + α(T₃ - T₀)

L₃/L₀ - 1 = α(T₃ - T₀)

T₃ - T₀ = (L₃/L₀ - 1)/α

T₃ = T₀ + (L₃/L₀ - 1)/α

substituting the values of the variables into the equation, we have

T₃ = 20 °C + (10.009 cm/10.0 cm - 1)/0.00001875 /°C

T₃ = 20 °C + (1.0009 - 1)/0.00001875 /°C

T₃ = 20 °C + 0.0009/0.00001875 /°C

T₃ = 20 °C + 48 °C

T₃ = 68 °C

8 0
2 years ago
an object weighing 15 newtons is lifted from the ground to a height of 0.22 meter what is the increase in the object's gravitati
kicyunya [14]
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5 0
2 years ago
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The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold
Liula [17]

Answer:

L_{o}=0.1224m

Explanation:

Given data

Force F=2 N

Length L=17 cm = 0.17 m

Spring Constant k=42 N/m

To find

Relaxed length of the spring

Solution

From Hooke's Law we know that

F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m

6 0
3 years ago
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