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kirill [66]
4 years ago
11

When switch S is open, the voltmeter across the battery reads 1.52V. When theswitch is closed , the voltmeter reading drops to 1

.37V and the ammeter reads 1.5A.Find the internal resistance of the battery.
Physics
1 answer:
Elis [28]4 years ago
5 0

Answer:

r = 0.1 Ω

Explanation:

We will use Ohm's Law in this question: V = IR, where I is the current and R is the resistance.

When the switch is open, the voltmeter reads 1.52 V, and there is no current in the circuit. We can deduce that the internal resistor in the battery causes 0.15 V to dissipate into heat, since the voltmeter reads 1.37 V when the switch is closed.

0.15 = (1.5)r\\r = 0.1~\Omega

You might be interested in
If the retina is 1.7 cm from the lens in the eye, how large is the image on the retina of a person of height 1.5 m standing 9.0
sineoko [7]

Answer:

The image height is 0.00283 m or 0.283 cm and is inverted due to the negative sign

Explanation:

u = Object distance =  9 m

v = Image distance = 1.7 cm (as the image is forming on the retina)

h_u= Object height = 1.5 m

Magnification

m=-\frac{v}{u}=\frac{h_v}{h_u}\\\Rightarrow -\frac{v}{u}=\frac{h_v}{h_u}\\\Rightarrow -\frac{0.017}{9}=\frac{h_v}{1.5}\\\Rightarrow h_v=-0.00283\ m

The image height is 0.00283 m or 0.283 cm and is inverted due to the negative sign

4 0
3 years ago
A particle P with speed 140 m s–1begins to decelerate uniformly at a certain instant while another particle Q starts from rest 6
Natasha2012 [34]

Answer:

i) The motion of both particles are shown on the same speed-time curve included

ii) Approximately 19.5 seconds

Explanation:

We are given that;

Initial velocity of particle, P = 140 m/s

Start time of particle P = 6 s before start time of particle Q

Position of particle Q when velocity is 25 m/s = 125 m

Therefore, from the equation of motion, we have for particle Q;

v² = u² + 2·a·s

Where:

v = Final velocity = 25 m/s

u = Initial velocity = 0 m/s

a = Acceleration

s = Distance covered = 125 m

Therefore;

25² = 0² + 2×a×125

Which gives a = 25²/(2×125) = 2.5 m/s²

The time taken for particle Q to reach 125 m is found from the relation;

s = u·t + 1/2·a·t²

Where:

t = Time of journey

Therefore;

125 = 0×t + 1/2×2.5×t²

Which gives 125 = 1.25 × t²

Hence, t² = 125/1.25 = 100

t = √(100) = 10 s

The equation for particle Q is v = 0 + 2.5×t

Hence, since particle P starts deceleration 6 seconds before the commencement of motion of particle Q, the amount of seconds after the commencement of deceleration of the first particle P that it takes for particle P to come to rest is found as follows;

Hence, at t = 6 + 10 = 16 seconds particle P speed = 25 m/s

From the equation of motion, for particle P (decelerating) we have

v = u - a·t

Where:

v = 25 m/s

u = 140 m/s

t = 16 s

Hence, 25 = 140 - a×16

∴ 16·a = 140 - 25 = 115

a = 115/16 = 7.1875 m/s²

Therefore, the time it takes before particle P comes to rest is found from the same equation of motion, where v = 0 as follows;

v = u - a·t

0 = 140 - 7.1875 × t

∴7.1875·t = 140

t = 140/7.1875 = 19.48 s ≈ 19.5 seconds.

4 0
3 years ago
An object starts at rest then accelerates at a rate of 5m/s^2 for 1 second and then 2m/s^2 for 2 seconds. What is the average ac
inn [45]

Acceleration = (change in speed) / (time for the change)

-- during the first second, the object increases its speed to

(5 m/s²) · (1 s) = 5 m/s .

-- During the next 2 seconds, the object increases its speed by

(2 m/s²) · ( s) = 4 m/s

So at the end of the whole 3 seconds, its speed is (5 m/s) + (4 m/s) = 9 m/s

-- Over the whole time, its speed has changed from zero to 9 m/s.

Acceleration = (change in speed) / (time for the change)

Acceleration = (9 m/s) / (3 sec)

<em>Acceleration = 3 m/s²</em>

7 0
3 years ago
At which point does a substance have the greatest amount of kinetic energy
kogti [31]
The hottest temp i think
8 0
3 years ago
Find the reaction supports at Ta and TB as shown in the loaded beam.
koban [17]

ANSWER

\begin{gathered} T_A=17.5N \\ T_B=12.5N \end{gathered}

EXPLANATION

First, we have to make a sketch of the direction of the moments of the forces about 12m from the left in the diagram:

The sum of upward forces must be equal to the sum of downward forces. This implies that:

\begin{gathered} T_A+T_B=20+10 \\ T_A+T_B=30N \end{gathered}

Also, the sum of clockwise moments must be equal to the counter-clockwise moments:

\begin{gathered} (20\cdot8)+(T_B\cdot4)=(T_A\cdot12) \\ 160+4T_B=12T_A \\ \Rightarrow12T_A-4T_B=160 \end{gathered}

From the first equation, make TA the subject of the formula:

T_A=30-T_B

Substitute that into the second equation:

\begin{gathered} 12(30-T_B)-4T_B=160 \\ 360-12T_B-4T_B=160 \\ -16T_B=160-360=-200 \\ T_B=\frac{-200}{-16} \\ T_B=12.5N \end{gathered}

Substitute that into the equation for TA:

\begin{gathered} T_A=30-12.5 \\ T_A=17.5N \end{gathered}

Therefore, the reaction supports at TA and TB are:

\begin{gathered} T_A=17.5N \\ T_B=12.5N \end{gathered}

5 0
2 years ago
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