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1 year ago

13

How are chargeable cells different from ordinary dry cells

1 answer:

topjm [15]1 year ago

6 0

Ordinary cells can convert chemical energy to electrical energy only, but rechargeable cells can also store electrical energy into chemical energy and vice versa. You will study more about it in your higher classes. secondary cells can be recharged and used again but dry cells cannot be recharged.

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Table 2.4 shows how the dispacement of a runner changed during a sprint race. Draw a dispacement-time graph to show this data, a

GalinKa [24]

4. Table 2.4 shows how the displacement of a runner changed
during a sprint race. Draw a displacement–time graph to show
this data, and use it to deduce the runner’s speed in the middle
of the race.
Table 2.4 Data for a sprinter during a race
Displacement
(m)
0 4 10 20 50 80 105
Time (s) 1 2 3 6 9 12

8 0

1 year ago

A rectangular coil of dimensions 5.40cm x 8.50cm consists of25 turns of wire. The coil carries a current of 15.0 mA.

Kazeer [188]

Answer:

(a) Magnetic moment will be $17.212\times 10^{-4}A-m^2$

(b) Torque will be $6.024\times 10^{-4}N-m$

Explanation:

We have given dimension of the rectangular 5.4 cm × 8.5 cm

So area of the rectangular coil $A=5.4\times 8.5=45.9cm^2=45.9\times 10^{-4}m^2$

Current is given as $i=15mA=15\times 10^{-3}A$

Number of turns N = 25

(A) We know that magnetic moment is given by $magnetic\ moment=NiA=25\times 45.9\times 10^{-4}\times 15\times 10^{-3}=17.212\times 10^{-4}A-m^2$

(b) Magnetic field is given as B = 0.350 T

We know that torque is given by $\tau =BINA=0.350\times 15\times 10^{-3}\times 25\times 45.9\times 10^{-4}=6.024\times 10^{-4}N-m$

4 0

1 year ago

Study the free body diagram above. Which scenario below can best be described with this free body diagram? A. a cup is at rest o

vekshin1

Answer: D

Explanation:

5 0

2 years ago

A cylinder is 0.10 m in radius and 0.20 in length. Its rotational inertia, about the cylinder axis on which it is mounted, is 0.

Bumek [7]

Answer: $5 rad/s^2$

Explanation:

Given

Radius of cylinder r=0.1 m

Length L=0.2 in.

Moment of inertia I=0.020 kg-m^2

Force F=1 N

We Know Torque is given by

$Torque =I\alpha =F\cdot r$

where $\alpha =angular\ acceleration$

$I\alpha =F\cdot r$

$0.02\cdot \alpha =1\cdot 0.1$

$\alpha =5 rad/s^2$

5 0

2 years ago

a plane travels 204 km, northeast in 15.0 minutes. It also increases elevation by 1.6 km, upward in the same amount of time. Wha

mina [271]

Answer:

230 m/s northeast, 1.8 m/s up

Explanation:

204 kilometres = 204000 metres

15.0 minutes = 900 seconds

Velocity = Distance / Time

= 204000 / 900

= 230 m/s northeast (to 2 sf.)

1.6km = 1600 metres

Velocity = 1600 / 900

= 1.8 m/s up (to 2 sf.)

7 0

1 year ago