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Gnesinka [82]
3 years ago
13

The compressor of an air conditioner draws an electric current of 16.2 A when it starts up. If the start-up time is 1.45 s long,

then how much electric charge passes through the circuit during this period
Physics
1 answer:
sp2606 [1]3 years ago
6 0

Answer:

Q = 23.49 C

Explanation:

We have,

Electric current drawn by the air conditioner is 16.2 A

Time, t = 1.45 s

It is required to find the electric charge passes through the circuit during this period. We know that electric current is defined as the electric charge flowing per unit time. So,

I=\dfrac{q}{t}\\\\q=It\\\\q=16.2\times 1.45\\\\q=23.49\ C

So, the charge of 23.49 C is passing through the circuit during this period.

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A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A
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Answer:

\omega_f = 585.37 \ rev/s

Explanation:

given,

mass of wheel(M) = 3 Kg

radius(r) = 35 cm

revolution (ω_i)=  800 rev/s

mass (m)= 1.1 Kg

I_{wheel} = Mr²

when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

I \omega_i = I' \omega_f

(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f

M\times 800 = ( M + m )\omega_f

3\times 800 = (3+1.1)\times \omega_f

2400 = (4.1)\times \omega_f

\omega_f = 585.37 \ rev/s

3 0
4 years ago
A coin 15.0 mm in diameter is placed 15.0 cm from a spherical mirror. The coin's image is 5.0 mm in diameter and is erect. Is th
s344n2d4d5 [400]

Answer:

15 cm

Explanation:

h_{o} = Diameter of the coin = 15 mm

h_{i} = Diameter of the image of coin =  5 mm

d_{o} = distance of the coin from mirror = 15 cm

d_{i} = distance of the image of coin from mirror = ?

Using the equation

\frac{d_{i}}{d_{o}} = \frac{- h_{i}}{h_{o}}

\frac{d_{i}}{15} = \frac{- (5)}{15}

d_{i} = - 5 cm

R = radius of curvature

Using the mirror equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{2}{R}

\frac{1}{15} + \frac{1}{- 5} = \frac{2}{R}

R = - 15 cm

6 0
3 years ago
Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30 second time interval. We take th
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Let say for every 5 s of time interval the speed will remain constant

so it is given as

v(mi/h)   16    21    23    26    33    30     28

now we have to convert the speed into ft/s as it is given that 1 mi/h = 5280/3600 ft/s

so here we will have

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now for each interval of 5 s we will have to find the distance cover for above interval of time

d = v \times t

d = (23.5 + 30.8 + 33.73 + 38.13 + 48.4 + 44 + 41.1) \times 5

d = 1298.1 ft

so here it will cover 1298.1 ft distance in 30 s interval of time

4 0
3 years ago
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