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Svetradugi [14.3K]
3 years ago
7

A portable television using a 12 volt, 3 Ah rechargeable battery can operate for a period of about 6 hours. What is the average

current drawn in amperes during this period
Physics
1 answer:
bezimeni [28]3 years ago
7 0

Answer:

0.5 A

Explanation:

Current: This is defined as the rate flow of electric charge round a circuit,

The S.I unit of current is Ampere (A). Mathematically,

Current (I) = C/t

I = C/t................  Equation 1

Where  C = capacity of the battery, t = time

Given: C= 3 Ah, t = 6 hours.

Substitute into equation 1

I = 3/6

I = 1/2

I = 0.5 A.

Hence the average current drawn = 0.5 A

You might be interested in
A bus took 8 hours to travel 639 km. For the first 5 hours, it
vladimir1956 [14]

Answer:

93 km/h

Explanation:

Given that a bus took 8 hours to travel 639 km. For the first 5 hours, it travelled at an average speed of 72 km/h

Let the first 5 hours journey distance = F

From the formula of speed,

Speed = distance/time

Substitute speed and time

72 = F/5

F = 72 × 5 = 360 km

The remaining distance will be:

639 - 360 = 279km

The remaining time will be:

8 - 5 = 3 hours

Speed = 279/3

Speed = 93 km/h

Therefore, the average speed for the remaining time of the journey is equal to 93 km/h

8 0
3 years ago
A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.
enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, KE_{0}  = 0.5 mu^{2}

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, KE_{f} = 0.5mv^{2}

a) Using the principle of energy conservation,

KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0
3 years ago
A lightning bolt transfers 6.0 coulombs of charge from a cloud to the ground in 2.0 x 10-3 second. what is the average current d
AlladinOne [14]
The current is defined as the amount of charge transferred through a certain point in a certain time interval:
I= \frac{Q}{\Delta t}
where
I is the current
Q is the charge
\Delta t is the time interval

For the lightning bolt in our problem, Q=6.0 C and \Delta t= 2.0 \cdot 10^{-3}s, so the average current during the event is
I= \frac{Q}{\Delta t} = \frac{6.0 C}{2.0 \cdot 10^{-3} s}=3000 A
4 0
3 years ago
A rocket sled accelerates from rest for a distance of 645 m at 16.0 m/s2. A parachute is then used to slow it down to a stop. If
inessss [21]

Answer:

the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

Explanation:

This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period

          v² = v₀² + 2 a₁ x

indicate that the initial velocity is zero

          v² = 2 a₁ x

let's calculate

          v = \sqrt {2 \ 15.0 \ 645}

          v = 143.666 m / s

now for the second interval let's find the distance it takes to stop

          v₂² = v² - 2 a₂ x₂

in this part the final velocity is zero (v₂ = 0)

         0 = v² - 2 a₂ x₂

         x₂ = v² / 2a₂

let's calculate

         x₂ = \frac{ 143.666^2 }{2 \  18.2}

         x₂ = 573 m

as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

3 0
3 years ago
On a very muddy football field, a 110kg linebacker tackles an 85kg halfback. Immediately before the collision, the linebacker is
Alchen [17]
<span>Using conservation of energy and momentum you can solve this question. M_l = mass of linebacker
M_ h = mass of halfback
V_l = velocity of linebacker
V_h = velocity of halfback

So for conservation of momentum,
rho = mv

M_l x V_li + M_h x V_hi = M_l x V_lf + M_h x V_hf

For conservation of energy (kinetic)
E_k = 1/2mv^2/ 1/2mV_li^2 + 1/2mV_{hi}^2 = 1/2mV_{lf}^2 + 1/2mV_{hf}^2

Where i and h stand for initial and final values.
We are already told the masses, \[M_l = 110kg\] \[M_h = 85kg\] and the final velocities \[V_{fi} = 8.5ms^{-1}\] and \[V_{ih} = 7.2ms^{-1} </span>
6 0
3 years ago
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