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timurjin
1 day 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:
inna [2.2K]1 day ago
7 0

Answer:

0.5 A

Explanation:

Current is measured as the flow rate of electric charge through a circuit,

with the standard unit being Ampere (A). Using the formula

Current (I) = C/t

I = C/t................  Equation 1

Where  C represents the battery's capacity, and t indicates time.

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

Plugging values into equation 1 yields

I = 3/6

I = 1/2

I = 0.5 A.

Thus, the average current utilized amounts to 0.5 A

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A 1.0 kg brick falls off a ledge of height 44m and lands on the ground at 3.0 s later.
Keith_Richards [2256]
vf^2 = 2ad
vf^2 = 2(9.81)(44m)
vf^2 = 863.28
vf = √863.28
vf = 29.4  - using equations of motion

ME = PE + KE
ME = mgh + 1/2mv^2
ME = (1)(9.81)(44) + 1/2(1)(3^2)
ME = 431.64 + 4.5
ME = 436.14 - applying the principle of energy conservation

hope this helps:)
7 0
2 days ago
You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w
Keith_Richards [2256]

Answer:

v₀ = 3.8 m/s

Explanation:

According to Newton's second law relating to the box:

∑F = m*a Formula (1)

∑F: the net force in Newton (N)

m: mass expressed in kilograms (kg)

a: acceleration measured in meters per second squared (m/s²)

Information known:

m = 2.1 kg, the mass of the box

d = 5.4m, the length of the roof

θ = 20° is the angle between the roof and the horizontal

μk = 0.51, the coefficient of kinetic friction between the box and the roof

g = 9.8 m/s², gravitational acceleration

Forces influencing the box:

The x-axis is oriented parallel to the box's movement on the roof, and the y-axis is oriented perpendicularly.

W: Weight of the box: directed vertically

N: Normal force: perpendicular to the roof's angle

fk: Frictional force: parallel to the direction along the roof

Calculating the weight of the box:

W = m*g = (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y components of weight:

Wx= Wsin θ=(20.58)*sin(20)°=7.039 N

Wy= Wcos θ=(20.58)*cos(20)°= 19.34 N

Finding the Normal force:

∑Fy = m*ay ay = 0

N-Wy = 0

N=Wy = 19.34 N

Calculating the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We substitute into Formula (1) to determine the box's acceleration:

∑Fx = m*ax ax=a: acceleration of the box

Wx-fk = (2.1)*a

7.039 - 9.86 = (2.1)*a

-2.821 = (2.1)*a

a=(-2.821)/(2.1)

a = -1.34 m/s²

Considering the box's Kinematics:

Since the box undergoes uniformly accelerated motion, we use the following to find the final speed of the box:

vf² = v₀² + 2*a*d Formula (2)

Where:

d refers to displacement = 5.4 m

v₀ is the initial speed

vf represents the final speed = 0

a is the box's acceleration = -1.34 m/s²

Plugging in the values into Formula (2):

0² = v₀² + 2*(-1.34)*(5.4)

2*(1.34)*(5.4) = v₀²

v_{o} =\sqrt{14.472}

v₀ = 3.8 m/s

7 0
29 days ago
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Yuliya22 [2420]

Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

It is known that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v are constant on both sides. Therefore we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2..............1

let the final pressure be P and the temperature T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}.............3

divide equation (2) by equation (3)

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, the left side temperature equals 1.48 times the right side temperature

6 0
12 days ago
Two planets having equal masses are in circular orbit around a star. Planet A has a smaller orbital radius than planet B. Which
serg [2593]

Answer:

Explanation:

To approach this problem, we need to understand two key concepts.

First, the gravitational force on an object in orbit equals its mass multiplied by centripetal acceleration.

Secondly, Newton's law of universal gravitation defines the force between two masses: Fg = mMG/r², where Fg denotes gravitational force, m and M signify the masses, G represents the gravitational constant, and r indicates the distance separating the two masses.

Thus:

Fg = m v²/r

mMG/r² = m v²/r

v² = MG/r

Potential energy for each planet is expressed as:

PE = mgr = m (MG/r²) r = mMG/r

Kinetic energy for each planet is computed as:

KE = 1/2 mv² = 1/2 m (MG/r) = 1/2 mMG/r

Total mechanical energy is calculated as:

ME = PE + KE = 3/2 mMG/r

Since both planets share the same mass, the only variable is their orbital radius. Consequently, Planet A, with a smaller radius, possesses greater potential, kinetic, and mechanical energy.

6 0
1 month ago
The spring in a retractable ballpoint pen is 1.8 cm long, with a 300 N/m spring constant. When the pen is retracted, the spring
Sav [2226]

Answer:

The pen requires 7.2 mJ of energy to extend.

Explanation:

Provided:

Length = 1.8 cm

Spring constant = 300 N/m

Initial compression = 1.0 mm

Additional compression = 6.0 mm

Total compression = 1.0 + 6.0 = 7.0 mm

We need to determine the energy needed

This energy is equivalent to the variation in spring potential energy

E=PE_{2}-PE_{1}

E=\dfrac{1}{2}kx_{2}^2-\dfrac{1}{2}kx_{1}^2

Substitute the values into the formula

E=\dfrac{1}{2}\times300\times(7.0\times10^{-3})^2-\dfrac{1}{2}\times300\times(1.0\times10^{-3})^2

E=0.0072\ J

E=7.2\ mJ

Therefore, a total of 7.2 mJ is needed to extend the pen.

7 0
1 month ago
Read 2 more answers
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