All categories

2 years ago

the headlamp of a car takes a current of 0.4 ampere from a 12 volt supply. the energy produced in 5 minutes is

1 answer:

5 0

Voltage (V) = 12 v
Current (I) = 0.4 A
Time (t) = 5min = 300sec

Power = Voltage x Current
= V x I
= 12 x 0.4
Power = 4.8wats = 4.8W

Power = Energy / Time
Energy = Power x Time
Energy = 4.8 x 300
Energy = 1440 joules
Final answer : 1440J

You might be interested in

How do we calculate the value of work?

Katarina [22]

The percent complete is calculated by dividing the quantity of material progressed at a point in time by the total quantity required for the project. The resulting percent is multiplied by the current agreed committed value of the material item to obtain the VOWD for that item.

7 0

3 years ago

Read 2 more answers

An electron moves in a straight path with a velocity of +500,000 m/s. It undergoes constant acceleration of +500,000,000,000 met

77julia77 [94]

Answer:

Distance travelled, d = 0.21 m

Explanation:

It is given that,

Initial velocity of electron, u = 500,000 m/s

Acceleration of the electron, a = 500,000,000,000 m/s²

Final velocity of the electron, v = 675,000 m/s

We need to find the distance travelled by the electron. Let distance travelled is s. Using third equation of motion as :

$v^2-u^2=2as$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{(675000\ m/s)^2-(500000\ m/s^2)^2}{2\times 500000000000\ m\s^2}$

s = 0.205 m

s = 0.21 m

So, the electron will travel a distance of 0.21 meters. Hence, this is the required solution.

6 0

3 years ago

What is the net force on a 116 kg woman free falling at 97 m/s who has reached terminal velocity.

natta225 [31]

It’s 545722 J , i think

4 0

3 years ago

The 0.8-Mg car travels over the hill having the shape of a parabola. If the driver maintains a constant speed of 9 m>s, deter

White raven [17]

The answer is incomplete. The complete question can be found in search engines. However, kindly find the complete question below.

Question

The 0.8-Mg car travels over the hill having the shape of a parabola. When the car is at point A, it is traveling at 9 m s and increasing its speed at . Determine both the resultant normal force and the resultant frictional force that all the wheels of the car exert on the road at this instant. Neglect the size of the car

Answer:

Recalling the fact that the statement can be related to that of geometry as represented in the diagram and illustration below

Therefore, referencing the question and calling the knowledge of geometry, we have:

dy / dx = - 0.00625x and

d²y / dx² = - 0.00625x .

Also, from the diagrammatic illustration, we can see that the slope angle tan θ at point A is given by

tan θ = dy / dx ║ ₓ = ₈₀ₙ = - 0.00625(80)

Therefore, tan θ = - 26.57°

also, if we consider the radius. The radius of curvature at point A is

ρ = [ 1 + (dx / dy )² ] ³⁺² / d² y / dx² = [ 1 + ( -0.00625x)² ] ³⁺²║ ₓ = ₈₀ₙ

Therefore, ρ = 223.61 m

Also, recalling the equation of Motion

We then apply the equation to Applying Eq. 13–8 with θ = 26.57° and ρ = 223.61 m,

we then have,

∑Ft = Mat; 800 (9.81) sin 26.57° - Ff = 800 (3)

Ff = 1109.73 N

= 1.11 kN

∑Fn = Man; 800(9.81) cos 26.57° - N = 800 (9² / 223.61)

N = 6729.67 N

= 6.73 kN

Therefore, the resultant normal force = 1.11 kN and the resultant minimal force = 6.73 kN

7 0

3 years ago

The only force acting on a 1.9 kg canister that is moving in an xy plane has a magnitude of 3.9 N. The canister initially has a

Hunter-Best [27]

Answer:

The work done on the canister is 15.34 J.

Explanation:

Given;

mass of canister, m = 1.9 kg

magnitude of force acting on x-y plane, F = 3.9 N

initial velocity of canister in positive x direction, $v_i$ = 3.9 m/s

final velocity of the canister in positive y direction, $v_j = 5.6 \ m/s$

The change in the kinetic energy of the canister is equal to net work done on the canister by 3.9 N.

ΔK.E = $W_{net}$

ΔK.E $= K.E_f -K.E_i$

The initial kinetic energy of the canister;

$K.E_i = \frac{1}{2} mv_i^2\\\\K.E_i = \frac{1}{2} m(\sqrt{v_i^2 +v_j^2 + v_z^2}\ )^2\\\\K.E_i = \frac{1}{2} *1.9(\sqrt{3.9^2 +0^2 + 0^2}\ )^2 = 14.45 \ J$

The final kinetic energy of the canister;

$K.E_f =\frac{1}{2} mv_j^2 \\\\K.E_f = \frac{1}{2} m(\sqrt{v_i^2 +v_j^2 + v_z^2}\ )^2\\\\K.E_f = \frac{1}{2} *1.9(\sqrt{0^2 +5.6^2 + 0^2}\ )^2 = 29.79 \ J$

ΔK.E = 29.79 J - 14.45 J

ΔK.E = $W_{net}$ = 15.34 J

Therefore, the work done on the canister is 15.34 J.

5 0

3 years ago

the headlamp of a car takes a current of 0.4 ampere from a 12 volt supply. the energy produced in 5 minutes is​

the headlamp of a car takes a current of 0.4 ampere from a 12 volt supply. the energy produced in 5 minutes is