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2 years ago

13

an incandescent lightbulb has an efficiency of 2.1% and a power of 60 w. how much light energy does the lighbulb produce in 1 se

cond

1 answer:

AveGali [126]2 years ago

5 0

Explanation:

Power output of the bulb:

0.021 × 60 W = 1.26 W

Energy produced by the bulb in 1 second:

E = Pt

E = (1.26 W) (1 s)

E = 1.26 J

Round as needed.

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A 2 kg ball is moving 3 m/s when it starts rolling up a hill.

AURORKA [14]

Answer:

the height reached is = 0.458 [m]

Explanation:

We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:

$Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]$

Replacing the values on the equation we have:

$Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\$

This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.

$Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\$

Now we have:

$h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]$

In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]

6 0

2 years ago

Two spheres are cut from a certain uniform rock. One has radius 4.50 cm. The mass of the other is five times greater. Find its r

Sonja [21]

Vol of sphere = 4/3 pi r^2.density of sphere = mass/volume.mass = densityxvolumesphere 1. mass = density x 4/3 pi 4.5^2sphere 2 5mass = density x 4/3 pi r^25=4/3 pi r^2 divided by 4/3 pi 4.5^25=r^2 divided by 4.5^25x4.5^2=r^2root(5x4.5^2)=r4.5 root 5 = r

4 0

2 years ago

Read 2 more answers

A basketball leaves a player's hands at a height of 2.00 m above the floor. The basket is 3.05 m above the floor. The player lik

shutvik [7]

Answer:(10.69, 11.436)

Explanation:

Given

initial height of ball is 2 m

height of basket is 3.05 m

Launching angle $=40^{\circ}$

$x =12\pm 0.27$

y=1.05

equation of trajectory of ball is given by

$y=xtan\theta -\frac{gx^2}{2u^2cos^2\theta }$

for x=12.27

$1.05=12.27\times tan40-\frac{g12.27^2}{2u^2cos^{2}40 }$

u=10.69

for x=11.73

$1.05=11.73\times tan40-\frac{g11.73^2}{2u^2cos^{2}40 }$

u=11.436 m/s

Thus range of speed is (10.69, 11.436)

3 0

2 years ago

A very long train is rolling at 4 m/s along a straight track. An observer is standing on the ground very dangerously close to th

hichkok12 [17]

Answer:

A. $\vec{r}=(6\frac{m}{s})t\ \ \hat{i}$

B. t = 50 s

Explanation:

A. The vectorial equation of the person who is getting closer to the other person is:

$\vec{r}=\vec{v}t$

r: position vector

v: speed vector = 6m/s i (if you consider the motion as a horizontal motion)

Then, you replace and obtain:

$\vec{r}=(6\frac{m}{s})t\ \ \hat{i}$

B. The time is:

$t=\frac{d}{v}$

d: distance to the observer = 300m

v: speed of the person on the car = 6.00 m/s

$t=\frac{300m}{6m/s}=50s$

4 0

2 years ago

A 5.00 kg object oscillates back and forth at the end ofa spring whose spring constant is 49.3 N/m. An obersever istraveling at

defon

Answer:

5.571 sec

Explanation:

angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s

Period To = 2π / angular frequency

Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got

T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec

t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been ( To / (√ (1 - (v²/c²))). where To = 2.00 sec

8 0

2 years ago