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3 years ago

Approximately how much electrical energy does a 5-W lightbulb convert to radiant and thermal energy in one hour?

Physics

1 answer:

k0ka [10]3 years ago

8 0

Answer:

18,000 j

Explanation:

the lightbulb dissipates 5W of power

P = ΔE / Δt

rearrange to solve for energy

ΔE = PΔt

P = 5W

t = 1 hour = 60 minutes = 3600 seconds

ΔE = 5 * 3600

ΔE = 18000 J

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[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r

blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b) it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity $V_{f}$ for an object traveling distance h taking the initial vertical component of velocity as $V_{i}$ the kinematics equation is written as

$V_{f}^{2}=V_{i}^{2}+2ah$ where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

$V_{f}^{2}=V_{i}^{2}+2gh$

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

$V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}$ = 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

$V_{f}^{2}=V_{i}^{2}+2gh$

Since we have final velocity of 25 m/s

$V_{i}^{2}=2gh-V_{f}^{2}$

$V_{i}=\sqrt{(V_{f}^{2}-2gh)}$

$V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}$ = 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

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3 years ago

The main difference between turbojets and rocket engines is the fact that

mafiozo [28]

Answer;

D. rocket engines are not dependent on oxygen from the air.

Explanation;

-Jet engines and rockets work on the same principle. They produce thrust through an internal pressure difference and, as explained by Newton’s Third Law of Motion, eject exhaust gases in an equal and opposite direction.

-The main difference between them is that jets get the oxygen to burn fuel from the air and rockets carry their own oxygen, which allows them to operate in space.

Additionally, Jet engines have two openings (an intake and an exhaust nozzle). Rocket engines only have one opening (an exhaust nozzle).

5 0

3 years ago

Read 2 more answers

An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

Viefleur [7K]

Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
where
$f=5 cm$ is the focal length
$p=10 cm$ is the distance of the object from the mirror
$q$ is the distance of the image from the mirror.

Rearranging the equation, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}$
so, the distance of the image from the mirror is $q=10 cm$ .

1b) The image height is given by the magnification equation:
$\frac{h_i}{h_o}=- \frac{p}{q}$
where $h_i$ is the heigth of the image and $h_o=1 cm$ is the height of the object. By rearranging the equation and using p and q, we find
$h_i=-h_o \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm$
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

3) As before, we find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

And now we can use the magnification equation to find the image height:
$\frac{h_i}{h_o}=- \frac{p}{q}$
Rearranging it, we find
$h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm$
and the negative sign means the image is inverted.

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3 years ago

A 200 kg object is initially at rest on a horizontal frictionless surface. At time t = 0 , a horizontal force of 100 N applied t

s344n2d4d5 [400]

Answer:

Explanation:

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3 years ago

What kind of radiation is emitted in the following nuclear reaction

miv72 [106K]

Beta emission is occurring in the given nuclear reaction.

Answer: Option B

<u>Explanation:</u>

In this equation, the reactant is the Thorium atom, which is reduced to palladium. As the atomic number get decreased by one, so an electron will be emitted. This process of emission of electrons by radiation or decaying the reactant nuclei to form a new product nuclei is termed as beta emission.

So, the electrons are generally termed as beta particles while the positrons are termed as positive beta particles. So this is a kind of radioactive reactions where the reactant changes to new element by releasing an electron and thus there is a change in the atomic number of the product by one.

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3 years ago

Approximately how much electrical energy does a 5-W lightbulb convert to radiant and thermal energy in one hour?​

Approximately how much electrical energy does a 5-W lightbulb convert to radiant and thermal energy in one hour?