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3 years ago

How many Joules of energy are required to run a 100W light bulb for one day?

Physics

1 answer:

Vlada [557]3 years ago

4 0

The following are the answers to the questions presented:

a. The joules of energy required to run a 100W light bulb for one day is 8640000J
b. The amount of coals that has to be burned to light that light bulb for one day is 0.96kg

The solution would be like this for this specific problem:

P=W/s→W=Pt=100W1day 24h/1day 3600s/1h=8640000J

W=30/100wm→m=100W/30w=100×8640000J/30×30×10in thepowerof6 J/kg=0.96kg

I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

You might be interested in

Doctor prescribed drug which is 500mg for patient change this unit into gram

ololo11 [35]

Answer:

.5 grams

Explanation:

1 gram is equal to 1000 milligrams (mg)

6 0

3 years ago

How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0

2 years ago

The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a

likoan [24]

Answer:

14869817.395 m

Explanation:

$\theta$ =22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

$22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians$

From Rayleigh Criterion

$\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m$

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0

2 years ago

022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati

svetoff [14.1K]

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

$v_f = 24.5 m/s$

Explanation:

Part 1)

initial speed of the ball upwards

$v_i = 24.5 m/s$

so maximum height of the ball is given by

$H = \frac{v_i^2}{2g}$

$H = \frac{24.5^2}{2(9.80)}$

$H = 30.6 m$

Part 2)

As we know that final speed will be zero at maximum height

so we will have

$v_f - v_i = at$

$0 - 24.5 = (-9.8)t$

$t = 2.5 s$

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

$v_f = 24.5 m/s$

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

$y = \frac{1}{2}at^2$

$4.88 = \frac{1}{2}a(1^2)$

$a = 9.76 m/s^2$

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

$a = 90 m/s^2$

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

$v_f = v_i + at$

$v_f = 0 + (90)(0.33)$

$v_f = 29.7 m/s$

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}(9.81)(2.3^2)$

$y = 25.95 m$

Part 2)

Distance traveled by it in horizontal direction is given as

$d = v_x t$

$d = 2.92 \times 2.3$

$d = 6.72 m$

6 0

3 years ago

In January 2006, astronomers reported the discovery of a planet comparable in size to the earth orbiting another star and having

Orlov [11]

Answer:

R = 5.28 103 km

Explanation:

The definition of density is

ρ = m / V

V = m /ρ

Where m is the mass and V the volume of the body

The volume of a sphere is

V = 4/3 π r³

Let's replace

4/3 π r³ = m / ρ

R =∛ ¾ m / ρ π

The mass of the planet is

M = 5.5 Me

R = ∛ ¾ 5.5 Me /ρ π

Let's reduce the density to SI units

ρ = 1.76 g / cm³ (1 kg / 10³ g) (10² cm / 1 m)³

ρ = 1.76 10³ kg / m³

Let's calculate

R = ∛ ¾ 5.5 5.97 10²⁴ / (1.76 10³ pi)

R = ∛ 0.14723 10²¹

R = 0.528 10⁷ m

R = 0.528 104 km

R = 5.28 103 km

8 0

3 years ago