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2 years ago

How many kilojoules is this?

Physics

1 answer:

Margaret [11]2 years ago

4 0

Answer:

How to convert joules to kilojoules

One joule is equal to thousandth of a kilojoule:

1J = 0.001kJ

The energy in kilojoules E(kJ) is equal to the energy in joules E(J) divided by 1000:

E(kJ) = E(J) / 1000

Example

Convert 500 joules to kilojoules.

The energy E in kilojoules (kJ) is equal to 500 joules (J) divided by 1000:

E(kJ) = 500J / 1000 = 0.5kJ

Joules to kilojoules conversion table

Energy (J) Energy (kJ)

1 J 0.001 kJ

2 J 0.002 kJ

3 J 0.003 kJ

4 J 0.004 kJ

5 J 0.005 kJ

6 J 0.006 kJ

7 J 0.007 kJ

8 J 0.008 kJ

9 J 0.009 kJ

10 J 0.01 kJ

20 J 0.02 kJ

30 J 0.03 kJ

40 J 0.04 kJ

50 J 0.05 kJ

60 J 0.06 kJ

70 J 0.07 kJ

80 J 0.08 kJ

90 J 0.09 kJ

100 J 0.1 kJ

200 J 0.2 kJ

300 J 0.3 kJ

400 J 0.4 kJ

500 J 0.5 kJ

600 J 0.6 kJ

700 J 0.7 kJ

800 J 0.8 kJ

900 J 0.9 kJ

1000 J 1 kJ

2000 J 2 kJ

3000 J 3 kJ

4000 J 4 kJ

5000 J 5 kJ

6000 J 6 kJ

7000 J 7 kJ

8000 J 8 kJ

9000 J 9 kJ

10000 J 10 kJ

100000 J 100 kJ

hope it help u

itz by Riddhi

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3 years ago

What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec

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Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

$E = \sigma T^{4}$ (1)

Where $\sigma$ is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

$\lambda max = \frac{2.898x10^{-3} m. K}{T}$ (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

$E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}$ (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

$T = \frac{2.898x10^{-3} m. K}{\lambda max}$ (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), $\lambda max$ (450 nm) will be express in meters:

$450 nm . \frac{1m}{1x10^{9} nm}$ ⇒ $4.5x10^{-7}m$

$T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}$

$T = 6440 K$

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

$700 nm . \frac{1m}{1x10^{9} nm}$ ⇒ $7x10^{-7}m$

$T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}$

$T = 4140 K$

Comparison:

$\frac{6440 K}{4140 K} = 1.55$

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

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3 years ago

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Answer:

Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.

Explanation:

Objects at terminal velocity, only under the influence of gravity, have maximized their speed and have an acceleration of zero. Thus, neither object is accelerating.

Recall Newton's second law: $\sum {\vec {F}}=m \vec {a}$

Since acceleration for each object is zero, the sum of the force acting on each of those objects must also be zero.

Since the only forces acting on the objects are gravity and the force of air friction, in order to zero out, the force of air friction must be equal in magnitude and opposite in direction to the force of gravity.

Recall that near the surface of the earth, $F_{gravity}=mg$ , so the Force of Gravity acting on an object is directly proportional to the object's mass. (A similar argument could be made even if this were not taking place on the surface of the earth, so long as the objects were the same distance from the object providing gravitational influence).

If the masses of the objects are different, the object with the greater mass will experience a larger force of gravity, and hence a larger force of air friction at terminal velocity.

Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.

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1 year ago

I'm completely lost in physics, the Kinematics and dynamics units.. My exam is coming up.. If a biker travelling at 6.4m/s sees

Alexeev081 [22]

D=rt
when biker A catches biker B, the time they've been riding is the same, so
t=t, or d/r=d/r
the rates are 6.4 and 4.7, so
d/6.4=d/4.7
biker B is 34m ahead, so
(d+34)/6.4=d/4.7
multiply both sides by 6.4*4.7:
4.7(d+34)=6.4d
4.7d+=6.4d+159.8
1.7d=159.8
d=94 meters

Another way to think of it is that biker A gains 1.7 meters on B every second (6.4-4.7=1.5), so the time it'll take for him to gain 34 meters is 34/1.7=20 seconds. In that time, biker B travels 4.7*20=94 meters

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2 years ago