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3 years ago

How to calculate kinetic energy given mass and velocity

1 answer:

3 0

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

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These UV photons can break chemical bonds in your skin to cause sunburn—a form of radiation damage. If the 305 nm radiation prov

astraxan [27]

Answer:

energy = 391.902 kJ /mol

Explanation:

given data

wavelength = 305 nm = 305 × $10^{-9}$ m

to find out

average energy

solution

we know speed of light is 3 × $10^{8}$ m/s

so we find frequency here first by speed of light formyla

speed = wavelength × frequency

3 × $10^{8}$ = 305 × $10^{-9}$ × frequency

frequency = 9.8360 × $10^{14}$ $s^{-1}$

so energy is

energy = hf

here h = 6.62 × $10^{-34}$ J-s

energy = 6.62 × $10^{-34}$ × 9.8360 × $10^{14}$

energy = 6.51 × $10^{-14}$ J

energy = 6.51 × $10^{-14}$ × $\frac{6.02*10^23}{1000}$ kJ/mol

energy = 391.902 kJ /mol

7 0

4 years ago

Acceleration on moon is approximately 1/6th of the Earth's; what would be the force of your egg drop package if dropped on the m

cricket20 [7]

Acceleration on earth=10m/s^2

Acceleration on moon:-

$\\ \rm\Rrightarrow 10\dfrac{1}{6}=1.6m/s^2$

Now

Mass=0.56kg

$\\ \rm\Rrightarrow F=ma$

$\\ \rm\Rrightarrow F=0.56(1.6)$

$\\ \rm\Rrightarrow F=0.896N$

8 0

3 years ago

Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te

alexgriva [62]

Answer:

4.9 minutes

Explanation:

Given; T(t) = Ce^-kt + Ts

Now;

T(t) = 190 degrees Fahrenheit

Ts = 60 degrees

To obtain C;

190 = Ce^0 + 60

190 - 60 = C

C = 130

Hence, to find k when t=11

172 = 130 e^-11k + 60

172 -60/130 = e^-k

e^-k = 0.86

ln(e^-k) = ln( 0.86)

-k = -0.15

k = 0.15

Hence at 122 degrees, t is;

T(t) = Ce^-kt + Ts

122 = 130e^-0.15t + 60

122 - 60/130 = e^-0.15t

0.477 = e^-0.15t

ln (e^-0.15t) = ln (0.477)

-0.15t = -0.74

t = 0.74/0.15

t = 4.9 minutes

5 0

3 years ago

The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on

sashaice [31]

To solve this problem we can apply the concept related to thermal expansion, including the analogy with resistance and final intensity.

The mathematical expression that describes the expansion of a material by a thermal process is given by

$R = R_0\alpha \Delta T$

Where

$R_0$ = Initial resistance

$\alpha =$ Thermal expansion coefficient

$\Delta T =$ Change in the temperature

If we want to directly obtain the final value of the resistance of the object, you would simply add the initial resistance to this equation - because at this moment we have the result of how much resistance changed, but not of its final resistance - So,

$R_f = R_0 + L_0\alpha \Delta T$

$R_f = R_0(1 + \alpha \Delta T)$

Re-arrange to find the change at the temperature,

$\Delta T=\frac{1}{\alpha}\frac{R_f}{R_0}-1}$

Since the resistance is inversely proportional to the current and considering that the voltage is constant then

$R \propto \frac{1}{I}$

Then,

$\Delta T=\frac{1}{\alpha}\frac{I_0}{I_f}-1}$

$\Delta T = \frac{1}{4.5*10^{-3}}(\frac{I_0}{I_0/8}-1)$

$\Delta T = \frac{1}{4.5*10^{-3}}(8-1)$

$\Delta T = 1555.5k$

(It is possible that there is a typing error and the value is not 4.5 but 4.3, so the closest approximate result would be 1627K and mark this as the correct answer)

6 0

3 years ago

An object is placed a great distance from a concave (converging) mirror. Where will light from the object be collected?

AveGali [126]

Answer: at the center of curvature of the mirror on the same side of the mirror as the object

A concave mirror has a reflective surface that is curved inwards. This type of mirrors reflects the light making it converge in a focal point, therefore they are used to focus the light.

This occurs because the light is reflected with different angles, since the normal to the surface varies from one point to another of the mirror.

Nevertheless, it is important to note the object must be within the radius of curvature of the mirror.

In addition, it is important to state clear the following:

-If the object is at a distance greater than the focal distance, a real and inverted image is formed that may be greater or less than the object.

-If the object is at a distance smaller than the focal distance, a virtual image is formed, right and larger than the object.

In this case the object is placed a great distance (it can be said at infinite) from the concave mirror, hence the image formed will be real, inverted and smaller than the object.

When we say real, it means the image is formed in the same side of the mirror as the object and the image can be seen on a screen.

5 0

3 years ago

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