What’s 4.5 inches into centimeters?

2. An electric motor transfers 2003 in 10.0 s. If it then transfers the same amount of energy in 8.0 s, has its

Marizza181 [45]

Answer:

decreased

Explanation:

because it transfered from 10.0s

to 8.0s

5 0

2 years ago

PLEASE ANSWER ASAP

Mekhanik [1.2K]

Answer:

yes h h jj h v gu i kj

Explanation:

jjejejejejj

8 0

3 years ago

One ball of mass 0.600kg travelling 9.00m/s to the right collides head on elastically with a second ball of mass 0.300kg travell

Alina [70]

Let m₁ and v₁ denote the mass and initial velocity of the first ball, and m₂ and v₂ the same quantities for the second ball. Momentum is conserved throughout the collision, so

m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'

where v₁' and v₂' are the balls' respective velocities after the collision.

Kinetic energy is also conserved, so

1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁')² + 1/2 m₂ (v₂')²

or

m₁ v₁² + m₂ v₂² = m₁ (v₁')² + m₂ (v₂')²

From the momentum equation, we have

(0.600 kg) (9.00 m/s) + (0.300 kg) (-8.00 m/s) = (0.600 kg) v₁' + (0.300 kg) v₂'

which simplifies to

10.0 m/s = 2 v₁' + v₂'

so that

v₂' = 10.0 m/s - 2 v₁'

From the energy equation, we have

(0.600 kg) (9.00 m/s)² + (0.300 kg) (-8.00 m/s)² = (0.600 kg) (v₁')² + (0.300 kg) (v₂')²

which simplifies to

67.8 J = (0.600 kg) (v₁')² + (0.300 kg) (v₂')²

or

226 m²/s² = 2 (v₁')² + (v₂')²

Substituting v₂' yields

226 m²/s² = 2 (v₁')² + (10.0 m/s - 2 v₁')²

which simplifies to

3 (v₁')² - (20.0 m/s) v₁' - 63.0 m²/s² = 0

Solving for v₁' using the quadratic formula gives two solutions,

v₁' ≈ -2.33 m/s or v₁' = 9.00 m/s

but the second solution corresponds to the initial conditions, so we omit that one.

Then the second ball has velocity

v₂' = 10.0 m/s - 2 (-2.33 m/s)

v₂' ≈ 14.7 m/s

6 0

2 years ago

Comparing ultraviolet photons to radio photons, we can say that the ultraviolet photons have ______ energy, ______ frequency, an

guajiro [1.7K]

Answer:

Energy of ultra violet photons < Energy of radio waves

Wavelength of ultra violet photons > Wavelength of radio waves

Explanation:

The increasing order of the wavelength is given by

Gamma rays

x rays

Ultra violet rays

Visible radiations

Infrared rays

micro waves

Radio waves

So, ultra violet rays has larger wavelength than radio waves.

As we know that the energy is inversely proportional to the wavelength.

So, the energy of radio waves is more than the energy of ultra violet waves.

Energy of ultra violet photons < Energy of radio waves

Wavelength of ultra violet photons > Wavelength of radio waves

7 0

3 years ago

The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:

kow [346]

Answer:

$\frac{d_{1}}{d_{2}}=0.36$

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

$\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]}$ (1)

So, the temperature of the first and the second star will be:

$T_{1}=3866.7 K$

$T_{2}=6444.4 K$

Now the relation between the absolute luminosity and apparent brightness is given:

$L=l\cdot 4\pi r^{2}$ (2)

Where:

L is the absolute luminosity
l is the apparent brightness
r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

$\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}$

So the relative distance between both stars will be:

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}}$ (3)

The Boltzmann Law says, $L=A\sigma T^{4}$ (4)

σ is the Boltzmann constant
A is the area
T is the temperature
L is the absolute luminosity

Let's put (4) in (3) for each star.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}$

As we know both stars have the same size we can canceled out the areas.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}$

$\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}$

$\frac{d_{1}}{d_{2}}=0.36$

I hope it helps!

5 0

3 years ago