Ask question

Ask question

All categories

elena-14-01-66 [18.8K]

3 years ago

13

How many kg are in 3000netons

1 answer:

Nat2105 [25]3 years ago

4 0

Answer: if i'm right it shloud be 1 US short ton is equal to 907.18474 kilograms (kg).

Explanation:

You might be interested in

A 40 kg skier starts at the top of a 12 meter high slope. At the bottom, she is traveling 10 miles. How much energy does she los

DaniilM [7]

2,704 J i believe. Not sure

7 0

3 years ago

Read 2 more answers

A physical pendulum consists of a uniform solid disk (of radius R 2.35 cm) supported in a vertical plane by a pivot located a di

lutik1710 [3]

Answer:

T = 0.3658

Explanation:

The expression to use to calculate the period is the following:

T = 2π √I/Mgd (1)

Where:

I: moment of Innertia of pendulum

g: gravity acceleration (9.81 m/s²)

d: distance of the pivot

M: mass of the disk.

Before we do anything, we will find first the moment of Innertia of the pendulum.

This can be calculated with the following expression:

I = 1/2 MR² + Md² (2)

At the moment we don't have the mass of the disk, but we don't need it, we will express I in function of M, and then, it will be canceled with the M of expression (1). Calculating M we have (Remember that the units of radius and distance should be in meter):

I = 1/2 M(0.0235)² + M(0.0175)²

I = (2.76x10^-4)M + (3.06x10^-4)M

I = (5.82x10^-4)M (3)

Now, we will replace this value in equation (1):

T = 2π √(5.82x10^-4)M / (9.81)*(0.0175)M ---> Here M cancels out

T = 2π √(5.82x10^-4) / (9.81)*(0.0175)

T = 2π * 0.0582

T = 0.3658 s

This is the period of the pendulum

7 0

3 years ago

Which is true of oxidation? PLS HELP FAST and thank you!

elena-14-01-66 [18.8K]

Answer:The Answer Is D

Explanation:

4 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.

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

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

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

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

I hope it helps!

5 0

3 years ago

A motorcycle of mass 250 kg drives around a circle with a centripetal acceleration of 3.4 m/s2. What is the centripetal force ac

Vlad [161]

Hope this helps and have a nice day!

8 0

3 years ago

Read 2 more answers