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2 years ago

Which is heavier , 1kg of steel or i kg of cork

Physics

2 answers:

goblinko [34]2 years ago

6 0

Answer:

Same weight so neither is heavier

LenaWriter [7]2 years ago

5 0

Since their masses are equal, their weights are equal whenever they're both on the same planet.

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Why don't atoms get too close?

Lera25 [3.4K]

Answer:

I would have to go with A, or maybe....yea A

3 0

2 years ago

Read 2 more answers

A light beam travels at 1.94×108 in quartz. The wavelength of the light in quartz is 355 .Part AWhat is the index of refraction

Alja [10]

A) 1.55

The speed of light in a medium is given by:

$v=\frac{c}{n}$

where

$c=3\cdot 10^8 m/s$ is the speed of light in a vacuum

n is the refractive index of the material

In this problem, the speed of light in quartz is

$v=1.94\cdot 10^8 m/s$

So we can re-arrange the previous formula to find n, the index of refraction of quartz:

$n=\frac{c}{v}=\frac{3\cdot 10^8 m/s}{1.94\cdot 10^8 m/s}=1.55$

B) 550.3 nm

The relationship between the wavelength of the light in air and in quartz is

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda$ is the wavelenght in quartz

$\lambda_0$ is the wavelength in air

n is the refractive index

For the light in this problem, we have

$\lambda=355 nm\\n=1.55$

Therefore, we can re-arrange the equation to find $\lambda_0$ , the wavelength in air:

$\lambda_0 = n\lambda=(1.55)(355 nm)=550.3 nm$

4 0

3 years ago

A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

s344n2d4d5 [400]

Answer:

At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.