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3 years ago

Use these words in a sentence proton neutron and isotope

Physics

1 answer:

Hunter-Best [27]3 years ago

4 0

"The proton and neutron have nothing to do with the isotope little billy"

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Two forces one of 12N and another of 5N act on a body in such away that they makes an angle of 90 digre with each other,what is

ELEN [110]

Answer:

Sorry I didn't know the answer

6 0

3 years ago

A boulder of mass 2000. kg breaks loose of a mountainside and plunges 200. m straight down into a lake at a temperature of 5.00◦

natka813 [3]

Answer:

DS = 13865.7[J/K]

Explanation:

We can calculate the energy of the rock, like the potential energy relative to the lake level. Which can be calculated by means of the following expression of the potential energy:

$E_{p}=m*g*h\\\\where:\\m = mass = 2000[kg]\\h = elevation = 200 [m]\\g = gravity = 9.81[m/s^2]$

Therefore:

$E_{p}=2000*9.81*200\\E_{p}=3924000 [J]\\$

This energy is transformed into thermal energy.

we shall remember that isothermal heat transfer processes are internally reversible, so the entropy change of a system during one of these processes can be determined, by the following expression.

$DS=\frac{Q}{T}\\ where:\\DS = entropy change [J/K]\\Q = Heat transfer [J]\\T = temperature [K]$

T = 5 + 278 = 283[K]

DS = 3924000 / 283

DS = 13865.7[J/K]

5 0

3 years ago

A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if

Dmitrij [34]

To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

$F_p = 6*10^{22}N$

$F_p = \frac{GMm}{R^2}$

$F_p = 9*10^{22}N$

Distance between planet and star

$r = \frac{R}{2}$

Gravitational force is

$F = \frac{GMm}{r^2}$

Applying the new distance,

$F = \frac{GMm}{(\frac{R}{2})^2}$

$F = 4\frac{GMm}{R^2}$

Replacing with the previous force,

$F = 4F_p$

Replacing our values

$F= 4(9*10^{22}N)$

$F = 36*10^{22}N$

Therefore the magnitude of the force on the star due to the planet is $36*10^{22}N$

5 0

3 years ago

You observe a hockey puck of mass 0.12 kg, traveling across the ice at speed 18.3 m/sec. The interaction of the puck and the ice

Galina-37 [17]

The stopping distance is 143.1 m

Explanation:

First of all, we have to find the acceleration of the hockey puck. This can be done by using Newton's second law of motion:

$\sum F =ma$

where

$\sum F = F_f = -0.14 N$ is the net force acting on the puck (the force of friction, negative because it acts in a direction opposite to the direction of motion)