When an unbalanced force acts on an object,

A ray of light bends upon entering a new medium. what is this phenomenon called?

Alina [70]

If C.) is meant to be refraction, it's C.)

3 0

3 years ago

Determine the amount of time for polonium-210 to decay to one fourth its original quantity. The half-life of polonium-210 is 138

ira [324]

Answer: 276 days

Explanation:

This problem can be solved using the Radioactive Half Life Formula:

$A=A_{o}.2^{\frac{-t}{h}}$ (1)

Where:

$A=\frac{1}{4}A_{o}$ is the final amount of the material

$A_{o}$ is the initial amount of the material

$t$ is the time elapsed

$h=138 days$ is the half life of polonium-210

Knowing this, let's substitute the values and find $t$ from (1):

$\frac{1}{4}A_{o}=A_{o}2^{\frac{-t}{138 days}}$ (2)

$\frac{A_{o}}{4A_{o}}=2^{\frac{-t}{138 days}}$ (3)

$\frac{1}{4}=2^{\frac{-t}{138 days}}$ (4)

Applying natural logarithm in both sides:

$ln(\frac{1}{4})=ln(2^{\frac{-t}{138 days}})$ (5)

$-1.386=-\frac{t}{138days}ln(2)$ (6)

Clearing $t$ :

$t=276days$ (7)

7 0

3 years ago

A 108 kg clock initially at rest on a horizontal floor requires a 639 N horizontal force to set it in motion. After the clock is

labwork [276]

Answer:

$\mu_s=0.60$

Explanation:

It is given that,

Mass of the clock, m = 108 kg

Force acting on it when it is in motion, $F=639\ N$

After the clock is in motion, a horizontal force of 521 N keeps it moving with a constant velocity, F' = 521 N

It is assumed to find the coefficient of between the clock and the floor. The force of friction is given by :

$F=\mu_smg$

$\mu_s=\dfrac{F}{mg}$

$\mu_s=\dfrac{639\ N}{108\ kg\times 9.8\ m/s^2}$

$\mu_s=0.60$

So, the coefficient of static friction between the clock and the floor is 0.6. Hence, this is the required solution.

5 0

3 years ago

is weigh-in time for the local under-85-kg rugby team. The bathroom scale used to assess eligibility can be described by Hooke’s

yulyashka [42]

Answer:

a) 156960 N/m

b) Yes the player can play on the team.

Explanation:

F = Force

a = Acceleration due to gravity = 9.81 m/s²

m = Mass

k = Spring constant

x = Deformation length of spring

F = ma

From Hooke's law

$F=kx\\\Rightarrow ma=kx\\\Rightarrow k=\frac{ma}{x}\\\Rightarrow k=\frac{120\times 9.81}{0.0075}\\\Rightarrow k=156960\ N/m$

The effective spring constant is 156960 N/m

When x = 0.48 cm = 0.0048 m

$m=\frac{kx}{a}\\\Rightarrow m=\frac{156960\times 0.0048}{9.81}\\\Rightarrow m=76.8\ kg$

The mass of the player is 76.8 kg which is less than the required mass limit of the players' which is 85 kg. So, the player is eligible to play.

5 0

3 years ago

What is peer review?...

Bond [772]

<span>computing or networking is a distributed application architecture that partitions tasks or work loads between peers. Peers are equally privileged, equipotent participants in the application. They are said to form a peer-to-peer network of nodes.</span>

6 0

3 years ago

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