An object has a mass of 8.00kg. What is the gravitational force on the object by the earth

Write down the DE of a simple harmonic oscillator (for example, a mass-spring system) Draw a diagram to show a mass spring syste

Sav [38]

Answer:

let m be the mass of the object, K be the force constant and Fs be the force by the spring on the mass.

7 0

3 years ago

What is the average velocity of the object?

Andrej [43]

Answer:the rate changes during the position of the object

Explanation:so there is no object that has the same rate but unless it is a specific one like a care but it changes during the position of the object

3 0

3 years ago

A 142 kg boulder is located near the edge of a cliff 25 m above the ground. What is the gravitational potential energy of the bo

jarptica [38.1K]

PE = mass * height * 9.81
PE = 142 * 25 * 9.81
PE = 34825.5 J

8 0

3 years ago

Read 2 more answers

A sinusoidal electromagnetic wave is propagating in vacuum. At a given point P and at a particular time, the electric field is i

olasank [31]

Answer:

a

The direction of the wave propagation is the negative z -axis

b

The amplitude of electric and magnetic field are $A_E= 3.35*10^5 V/m$ ,

$A_M= 1.12 *10^{-3} T$ respectively

Explanation:

According to right hand rule, your finger (direction of electric field) would be pointing in the positive x-axis i.e towards your right let your palms be face toward the direction of the magnetic field i.e negative y-axis (toward the ground ) Then anywhere your thumb stretched out is facing is the direction of propagation of the wave here in this case is the negative z -axis

The Intensity of the wave is mathematically represented as

$I = \frac{1}{2} c \epsilon _O E_{rms}^2$

Given that $I = 7.43 \frac{kW}{cm^2} = 7.43 \frac{*10^3}{*10^-{4} }= 7.43*10^7 \frac{W}{m^2}$

Making $E_{rms}$ the subject we have

$E_{rms} = \sqrt{\frac{I}{0.5*c*\epsilon_o} }$

Substituting values as given on the question

$E_{rms} = \sqrt{\frac{7.43 *10^7[\frac{W}{m^2} ]}{0.5 * 3.08*10^8 *8.85*10^{-12}} }$

$= 2.37*10^5 \ V/m$

The amplitude of the electric field is mathematically represented as

$A_E = \sqrt{2} * E_{rms}$

$= \sqrt{2} * 2.37*10^5$

$A_E= 3.35*10^5 V/m$

The amplitude of the magnetic field is mathematically represented as

$A_M = \frac{A_E}{c}$

Substituting value

$A_M = \frac{3.35 *10^5}{3.0*10^8}$

$A_M= 1.12 *10^{-3} T$

7 0

3 years ago

Scuba divers are warned that if they must make a rapid ascent, they should exhale on the way up. If a diver rapidly ascends to t

Whitepunk [10]

Air for a diver comes out of a high pressure tank at - Same- pressure compared to the water around the diver (metered by the regulator).

This means the lungs are inflated with - Highly pressurized- gas.

This does not adversely affect the diver when deep underwater, because the entire environment around the diver is at -Same - pressure.

If the diver suddenly surface, the air in the alveoli in the lungs will still be at - a higher - pressure compared to the air around the diver, which will be at - a lower - pressure.

The gas in the diver's lungs will - expand - and can damage the alveoli.

5 0

3 years ago