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vova2212 [387]
3 years ago
8

Describe using examples how objects can be at rest and in motion simultaneously

Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
5 0
An object can be at rest and still be in motion because the earth is always in motion.

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Giving brainliest!! if you answer correctly :) (30pts)
AnnZ [28]

Answer:

32000joule.

Explanation:

given

mass. (m)=160kg

speed (v)=20m/s

now

kinetic energy =1/2 (mv²)=1/2 ×{160×20²}=32000joule.

8 0
3 years ago
Read 2 more answers
4. show your understanding of electric and magnetic forces<br> below.
Vitek1552 [10]

Answer:

Explanation:

Electric forces exist among stationary electric charges; both electric and magnetic forces exist among moving electric charges. ... The magnetic force between two moving charges may be described as the effect exerted upon either charge by a magnetic field created by the other.

8 0
2 years ago
Water boiling<br><br> Which one is shown? (Look at pic)
Mkey [24]

Answer:

conduction.

Explanation:

Hoped I helped! Im Eve btw have a great day and consider marking this brainliest if you do thank you in advanced!

4 0
3 years ago
0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained a
klio [65]

Answer:

The boiling point temperature of this substance when its pressure is 60 psia is  480.275 R

Explanation:

Given the data in the question;

Using the Clapeyron equation

(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}

(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }

where h_{fg is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu

T is the temperature ( 15 + 460 )R

m is the mass of water ( 0.5 Ibm )

V_{fg is specific volume ( 1.5 ft³ )

we substitute

(\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5}) / ( (15+460)\frac{1.5}{0.5})  

(\frac{dP}{dT} )_{sat } = 272.98 Ibf-ft²/R

Now,

(\frac{dP}{dT} )_{sat } = (\frac{P_2 - P_1}{T_2 - T_1})_{sat

where P₁ is the initial pressure ( 50 psia )

P₂ is the final pressure ( 60 psia )

T₁ is the initial temperature ( 15 + 460 )R

T₂ is the final temperature = ?

we substitute;

T_2 = ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}

T_2 = 475 + 5.2751\\

T_2 = 480.275 R

Therefore, boiling point temperature of this substance when its pressure is 60 psia is  480.275 R

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3 years ago
Which of the following are not units of frequency: Hertz. m/s. waves/s
ra1l [238]

The answer is waves.

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