Answer is: <span>the mass of the body is 35 kg.
</span>Newton's second law of motion: F = m · a.
F - <span> force applied.
m - </span><span>mass of the object receiving the force.
</span>a<span> - the acceleration of the object.
</span>F = 350 N.
a = 10 m/s².
m = ?
m = F ÷ a.
m = 350 N ÷ 10 m/s².
m = 35 kg.
<span>One newton is the force needed to accelerate one kilogram of mass at the rate of one metre per second squared in direction of the applied force.</span>
The temperature scale used in science is the Kelvin scale. The correct option among all the options given in the question is option "a". This is the standard scale used for measuring in astronomical and thermodynamics fields. The other scales can create complications as different countries use different type of scales.
The farther apart the two objects, the weaker the gravitational attraction between them.
Answer:75 percent
Explanation:so in order tro fin d thge efficiency i used the forumla ,efficency=useful output energy/input energyx100%,in order to use this formula i needed the output,which i found by subtracting the input energy with wasted energy,that gave me the output,and after founding the output,i put that into the formula,
output energy=input energy - wasted energy
output energy=6000j-1500j
output energy=4500
put that into the formula
efficiency =output energy/input enrgy x100%
efficiency=4500/6000 multiplied by 100%
efficiency=0.75x100%
efficiency=75%
Work,
in thermodynamics, is the amount of energy that is transferred from one system
to another system without transfer of entropy. It is equal to the external
pressure multiplied by the change in volume of the system. It is expressed as
follows:<span>
W = PdV
Integrating and assuming that P is not affected
by changes in V or it is constant, then we will have:
W = P (V2 - V1)
Substituting the given values:
P = 1.0 atm = 101325 Pa
(V2 - V1) = 0.50 L =
W = 101325 N/m^3 ( 0.50) (1/1000) m^3
W = 50.66 N-m or 50.66 J
<span>
So, in the expansion process about 50.66 J of work is being done.
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