247.11 Kelvin, you have to convert the 34 Fahrenheit to get to the answer that you need.
The net force on q2 will be 1.35 N
A force in physics is an effect that has the power to alter an object's motion. A mass-containing object's velocity can vary, or accelerate, as a result of a force. Intuitively, a push or a pull can also be used to describe force. Being a vector quantity, a force has both magnitude and direction.
Given Particles q1, q2, and q3 are in a straight line. Particles q1 = -5.00 x 10-6 C,q2 = +2.50 x 10-6 C, and q3 = -2.50 x 10-6 C. Particles q₁ and q2 are separated by 0.500 m. Particles q2 and q3 are separated by 0.250 m.
We have to find the net force on q2
At first we will find Force due to q1
F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.5²
F = 450 × 10⁻³
F₁ = 0.45 N (+)
Now we will find Force due to q2
F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.25²
F = 1800 × 10⁻³
F₂ = 1.8 N (-)
So net force (F) will be
F = F₂ - F₁
F = 1.8 - 0.45
F = 1.35 N
Hence the net force on q2 will be 1.35 N
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A colloid I think. Don’t hold it against me if I’m wrong my dude.
To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.
The vector product between the Force and the radius allows us to obtain the torque, in this way,





Therefore the torque on the particle about the origen is 50k
PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,





Therefore the angle between the ratio and the force is 103.88°
Answer:
The specific heat capacity is q_{L}=126.12kJ/kg
The efficiency of the temperature is n_{TH}=0.67
Explanation:
The p-v diagram illustration is in the attachment
T_{H} means high temperature
T_{L} means low temperature
The energy equation :
= R*
in(
/
)



The specific heat capacity:
=q_{h}*(T_{L}/T_{H})
q_{L}=378.36 * (400/1200)
q_{L}=378.36 * 0.333
q_{L}=126.12kJ/kg
The efficiency of the temperature will be:
=1 - (
/
)
n_{TH}=1-(400/1200)
n_{TH}=1-0.333
n_{TH}=0.67