Answer:
Let I and j be the unit vector along x and y axis respectively.
Electric field at origin is given by
E= kq1/r1^2 i + kq2/r2^2j
= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)
= (2.88i + 1.44j)*10^-3 N/C
Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3
F= (1.382 i + 0.691 j) *10^-21
Goodluck
Explanation:
Answer:
See explanation below.
Explanation:
The internal energy (kinetic energy) of the milk particles increases thus warming the milk up, while the internal energy of the coffee particles reduce their initial average speed as they are imparting energy to the slow milk particles.
Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Answer:
Heat Input = Work Output (at 100% efficiency)
ΔQ = ΔW
(you cannot get something for nothing)
