Answer:
The work done by the applied force is 259.22 J.
Explanation:
The work done by the applied force is given by:
Where:
F: is the applied horizontal force = 108.915 N
d: is the distance = 2.38 m
Hence, the work is:
Therefore, the work done by the applied force is 259.22 J.
I hope it helps you!
Hello!
Since the two weights are <em>off</em> the table, the block will move towards letter F.
I hope this helps :))
Answer:
E = (-3.61^i+1.02^j) N/C
magnitude E = 3.75N/C
Explanation:
In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:
(1)
Where the minus sign means that the electric field point to the charge.
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
q = -4.28 pC = -4.28*10^-12C
r: distance to the charge from the point P
The point P is at the point (0,9.83mm)
θ: angle between the electric field vector and the x-axis
The angle is calculated as follow:
The distance r is:
You replace the values of all parameters in the equation (1):
The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C
Answer:
1 W = 1 J / sec Definition of watt is 1 joule / sec
So if a bulb uses 75 J / sec it must use
75 J/s * 60 sec / min = 4500 J/min energy used by bulb
If bulb is 15% efficient then the light delivered is
P = 4500 J / min * .15 = 675 J / min
Answer:
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