Answer:
A
Explanation:
since its the tip, you will feel it the most there.
Radio waves are the waves with the lowest energy in the electromagnetic spectrum. X-rays and gamma rays are the highest. Sound is not part of the electromagnetic spectrum.
Answer:
Explanation:
Due to first charge , electric field at origin will be oriented towards - ve of y axis.
magnitude
Ey = -8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j
= - 31.6 j N/C
Due to second charge electric field at origin
= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.2²+ .6²
= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.8
= 18 N/C
It is making angle θ where
Tanθ = .6 / 1.2
= 26.55°
this field in vector form
= - 18 cos 26.55 i - 18 sin26.55 j
= - 16.10 i - 8.04 j
Total field
= - 16.10 i - 8.04 j + ( - 31.6 j )
= -16.1 i - 39.64 j .
Ex = - 16.1 i
Ey = - 39.64 j .
The possible magnitude for the force of static friction on the stationary cart is 72.1 N.
The given parameters:
- <em>Applied force on the cart, F = 72.1 N</em>
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Based on Newton's second law of motion, the force applied to object is directly proportional to the product of mass and acceleration of the object.
F = ma
Static frictional force is the force resisting the motion of an object at rest.

where;
is the frictional force

Thus, the possible magnitude for the force of static friction on the stationary cart is 72.1 N.
Learn more about Newton's second law of motion: brainly.com/question/25307325
Given Information:
Power = P = 100 Watts
Voltage = V = 220 Volts
Required Information:
a) Current = I = ?
b) Resistance = R = ?
Answer:
a) Current = I = 0.4545 A
b) Resistance = R = 484 Ω
Explanation:
According to the Ohm’s law, the power dissipated in the light bulb is given by

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.
Re-arranging the above equation for current I yields,

Therefore, 0.4545 A current is flowing through the light bulb.
According to the Ohm’s law, the voltage across the light bulb is given by

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.
Re-arranging the above equation for resistance R yields,

Therefore, the resistance of the bulb is 484 Ω