I believe that fidget toys should not be brought back in school, reason being is because fidget spinners were originally made for students with example: Autism . Who actually needed them. Now that people who don't need them are buying them and playing with them in the classroom, it causes a complete distraction and is not being used for its original purpose whatsoever .
Answer:
B.The force of friction between the block and surface will decrease.
Explanation:
The force of friction is given by
where 
is the coefficient of friction and 
is the normal force.
When the student pulls on the block with force 
at an angle 
, the normal force on the block becomes
and hence the frictional force becomes
.
Now, as we increase , 
increases which as a result decreases the normal force 
, which also means the frictional force decreases; Hence choice B stands true.
<em>P.S: Choice D is tempting but incorrect since the weight </em>
<em> is independent of the external forces on the block. </em>
Answer:
Explanation:
For parallel inductors ,
For series combination
Total inductance
= 16.67 + 20
= 36.67 mH .
reactance of total inductance at 300 kHz
= ω
where ω is angular frequency
= 2πf
= 2 x 3.14 x 300 x 10³ x 36.67 x 10⁻³
= 69.1 x 10³ ohm
Total rms current = Vrms / reactance
= 60 / 69.1 x 10³ A
= .87 x 10⁻³ A
= .87 mA
Answer:
Angle θ = 30.82°
Explanation:
From Malus’s law, since the intensity of a wave is proportional to its amplitude squared, the intensity I of the transmitted wave is related to the incident wave by; I = I_o cos²θ
where;
I_o is the intensity of the polarized wave before passing through the filter.
In this question,
I is 0.708 W/m²
While I_o is 0.960 W/m²
Thus, plugging in these values into the equation, we have;
0.708 W/m² = 0.960 W/m² •cos²θ
Thus, cos²θ = 0.708 W/m²/0.960 W/m²
cos²θ = 0.7375
Cos θ = √0.7375
Cos θ = 0.8588
θ = Cos^(-1)0.8588
θ = 30.82°
Answer:
Explanation:
This is a simple gravitational force problem using the equation:
where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.
Filling in:
I'm going to do the math on the top and then on the bottom and divide at the end.
and now when I divide I will express my answer to the correct number of sig dig's:
6.45 × 10¹⁶ N
