Answer:
0.72
Explanation:
= Time period of oscillation = 1.5 s
Angular frequency is given as
= Amplitude of oscillation = 40 cm = 0.40 m
= Coefficient of static friction = ?
= acceleration of the block
= mass of the block
Maximum acceleration of the block is given as
frictional force is given as
As per newton's second law
Answer:
Zero
Explanation:
The work done by a force on an object is given by:
where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and the displacement of the object
In this situation, the force is the force of gravity acting on the satellite. This force always points towards the centre of the trajectory, so it is always perpendicular to the direction of motion of the satellite (since the orbit is circular), so 
and 
. Therefore, the work done by gravity is also zero.
The variable you can change in an experiment is ( B ) A dependent variable
To verify the identity, we can make use of the basic trigonometric identities:
cot θ = cos θ / sin θ
sec θ = 1 / cos <span>θ
csc </span>θ = 1 / sin θ<span>
Using these identities:
</span>cot θ ∙ sec θ = (cos θ / sin θ ) (<span> 1 / cos </span><span>θ)
</span>
We can cancel out cos <span>θ, leaving us with
</span>cot θ ∙ sec θ = 1 / sin θ
cot θ ∙ sec θ = = csc <span>θ</span>
V=IR
V=15x11
V=165ohms
I don’t quite remember the unit
Ohms law
