To solve this problem we will use the vector concept given by the cross product between two perpendicular vectors and which results in a vector perpendicular to these two. From the definition of the Magnetic Force we have to
From the property of cross product the magnetic force should point in the direction perpendicular to the plane containing the vectors v and B.
The direction of velocity is north, and the direction of the magnetic force is northeast.
This cannot be the case, as the direction of magnetic force is not perpendicular to the direction of velocity of the charge.
Therefore the correct option for the direction of the magnetic field is <em>"This situation cannot exist because of the relative orientations of the velocity and force vectors" </em>
Velocity ... m/s (meters per second) and angle
Acceleration ... m/s^2 and angle
Force ... Newton (kg-m/s^2) and angle
Mass ... kilogram
Answer:D
Explanation:
Given
mass of object 
Distance traveled 
velocity acquired 
conserving Energy at the moment when object start falling and when it gains 12 m/s velocity
Initial Energy
Final Energy
where 
is friction work if any
Since Friction is Present therefore it is a case of Open system and net external Force is zero
An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .
Incomplete question.The complete question is here
What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade? The outer edge of the blade is 21 cm from the center of the blade, and the mass of the outer portion is 7.7 g. Even though the blade is 21cm long, the last 2cm should be treated as if they were at a point 20cm from the center of rotation.
Answer:
F= 0.034 N
Explanation:
Given Data
Outer=2 cm
Edge of blade=21 cm
Mass=7.7 g
Length of blade=21 cm
The last 2cm is treated as if they were at a point 20cm from the center of rotation
To Find
Force=?
Solution
Convert the given frequency to angular frequency
ω = 45 rpm * (2*pi rad / 1 rev) * (1 min / 60 s)
ω= 3/2*π rad/sec
Now to find centripetal force.
F = m×v²/r
F= m×ω²×r
Put the data
F = 0.0077 kg × (3/2×π rad/sec
)²× 0.20 m
F= 0.034 N
When block is pushed upwards along the inclined plane
the net force applied on the block will be given as
here we know that
m = 75 kg
now plug in all values into this
now for finding the power is given as
