Answer:
Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s
x-component = -1.50 m/s
y-component = 3.90 m/s
Explanation:
Relative velocity of a body A relative to another body B, Vab, is given as
Vab = Va - Vb
where
Va = Relative velocity of body A with respect to another third body or frame of reference C
Vb = Relative velocity of body B with respect to that same third body or frame of reference C.
So, relative velocity can be given further as
Vab = Vac - Vbc
Velocity of Newton relative to Daniel = Vnd = 3.90 m/s due north = (3.90ĵ) m/s in vector form.
Velocity of Newton relative to Pauli = Vnp = 1.50 m/s due East = (1.50î) m/s in vector form
What is Pauli's velocity relative to Daniel?
Vpd = Vp - Vd
(Pauli's velocity relative to Daniel) = (Pauli's velocity relative to Newton) - (Daniel's velocity relative to Newton)
Vpd = Vpn - Vdn
Vpn = -Vnp = -(1.50î) m/s
Vdn = -Vnd = -(3.90ĵ) m/s
Vpd = -1.50î - (-3.90ĵ)
Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s
Hope this Helps!!!!
If the machine is 100% efficient, then its
Mechanical Advantage is (450/50) = 9 .
If the machine is less than 100% efficient,
then the MA is more than 9 .
The object will move if the forces are unbalanced.
Newtons second tells you that when a net force (the unbalanced force) is applied to and object it will produce an acceleration (movement) in direct proportion to the force and in inverse proportion to the mass of the object.
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
The vector force on the unit positive charge placed at any location in the field defines the strength of the electric field at that point. The charge used to determine field intensity (field strength) is known as the test charge. Now, a field line is defined as a line to which the previously mentioned field strength vectors are tangents at the relevant places. When we study positive charge field lines, the field strength vectors point away from the positive charge. If there is a negative charge anywhere in the vicinity, the field lines that began from the positive charge will all terminate at the negative charge if the value of the negative charge is the same as the value of the positive charge. Remember that the number of field lines originating from positive charge is proportional to the charge's value, and similarly, the number of field lines terminating at negative charge is proportionate to the charge's value. As a result, if all charges are equivalent, all lines originating from the positive charge terminate at the negative charge. If the value of the positive charge is greater than the value of the negative charge, the number of lines ending at the negative charge will be proportionally less than the number of lines beginning at the positive charge. The remaining lines that do not end at the negative charge will go to infinity. If the positive charge is less, all lines from it terminate at a negative charge, and any other reasonable number of ines terminate at a negative charge from infinity. We should also keep in mind that the number of lines that run perpendicular to the field direction across a surface of unit area is proportional to the field strength at that location. As a result, lines are dense in the strong field zone and sparse in the low intensity region.
