Answer:
Statement:
The electric current passing through a conductor is directly proportional to the potential difference across its ends provided temperature and other physical conditions remain constant.
Explanation:
Current is directly proportional to voltage loss through a resistor. That is, if the current doubles, then so does the voltage. To make a current flow through a resistance there must be a voltage across that resistance. Ohm's Law shows the relationship between the voltage (V), current (I) and resistance (R).
V∝I or I∝V⇒V=IR.
The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.
To find the answer, we need to know about the tension.
<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
- Let's draw the free body diagram of the system using the given data.
- From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
- For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.
- We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.
- To find Ny, we need to find the tension T.
- For this, we can equate the net horizontal force.
- Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,
- Thus, the magnitude of the force that the beam exerts on the hi.nge will be,
Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.
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Answer:
Tension= 21,900N
Components of Normal force
Fnx= 17900N
Fny= 22700N
FN= 28900N
Explanation:
Tension in the cable is calculated by:
Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium
FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)
Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)
Ftorque= 2/3FBcostheta+ 4/3FWcostheta
Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°
Ftorque= 21900N
b) components of Normal force
Efx=FNx-FTcos(90-theta)=0 static equilibrium
Fnx=21900cos(90-55)=17900N
Fy=FNy+ FTsin(90-theta)-FB-FW=0
FNy= -FTsin(90-55)+FB+FW
FNy= -21900sin(35)+(1350+2250)×9.81=22700N
The Normal force
FN=sqrt(17900^2+22700^2)
FN= 28.900N
Answer:
0.00970 s
Explanation:
The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field
Force due to magnetic field = qvB sin θ
q = charge on the particle = 5.4 μC
v = velocity of the charge
B = magnetic field strength = 2.7 T
θ = angle between the velocity of the charge and the magnetic field = 90°, sin 90° = 1
F = qvB
Centripetal force responsible for circular motion = mv²/r = mvw
where w = angular velocity.
The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field
mvw = qvB
mw = qB
w = (qB/m) = (5.4 × 10⁻⁶ × 2.7)/(4.5 × 10⁻⁸)
w = 3.24 × 10² rad/s
w = 324 rad/s
w = (angular displacement)/time
Time = (angular displacement)/w
Angular displacement = π rads (half of a circle; 2π/2)
Time = (π/324) = 0.00970 s
Hope this Helps!!!
