Answer:
The circuit is in series connection,
the same current will flow through i.e
I1 = I2 = I3 = 2A.
Explanation:
We'll begin by calculating the total resistance in the circuit. This is shown below:
R1 = 4Ω
R2 = 3Ω
RT =..?
Total resistance in series can be obtained as follow:
RT = R1 + R2
RT = 4 + 3
RT = 7Ω
Next we shall determine the total current flowing in the circuit:
Voltage (V) = 14V
Resistance (R) = 7Ω
Current (I) =..?
V = IR
14 = I x 7
Divide both side by 7
I = 14/7
I = 2A.
Since the circuit is in series connection,
the same current will flow through i.e
I1 = I2 = I3 = 2A.
Answer:
a) 0.64 b) 2.17m/s^2 c) 8.668joules
Explanation:
The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,
Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move
Frictional force = mgsin20o + 5N = 6.71+5N = 11.71
The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44
Coefficient of static friction = 11.71/18.44= 0.64
Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)
b) coefficient of kinetic friction = frictional force/ normal force
Fr = 0.4* mgcos 20o = 7.375N
F due to motion = ma = total force - frictional force
Ma = 11.71 - 7.375 = 4.335
a= 4.335/2(mass of the block) = 2.17m/s^2
C) work done = net force *distance = 4.335*2= 8.67Joules
Answer:
![x=9m](https://tex.z-dn.net/?f=x%3D9m)
Explanation:
According to Newton's second law, we have:
![-F_f=ma(1)](https://tex.z-dn.net/?f=-F_f%3Dma%281%29)
The force of friction is opposite to the motion of the block.
Using the following kinematic equation, we can calculate the distance traveled by the block before it stops(
).
![v_f^2=v_0^2+2ax](https://tex.z-dn.net/?f=v_f%5E2%3Dv_0%5E2%2B2ax)
Solving for x and replacing a from (1):
![x=\frac{v_f^2-v_0^2}{2a}\\x=\frac{0-v_0^2}{2(\frac{-F_f}{m})}\\x=\frac{mv_0^2}{2F_f}\\x=\frac{10kg(6\frac{m}{s})^2}{2(20N)}\\x=9m](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bv_f%5E2-v_0%5E2%7D%7B2a%7D%5C%5Cx%3D%5Cfrac%7B0-v_0%5E2%7D%7B2%28%5Cfrac%7B-F_f%7D%7Bm%7D%29%7D%5C%5Cx%3D%5Cfrac%7Bmv_0%5E2%7D%7B2F_f%7D%5C%5Cx%3D%5Cfrac%7B10kg%286%5Cfrac%7Bm%7D%7Bs%7D%29%5E2%7D%7B2%2820N%29%7D%5C%5Cx%3D9m)
Answer:. The acceleration of the object is directly proportional to the force and inversely proportional to the mass.
Explanation:
According to Newton s Second Law of Motion, also known as the Law of Force and Acceleration, a force upon an object causes it to accelerate according to the formula net force = mass x acceleration.
Answer:
The speed of the rod is 2.169 m/s.
Explanation:
Given that,
Mass = 0.100 kg
Current = 15.0 A
Distance = 2 m
Length = 0.550 m
Kinetic friction = 0.120
(a). We need to calculate the magnetic field
Using relation of frictional force and magnetic force
![F_{f}=F_{B}](https://tex.z-dn.net/?f=F_%7Bf%7D%3DF_%7BB%7D)
![\mu mg=Bli](https://tex.z-dn.net/?f=%5Cmu%20mg%3DBli)
![B=\dfrac{\mu mg}{li}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B%5Cmu%20mg%7D%7Bli%7D)
Where, l = length
i = current
m = mass
Put the value into the formula
![B=\dfrac{0.120\times0.1\times9.8}{0.550\times15.0}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B0.120%5Ctimes0.1%5Ctimes9.8%7D%7B0.550%5Ctimes15.0%7D)
![B=0.01425\ T](https://tex.z-dn.net/?f=B%3D0.01425%5C%20T)
![B=1.425\times10^{-2}\ T](https://tex.z-dn.net/?f=B%3D1.425%5Ctimes10%5E%7B-2%7D%5C%20T)
(b). If the friction between the rod and rail is reduced zero.
So, ![f_{f}=0](https://tex.z-dn.net/?f=f_%7Bf%7D%3D0)
We need to calculate the acceleration
Using formula of force
![F_{net}=f_{f}+F_{B}](https://tex.z-dn.net/?f=F_%7Bnet%7D%3Df_%7Bf%7D%2BF_%7BB%7D)
![F_{net}=0+Bil](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D0%2BBil)
![ma=Bil](https://tex.z-dn.net/?f=ma%3DBil)
![a=\dfrac{Bil}{m}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7BBil%7D%7Bm%7D)
Put the value into the formula
![a=\dfrac{1.425\times10^{-2}\times15\times0.55}{0.1}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B1.425%5Ctimes10%5E%7B-2%7D%5Ctimes15%5Ctimes0.55%7D%7B0.1%7D)
![a=1.176\ m/s^2](https://tex.z-dn.net/?f=a%3D1.176%5C%20m%2Fs%5E2)
We need to calculate the speed of the rod
Using equation of motion
![v^2=u^2+2as](https://tex.z-dn.net/?f=v%5E2%3Du%5E2%2B2as)
Put the value into the formula
![v^2=0+2\times1.176\times2](https://tex.z-dn.net/?f=v%5E2%3D0%2B2%5Ctimes1.176%5Ctimes2)
![v^2=\sqrt{4.704}\ m/s](https://tex.z-dn.net/?f=v%5E2%3D%5Csqrt%7B4.704%7D%5C%20m%2Fs)
![v=2.169\ m/s](https://tex.z-dn.net/?f=v%3D2.169%5C%20m%2Fs)
Hence, The speed of the rod is 2.169 m/s.