Answer:
applied force
Explanation:
any force where you push or pull is always applied force.
Answer:
He needs 1.53 seconds to stop the car.
Explanation:
Let the mass of the car is 1500 kg
Speed of the car, v = 20.5 m/s
He will not push the car with a force greater than, 
The impulse delivered to the object is given by the change in momentum as :
So, he needs 1.53 seconds to stop the car. Hence, this is the required solution.
The complete question is;
A circular coil consists of N = 410 closely winded turns of wire and has a radius R = 0.75 m. A counterclockwise current I = 2.4 A is in the coil. The coil is set in a magnetic field of magnitude B = 1.1 T.
a. Express the magnetic dipole moment μ in terms of the number of the turns N, the current I, and radius
R.
b. Which direction does μ go?
Answer:
A) μ = 1738.87 A.m²
B) The direction of the magnetic moment will be in upward direction.
Explanation:
We are given;
The number of circular coils;
N = 410
The radius of the coil;R = 0.75m
The current in the coils; I = 2.4 A
The strength of magnetic field;
B =1.1T
The formula for magnetic dipole moment is given as;
μ = NIA
Where;
N is number of turns
I is current
A is area
Now, area; A = πr²
So, A = π(0.75)²
Thus,plugging in relevant values, the magnetic dipole moment is;
μ = 410 * 2.4 * π(0.75)²
μ = 1738.87 A.m²
B) According to Fleming's right hand rule, the direction of the magnetic moment comes out to be in upward direction.
Answer:
because they are found freely in nature uncombined so they are highly reactive with other elements
The Period of the resulting shm will be T=39.7
<u>Explanation:</u>
<u>Given data</u>
m=3kg
d=.06m
k=1200 N/m
Θ=3 °
T=?
we have the formulas,
I = (1/6)Md2
F = ma
F = -kx = -(mω2x)
k = mω2 τ = -d(FgsinΘ)
T=2 x 3.14/ √(m/k)
Solution for the given problem would be,
F=-Kx (where x= dsin Θ)
F=-k dsin Θ
F=-(1200)(.06)sin(3 °)
F=-10.16N
<u>By newton's second law.</u>
F = ma
a= F/m
a=(-10.16N)/3
a=3.38
<u>using the k=mω value</u>
k=mω
ω=k/m
ω=1200/3
ω=400
<u>Using F = -kx value</u>
x = F/-k
x=(-10.16)/1200
x=0.00847m
<u>Restoring the torque value </u>
τ = -dmgsinΘ where( τ = Iα so.).. Iα = -dmgsinΘ α = -(.06)(4)α =
α =(.06)(4)(9.81)sin(4°)
α=-1.781
<u>Rotational to linear form</u>
a = αr
r = .1131 m
a=-1.781 x .1131 m
a=-0.2015233664
<u>Time Period</u>
T=2 x 3.14/ √(m/k)
T=6.28/√(3/1200)
T=6.28/0.158
T=39.7
