Answer:81.235N
Explanation:
Work=88J
theta=10°
distance=1.1 meters
work=force x cos(theta) x distance
88=force x cos10 x 1.1 cos10=0.9848
88=force x 0.9848 x 1.1
88=force x 1.08328
Divide both sides by 1.08328
88/1.08328=(force x 1.08328)/1.08328
81.235=force
Force=81.235
Answer:
Magnitude of y-component=7 units
Direction= angle of vector A with x-axis= 
Explanation:
Explained solution is in the picture attached
I don't know what the exact word is, but I do know that the bigger an objects mass is the more it will attract other objects toward it, mainly smaller objects with less mass. it might be gravity or something around those lines....is it a multiple choice question?
Answer:
Incomplete questions
Let assume we are asked to find
Calculate the induced emf in the coil at any time, let say t=2
And induced current
Explanation:
Flux is given as
Φ=NAB
Where
N is number of turn, N=1
A=area=πr²
Since r=2cm=0.02
A=π(0.02)²=0.001257m²
B=magnetic field
B(t)=Bo•e−t/τ,
Where Bo=3T
τ=0.5s
B(t)=3e(−t/0.5)
B(t)=3exp(-2t)
Therefore
Φ=NAB
Φ=0.001257×3•exp(-2t)
Φ=0.00377exp(-2t)
Now,
Induce emf is given as
E= - dΦ/dt
E= - 0.00377×-2 exp(-2t)
E=0.00754exp(-2t)
At t=2
E=0.00754exp(-4)
E=0.000138V
E=0.138mV
b. Induce current
From ohms laws
V=iR
Given that R=0.6Ω
i=V/R
i=0.000138/0.6
i=0.00023A
i=0.23mA
Answer:
62.8 μC
Explanation:
Here is the complete question
The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?
Solution
The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ
So, Q = ∫∫∫ρdV
Q = ∫∫∫ρr²sinθdθdrdΦ
Q = ∫∫∫(0.2r²)r²sinθdθdrdΦ
Q = ∫∫∫0.2r⁴sinθdθdrdΦ
We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π
So, Q = ∫∫∫0.2r⁴sinθdθdrdΦ
Q = ∫∫∫0.2r⁴[∫sinθdθ]drdΦ
Q = ∫∫0.2r⁴[-cosθ]drdΦ
Q = ∫∫0.2r⁴-[cosπ - cos0]drdΦ
Q = ∫∫∫0.2r⁴-[-1 - 1]drdΦ
Q = ∫∫0.2r⁴-[- 2]drdΦ
Q = ∫∫0.2r⁴(2)drdΦ
Q = ∫∫0.4r⁴drdΦ
Q = ∫0.4r⁴dr∫dΦ
Q = ∫0.4r⁴dr[Φ]
Q = ∫0.4r⁴dr[2π - 0]
Q = ∫0.4r⁴dr[2π]
Q = ∫0.8πr⁴dr
Q = 0.8π∫r⁴dr
Q = 0.8π[r⁵/5]
Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]
Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]
Q = 0.8π[0.025 m⁵ - 0 m⁵]
Q = 0.8π[0.025 m⁵]
Q = (0.02π mC/m⁵) m⁵
Q = 0.0628 mC
Q = 0.0628 × 10⁻³ C
Q = 62.8 × 10⁻³ × 10⁻³ C
Q = 62.8 × 10⁻⁶ C
Q = 62.8 μC
