Answer:
M_c = 100.8 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two cases shown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *144
M_c = 2.5*( 42 / 150 ) *144
M_c = 100.8 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
Work done = force x distance = 40 x 2 = 80 Joules.
Answer:
Explanation:
Intensity of light is inversely proportional to distance from source
I ∝ 1 /r² where I is intensity and r is distance from source . If I₁ and I₂ be intensity at distance r₁ and r₂ .
I₁ /I₂ = r₂² /r₁²
If r₂ = 4r₁ ( given )
I₁ / I₂ = (4r₁ )² / r₁²
= 16 r₁² / r₁²
I₁ / I₂ = 16
I₂ = I₁ / 16
So intensity will become 16 times less bright .
"16 times " is the answer .
Answer:
7.28×10⁻⁵ T
Explanation:
Applying,
F = BILsin∅............. Equation 1
Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.
make B the subject of the equation
B = F/ILsin∅.................. Equation 2
From the question,
Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°
Substitute these values into equation 2
B = 0.16/(68×34×sin72°)
B = 0.16/(68×34×0.95)
B = 0.16/2196.4
B = 7.28×10⁻⁵ T
Answer:
The correct answer is "6666.67 N".
Explanation:
The given values are:
Mass,
m = 0.100
Relative speed,
v = 4.00 x 10³
time,
t = 6.00 x 10⁻⁸
As we know,
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
