Answer:
Your question but don't know how to solve it like a dog
<em>D. Less luminous than those on the right []</em>
Answer:
Explanation:
λ=c x²
c = λ / x²
λ is mass / length
so its dimensional formula is ML⁻¹
x is length so its dimensional formula is L
c = λ / x²
= ML⁻¹ / L²
= ML⁻³
B )
We shall find out the mass of the rod with the help of given expression of mass per unit length and equate it with given mass that is M
The mass in the rod is symmetrically distributed on both side of middle point.
we consider a small strip of rod of length dx at x distance away from middle point
its mass dm = λdx = cx² dx
By integrating it from -L to +L we can calculate mass of whole rod , that is
M = ∫cx² dx
= [c x³ / 3] from -L/2 to +L/2
= c/3 [ L³/8 + L³/8]
M = c L³/12
c = 12 M L⁻³
C ) Moment of inertia of rod
∫dmx²
= ∫λdxx²
= ∫cx²dxx²
= ∫cx⁴dx
= c x⁵ / 5 from - L/2 to L/2
= c / 5 ( L⁵/ 32 +L⁵/ 32)
= (2c / 160)L⁵
= (c / 80) L⁵
= (12 M L⁻³/80)L⁵
= 3/20 ML²
=
=
Answer:
323.4 N
Explanation:
Given that,
The coefficient of static friction, ![\mu_s=0.63](https://tex.z-dn.net/?f=%5Cmu_s%3D0.63)
The coefficient of kinetic friction, ![\mu_k=0.49](https://tex.z-dn.net/?f=%5Cmu_k%3D0.49)
We need to find the frictional force acting on the player. As the baseball player is sliding into home base, we will use the coefficient of kinetic friction to find the frictional force.
![F=\mu _kN\\\\=0.49\times 660\\\\=323.4\ N](https://tex.z-dn.net/?f=F%3D%5Cmu%20_kN%5C%5C%5C%5C%3D0.49%5Ctimes%20660%5C%5C%5C%5C%3D323.4%5C%20N)
So, the frictional force acting on the player is 323.4 N.
Answer:
The answer to your question is: D) Ф₂ = 49.71°
Explanation:
Data
n₁ = 1.33
Ф₁ = 35°C
n₂ = 1
Ф₂ = sin⁻¹ (n₁ sinФ₁/n₂)
Process
Substitution
Ф₂ = sin⁻¹ (n₁ sinФ₁/n₂)
Ф₂ = sin⁻¹ (1.33 sin 35/1)
Ф₂ = sin⁻¹ (1.33 x 0.574/ 1)
Ф₂ = sin⁻¹ ( 0.7628 / 1)
Ф₂ = sin⁻¹ (0.7628)
Ф₂ = 49.71°