Answer:
The value of R is
.
(B) is correct option.
Explanation:
Given that,
In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression

We need to calculate this for value of R


So, The nearest option of the value of R is 
Hence, The value of R is
.
Answer:
h = 2 R (1 +μ)
Explanation:
This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the
let's use the mechanical energy conservation agreement
starting point. Lower, just at the curl
Em₀ = K = ½ m v₁²
final point. Highest point of the curl
= U = m g y
Find the height y = 2R
Em₀ = Em_{f}
½ m v₁² = m g 2R
v₁ = √ 4 gR
Any speed greater than this the body remains in the loop.
In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law
X axis
-fr = m a (1)
Y Axis
N - W = 0
N = mg
the friction force has the formula
fr = μ N
fr = μ m g
we substitute 1
- μ mg = m a
a = - μ g
having the acceleration, we can use the kinematic relations
v² = v₀² - 2 a x
v₀² = v² + 2 a x
the length of this zone is x = 2R
let's calculate
v₀ = √ (4 gR + 2 μ g 2R)
v₀ = √4gR( 1 + μ)
this is the speed so you must reach the area with fricticon
finally have the third part we use energy conservation
starting point. Highest on the ramp without rubbing
Em₀ = U = m g h
final point. Just before reaching the area with rubbing
= K = ½ m v₀²
Em₀ = Em_{f}
mgh = ½ m 4gR(1 + μ)
h = ½ 4R (1+ μ)
h = 2 R (1 +μ)
Answer:
The moment (torque) is given by the following equation:
![\vec{\tau} = \vec{r} \times \vec{F}\\\vec{r} \times \vec{F} = \left[\begin{array}{ccc}\^{i}&\^j&\^k\\r_x&r_y&r_z\\F_x&F_y&F_z\end{array}\right] = \left[\begin{array}{ccc}\^{i}&\^j&\3k\\0.23&0.04&0\\150&260&0\end{array}\right] = \^k((0.23*260) - (0.04*150)) = \^k (53.8~Nm)](https://tex.z-dn.net/?f=%5Cvec%7B%5Ctau%7D%20%3D%20%5Cvec%7Br%7D%20%5Ctimes%20%5Cvec%7BF%7D%5C%5C%5Cvec%7Br%7D%20%5Ctimes%20%5Cvec%7BF%7D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5C%5E%7Bi%7D%26%5C%5Ej%26%5C%5Ek%5C%5Cr_x%26r_y%26r_z%5C%5CF_x%26F_y%26F_z%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5C%5E%7Bi%7D%26%5C%5Ej%26%5C3k%5C%5C0.23%260.04%260%5C%5C150%26260%260%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5C%5Ek%28%280.23%2A260%29%20-%20%280.04%2A150%29%29%20%3D%20%5C%5Ek%20%2853.8~Nm%29)
Explanation:
The cross-product between the distance and the force can be calculated using the method of determinant. Since the z-components are zero, it is easy to calculate.
Answer:
The moment of inertia is 0.7500 kg-m².
Explanation:
Given that,
Mass = 2.2 kg
Distance = 0.49 m
If the length is 1.1 m
We need to calculate the moment of inertia
Using formula of moment of inertia

Where, m = mass of rod
l = length of rod
x = distance from its center
Put the value into the formula


Hence, The moment of inertia is 0.7500 kg-m².
Answer:
speed of boat as

river speed is given as

Explanation:
When boat is moving down stream then in that case net resultant speed of the boat is given as
since the boat and river is in same direction so we will have

Now when boat moves upstream then in that case the net speed of the boat is opposite to the speed of the river
so here we have

as we know when boat is in downstream then in that case it covers 24 miles in 2 hours

also when it moves in upstream then it covers same distance in 3 hours of time



so we have speed of boat as

river speed is given as
