Answer:
3.944 x 10⁴ K.
Explanation:
Let the surface temperature of star be T in absolute scale.
Total thermal radiation from its surface can be calculated with the help of
stefan's formula as follows
E = σ A T⁴
σ is a constant called stefan's constant and is equal to 5.67x 10⁻⁸ W m⁻²K⁻⁴
A is area of surface of star and T is the surface temperature.
Putting the given values
E = 5.67 X 10⁻⁸ X 4 X 3.14 X ( 5 X 10⁸ )² X T⁴
= 1780.38 X 10⁸ T⁴
At a distance of 2.5 x 10¹³ m intensity will be calculated by dividing E with area of sphere having radius equal to distance , so
Intensity at given point
= E / 4πd²
= 
= 22.68 T⁴ X 10⁻¹⁸
Putting the value of given intensity at that point
.055 = 22.68 x 10⁻¹⁸ T⁴
T⁴ = 24.25 X 10¹⁷
= 242.5 X 10¹⁶
T = 3.944 x 10⁴ K.
Answer:
Explanation:
The magnitude of the acceleration makes an angle of 30° with the tangential velocity.
Resolving the acceleration to tangential and radial acceleration
at = aCos30 = √3a/2
ar = aSin30 = ½a
a = 2•ar
Then, the tangential acceleration is the linear acceleration, so the relationship between the tangential acceleration and angular acceleration is given as:
at = Rα
Then, α = at/R
since at = √3a/2
Then, α = √3 at/2R, equation 1
The radial acceleration is given as
ar = ω²R
Note that, at² + ar² = a²
at = √(a²-ar²)
Back to equation 1
α = √3 at/2R
α = √3√(a²-ar²)/2R
α = √3√(a²-(w²R)²)/2R
α = √3(a²-w⁴R²) / 2R
Also, a = 2•ar = 2w²R
Then,
α = √3((2w²R)²-w⁴R²) / 2R
α = √3(4w⁴R²-w⁴R²) / 2R
α = √3(3w⁴R²) / 2R
α = √9w⁴R² / 2R
α = 3w²R / 2R
α = 3w²/2
Answer:
1768 N
Explanation:
We can solve the problem by using Newton's second law:
where
F is the net force acting on an object
m is the mass of the object
a is its acceleration
In this problem, we have a car of mass
m = 884 kg
And its acceleration is
Substituting into the equation, we find the net force on the car:
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