Answer:
Although this question is not complete, I would give a general solution to this kind of problems.
If y(t) describes the position of a body with time such that
y(t) = at^(n) + bt^(m) + C
Then
V(t) = dy(t)/dt = ant^(n-1) + bmt^(m-1)
Explanation:
As an example supplies the position of a particle is given by
y(t) = 4t³- 3t² + 9
V(t) = 4x3t²- 3x2t¹
V(t) = d(t)/dt = 12t² - 6t.
Another example,
If y(t) = 15t³ - 2t² + 30t -80
V(t) = d(t)/dt = 15x3t² - 4t +30 = 45t² + 4t + 30.
Basically, in the equations above the powers of t reduces by one when computing the velocity function from y(t) by differentiation (calculating the derivative of y(t)). The constant term C (9 and 80 in the functions of y(t) in examples 1and 2 above) reduces to zero because the derivative of a constant (and ordinary number without the t attached to it) is always zero.
One last example,
y(t) = 2t^6 -3t²
V(t) = d(t)/dt = 12t^5 - 6t
Answer:
Explanation:
Let us calculate gallon used in one hour .
It travels 70 miles in one hour
in 70 miles it uses 70 / 34 gallons of fuel
70 / 34 gallons = 70 / 34 x 3.7854 kg
= 7.8 kg
heat generated = 7.8 x 44 x 10⁶ J
= 343.2 x 10⁶ J
This is heat generated in one hour
heat generated in one second = 343.2 x 10⁶ / 60 x 60 J/s
= 95.33 x 10³ J /s
= 95.33 kW.
Answer:
elastic partial width is 2.49 eV
Explanation:
given data
ER E = 250 eV
spin J = 0
cross-section magnitude σ = 1300 barns
peak P = 20ev
to find out
elastic partial width W
solution
we know here that
σ = λ²× W / ( E × π × P ) ...................1
put here all value
σ = (0.286)² × W × 
/ ( 250 × π × 20 )
1300 × 
= (0.286)² × W × / ( 250 × π × 20 )
solve it and we get W
W = 249.56 × 
so elastic partial width is 2.49 eV
If you know what dark matter is, my friend, then there are several thousand Astrophysicists and Cosmologists at universities around the world who would pay dearly for the opportunity to buy you a cup of coffee, and chat while you sip it.
Answer:
D
Explanation:
From the information given:
The angular speed for the block 
Disk radius (r) = 0.2 m
The block Initial velocity is:
Change in the block's angular speed is:
However, on the disk, moment of inertIa is:
The time t = 10s
∴
Frictional torques by the wall on the disk is:
Finally, the frictional force is calculated as:
