Old temperature = 283 K.
New temp = 323 K.
(323/283) x (325 kPa) = 371 kPa.
Answer:
Option (D)
Explanation:
The velocity at which blood flows in the blood vessels is inversely proportional to the total cross-sectional area of the blood vessels present in the body. This means that if the cross sectional area of the vessels low, then there will be high rate of blood flow, and vice versa. This blood flow is minimum in the case of capillaries, where it gets enough time for the exchanging of essential nutrients as well as gases.
Thus, the correct answer is option (D).
Answer:
Distance travel by go-cart = 500 meter
Explanation:
Given:
Speed of go cart = 25 m/s
Time travel = 20 seconds
Find:
Distance travel by go-cart
Computation:
Distance = Speed x time
Distance travel by go-cart = Speed of go cart x Time travel
Distance travel by go-cart = 25 x 20
Distance travel by go-cart = 500 meter
Answer:
Explanation:
For calculating resistance of a conductor , the formula is
R = ρ l / A , ρ is specific resistance , l is length and A is cross sectional area of wire.
For first wire length is l₁ , area is A₁ resistance is R₁, for second resistance is R₂ , length is l₂ and area is A₂
Given , l₁ = 2l₂ , A₁ = 4A₂ , area is proportional to square of thickness.
R₁ / R₂ = I₁A₂ / I₂A₁
= 2l₂ x A₁ / 4 I₂A₁
= 1 / 2
2R₁ = R₂
Power = V² / R
Ratio of power = (V² / R₁) x (R₂ / V²)
= R₂ / R₁
= 2 .
Answer:

#Where
is in meters and
in seconds.
Explanation:
Given that :
From
we have:

From
we have that:

Now,given that the initial value problem is given by:

Hence,the position of u at time t is given by
,
in meters,
in seconds.