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ladessa [460]
3 years ago
8

(a) The volume flow rate in an artery supplying the brain is 3.4 x 10-6 m3/s. If the radius of the artery is 5.6 mm, determine t

he average blood speed.(b) Find the average blood speed as a constriction in the artery if the constriction reduces the radius by a factor of 2. Assume that the volume flow rate is the same as that in part (a).

Physics
1 answer:
UNO [17]3 years ago
3 0

Answer: a) speed = 3.45 × 10^-2 m/s

b) speed = 1.38 × 10^-1 m/s

Explanation: shown in the attachment

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Lions can run at speeds up to approximately 80.0 km / h. A hungry 109 kg lion running northward at top speed attacks and holds o
sukhopar [10]

Answer:

17.34 m/s

Explanation:

Given:

Mass of lion (m₁) = 109 kg

Initial speed of lion (v₁) = 80.0 km/h (Northward direction)

Mass of gazelle (m₂) = 39.0 kg

Initial speed of gazelle (v₂) = 78.5 km/h (Eastward direction)

Final velocity of both lion and gazelle is, v_f=?

First, let us convert the speeds from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

Therefore,

v_1= 80.0\ km/h=80\times \frac{5}{18}=22.22\ m/s\\\\v_2=78.5\ km/h=78.5\times \frac{5}{18}=21.81\ m/s

Now, the concept of conservation of total momentum is used here as this is a case of perfectly inelastic collision. In inelastic collision, the masses move together with same velocity after collision.

Here, as the lion and gazelle are moving in directions at right angles to each other, the vector sum of their momentums will give the net initial momentum of the system.

So, initial momentum is given as:

P_i=\sqrt{P_1^2+P_2^2}\\\\Where,\\\\P_1\to initial\ momentum\ of\ lion\\P_2\to initial\ momentum\ of\ gazelle

Now, we calculate P₁ and P₂.

P_1=m_1v_1=(109\ kg)(22.22\ m/s) = 2421.98\ Ns\\\\P_2=m_2v_2=(39\ kg)(21.81\ m/s) = 850.59\ Ns

Therefore, the net initial momentum of the system is given as:

P_i=\sqrt{(2421.98)^2+(850.59)^2}=2567\ Ns

The final momentum of the system is given as:

P_f=(m_1+m_2)(v_f)\\\\P_f=(109+39)v_f\\\\P_f=148v_f

From the law of conservation of momentum, the final momentum is equal to the initial momentum. So,

P_f=P_i\\\\148v_f=2567\\\\v_f=\frac{2567}{148}=17.34\ m/s

Therefore, the final speed of the lion-gazelle system is 17.34 m/s

3 0
3 years ago
Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsice force
Dmitriy789 [7]

Answer:

Q = 1.095 x 10^-9 C

Let the force experienced by the top piece of tape be F

F = kQ²/r²

r = distance between the two pieces tape = 1.00cm = 1.00 x 10^ -2 m

1/4(pi)*Eo = k = 8.99 x 10^9 Nm²/C²

The electric force of repulsion between the two charges and the weight of the top piece of tape are equal so

F = kQ²/r² = mg

Where m is the mass of the top piece of tape and g is the acceleration due to gravity

On re-arranging the equation above,

Q² = mgr²/k

Q² = ((11.0 x 10^-6) x 9.8 x (1.00x10^-2)²)/(8.99 x 10^9)

Q = 1.095x10^-9 C

Explanation:

The charge Q on both pieces of tape are equal and both act with a force of repulsion on each other.

The force of repulsion between both tapes pushes the top piece of tape upwards. The weight of the top piece of tape acts vertically downward. Since the top tape is in a position of equilibrium, the two forces acting on the top piece of tape must be equal to each other. This assumption is backed up by newton's first law of motion which states that the summation of all forces acting on a body at rest must be equal to zero. That is

Fe (electric force) - Fg (gravitational force) = 0

Fe = Fg

kQ²/r² = mg

On substituting the respective values for all variables except Q and rearranging the equation Q = 1.09 x 10^-9

6 0
4 years ago
The age of a piece of wood from an archeological site is to be determined using the Carbon-14 method. The activity of the sample
Bezzdna [24]

Answer:

4589.05 year

Explanation:

The relation of activity is given by A=A_0e^{\lambda t} we have given \frac{A}{A_0}=0.574

Half life T=5730\ years

We know that half life T=\frac{0.693}{\lambda }

\lambda =\frac{0.693}{5730}=1.2094\times 10^{-4}\ per\ year

So 0.574=e^{-1.2094\times 10^{-4}t}

e^{1.2094\times 10^{-4}t}=\frac{1}{0.574}=1.742

{1.2094\times 10^{-4}t}=ln1.742=0.555

t=\frac{0.555}{1.2094\times 10_{-4}}=4589.05\ year

7 0
3 years ago
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