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4 years ago

8

How can fluid friction be decreased?

1 answer:

fgiga [73]4 years ago

5 0

Answer: by giving special shapes called streamlined shape to the objects

Explanation:

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The radius of the aorta is about 1.4cm , and the blood passing through it has a speed of about 40cm/s .

Lapatulllka [165]

For the average speed of blood flow in the major arteries of the body is mathematically given as

v2 = 117.29m/s

<h3>What is the average speed of blood flow in the major arteries of the body?</h3>

Generally, the equation for the average speed is mathematically given as

A1 v1 = A2 v2

(pi r1^2) v1 = A2 v2

(3.14x(1.4)^2 )x 40 = (2.1) xV2

v2 = 117.29m/s

In conclusion, the average speed of blood flow

v2 = 117.29m/s

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2 years ago

A square block of steel with volume 10 cm3 and mass of 75 g is cut precisely in half. The density of the two smaller pieces is n

dem82 [27]

A. Density only depends on the substance. It doesn't matter whether you have a little chip of it or a supertanker full of it ... the density doesn't change.

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3 years ago

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How long is a light year?

VashaNatasha [74]

A light year is = to 9.461 × 1015 metres

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4 years ago

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A batter hits a ball at 35 degrees above the horizontal and it is caught 4 seconds later 100 meters from home plate. What is the

Mamont248 [21]

Answer:

$u=(26.5 i + 18.5j) m/s$

Explanation:

The range of a projectile is given by the formula

$d=\frac{u^2}{g} sin 2\theta$

where in this case, we have

d = 100 m is the range

u is the initial speed (the magnitude of the initial velocity)

g = 9.8 m/s^2 is the acceleration of gravity

$\theta = 35^{\circ}$ is the angle of projection

Solving for u, we find:

$u=\sqrt{\frac{dg}{sin 2\theta}}=\sqrt{\frac{(100)(9.8)}{sin(2\cdot 35^{\circ})}}=32.3 m/s$

Now we can easily find the components of the initial velocity:

$u_x = u cos \theta = (32.3)(cos 35^{\circ})=26.5 m/s\\u_y = u sin \theta = (32.3)(sin 35^{\circ})=18.5 m/s$

So, the initial velocity of the ball is

$u=(26.5 i + 18.5 j) m/s$

where i and j are the unit vector indicating the horizontal and vertical direction.

7 0

3 years ago

A 1800 kg car moves along a horizontal road at speed v₀ = 18.4 m/s. The road is wet, so the static friction coefficient between

slamgirl [31]

Answer:

The shortest possible distance is $|s| = 91.9 \ m$

Explanation:

From the question we are told that

The mass of the car is $m = 1800 \ kg$

The speed along the horizontal road is $v_o =u = 18.4 \ m/s$

The static friction coefficient is $\mu_s = 0.188$

The kinetic friction coefficient is $\mu_k = 0.1316$

Generally the static frictional force acting on the car is mathematically represented as

$F_f = m * g * \mu_s$

Generally the force propelling the car is mathematically represented as

$F = m * a$

Here a is the maximum acceleration

at the point which the car stops ,

$F = F_f$

=> $m * g * \mu_s = ma$

=> $g * \mu_s =a$

=> $a = 9.8 * 0.188$

=> $a = 1.8424 \ m/s^2$

Generally from kinematic equation

$v^2 = u^2 + 2as$

Here v is the final velocity of the car and the value is zero given that the car comes to rest

So

$0^2 = 18.4^2 + 2* 1.8424 s$

=> $s = - \frac{18.4^2}{2 * 1.8424}$

=> $|s| = 91.9 \ m$

6 0

3 years ago