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1 year ago

The long structure of small intestine is accommodated in small space within our body. Comment.

1 answer:

6 0

The long structure of small intestine is accommodated in small space within our body because of extensive coiling. the small intestine is highly coiled structure and thus can easily be fixed in a small space.

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The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of

motikmotik

Explanation:

It is given that, the force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the car's speed such that,

$F\propto \dfrac{mgv^2}{r}$

$F=\dfrac{kmgv^2}{r}$

mg is the weight of the car

r is the radius of the curve

v is the speed of the car

Case 1.

F = 640 pounds

Weight of the car, W = mg = 2600 pound

Radius of the curve, r = 650 ft

Speed of the car, v = 40 mph

$640=\dfrac{k(2600)(40)^2}{650}$

k = 0.1

Case 2.

Radius of the curve, r = 750 ft

Speed of the car, v = 30 mph

$F=\dfrac{0.1\times 2600\times (30)^2}{750}$

F = 312 N

Hence, this is the required solution.

6 0

3 years ago

If the sprinter accelerates at that rate for a distance of 15 m, and then maintains the velocity he has at that point for the re

ahrayia [7]

Answer:

The time for the entire race is 11.39 sec.

Explanation:

Given that,

Distance = 15 m

Remainder distance = 100 m

Suppose A sprinter begins a race with an acceleration of 3.4 m/s².

We need to calculate the time

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value in the equation

$15=0+\dfrac{1}{2}\times3.4\times t^2$

$t^2=\dfrac{30}{3.4}$

$t=2.97\ sec$

We need to calculate the final velocity of sprinter

Using equation of motion again

$v=u+at$

Put the value into the formula

$v=0+3.4\times2.97$

$v=10.09\ m/s$

We need to calculate the distance covers by sprinter

$d=100-15=85\ m$

The sprinter need to covers only 85 m.

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t'=\dfrac{85}{10.09}$

$t'=8.42\ sec$

We need to calculate the time for the entire race

$t''=t+t'$

Put the value into the formula

$t''=2.97+8.42$

$t''=11.39\ sec$

Hence, The time for the entire race is 11.39 sec.

4 0

3 years ago

Find the length of the third side of the triangle.

AlladinOne [14]

Answer:

Image result for What is the length of the hypotenuse of the triangle below

One way to solve this is to use Pythagorean theorem: the square of one leg of triangle plus square of other leg of the triangle equals c the hypotenuse (longest side of triangle). You might see this as the formula a^2 + b^2 = c^2, where a and b are the legs and c is the hypotenuse.Nov 23, 2016

Explanation:

4 0

3 years ago

The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.10. The mass of the object i

denis-greek [22]

Answer:

Explanation:

1) Force Friction = Normal Force * Coefficient of Friction

Force Friction = Mass * Gravity * Coefficient of Friction

2) F = ma

Force = mass * acceleration

Force Friction (from #1) = mass * acceleration

acceleration = Force Friction / Mass

6 0

2 years ago

A car with a mass of 850kg is moving at a speed of 72km/h when colliding with a concrete wall until it stops. After the collisio

Sergeeva-Olga [200]

Answer:

Explanation:

The vehicle is experiencing a large force created by the concrete wall.

Equation

vf^2 = vi^2 + 2*a * d

Givens

vf = 0 The car eventually does stop.

vi = 72 km/hr * [ 1000 m/ km] * [1 hour / 3600 seconds]

vi = 20 meters / second

a = ?

m = 850 kg

Solution

vf^2 = vi^2 + 2a*d

0 = 20 m/s + 2* 2 *a

-20 m/s = 4a

-20/4 = a

a = - 5 m/s^2 The minus sign tells you the vehicle is slowing down. It sure should be.

Force = m * a

F = - 850 * (-5)

F = - 4250 N

The car provides a 4250 N force on it going east to west and a 4250 N force going from west to east provided by the concrete wall.

8 0

3 years ago