The energy lost to friction is destroyed.TrueFalse

What is the main difference between speed and velocity?

sertanlavr [38]

Speed is the rate something is moving, while velocity is the rate something is going in a given direction.

5 0

3 years ago

You considered the situation of moving a couch at a constant speed. You concluded that to move the couch at a constant speed, yo

Dmitry [639]

Answer:

The couch will remain stationary

Explanation:

Friction is the force that resist/opposes the motion. It is the force that has to e overcome before an object can move.

Since the frictional force of 50 N is required before the couch can move, applying 25 N can not overcome the frictional force of 50 N, so the couch will remain stationary.A force greater than 50 will be require to set the couch in motion

8 0

4 years ago

The radius of the aorta is «10 mm and the blood flowing through it has a speed of about 300 mm/s. A capillary has a radius of ab

stealth61 [152]

Answer:

(I). The effective cross sectional area of the capillaries is 0.188 m².

(II). The approximate number of capillaries is $3.74\times10^{9}$

Explanation:

Given that,

Radius of aorta = 10 mm

Speed = 300 mm/s

Radius of capillary $r=4\times10^{-3}\ mm$

Speed of blood $v=5\times10^{-4}\ m/s$

(I). We need to calculate the effective cross sectional area of the capillaries

Using continuity equation

$A_{1}v_{1}=A_{2}v_{2}$

Where. v₁ = speed of blood in capillarity

A₂ = area of cross section of aorta

v₂ =speed of blood in aorta

Put the value into the formula

$A_{1}=A_{2}\times\dfrac{v_{2}}{v_{1}}$

$A_{1}=\pi\times(10\times10^{-3})^2\times\dfrac{300\times10^{-3}}{5\times10^{-4}}$

$A_{1}=0.188\ m^2$

(II). We need to calculate the approximate number of capillaries

Using formula of area of cross section

$A_{1}=N\pi r_{c}^2$

$N=\dfrac{A_{1}}{\pi\times r_{c}^2}$

Put the value into the formula

$N=\dfrac{0.188}{\pi\times(4\times10^{-6})^2}$

$N=3.74\times10^{9}$

Hence, (I). The effective cross sectional area of the capillaries is 0.188 m².

(II). The approximate number of capillaries is $3.74\times10^{9}$

3 0

3 years ago

A sample contains 20 kg of radioactive material. The decay constant of the material is 0.179 per second. If the amount of time t

Nataly [62]

Answer:

Explanation:

Given

N0 = 20kg (original substance)

decay constant λ = 0.179/sec

time t = 300s

We are to find N(t)

Using the formula;

n(t) = N0e^-λt

Substitute the given values

N(t) = 20e^-(0.179)(300)

N(t) = 20e^(-53.7)

N(t) = 20(4.7885)

N(t) =143.055

To know how much of the original material that is active, we will find N(t)/N0 = 143.055/20 = 7.152

About 7 times the original material is still radioactive

4 0

3 years ago

PLEASE HELP In a system, when potential energy decreases, then entropy also decreases.

expeople1 [14]

False it never decreases, it increases or remain the same.

7 0

3 years ago

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