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4 years ago

How much power should a braked 1.5t car have to be braked to reduce its speed from 30m / s to 10m / s in 5s?

Physics

1 answer:

grandymaker [24]4 years ago

7 0

Answer:

-120000 W

Explanation:

Power = change in energy / time

P = ΔE / t

P = (½ mv₂² − ½ mv₁²) / t

P = m (v₂² − v₁²) / (2t)

Given m = 1.5 t = 1500 kg, v₂ = 10 m/s, v₁ = 30 m/s, and t = 5 s:

P = (1500 kg) ((10 m/s)² − (30 m/s)²) / (2 × 5 s)

P = -120000 W

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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 $3.74\times10^{9}$

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3 years ago

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Alisiya [41]

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3 years ago

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Gravity will have a greater effect on an object with

prohojiy [21]

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A force of 58 newtons is applied to a crate at an angle of 32° with the horizontal. What is the value of force acting in the y–d

Mumz [18]

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3 years ago

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A person who weighs 800N on the earth's surface will weigh 200N at what height above the earth

Marina86 [1]

Answer: 6,400 km

Explanation:

The weight of a person is given by:

$W=mg$

where m is the mass of the person and g is the acceleration due to gravity. While the mass does not depend on the height above the surface, the value of g does, following the formula:

$g=\frac{GM}{r^2}$

where

G is the gravitational constant

M is the Earth's mass

r is the distance of the person from the Earth's center

The problem says that the person weighs 800 N at the Earth's surface, so when r=R (Earth's radius):

$800 N= W=mg=m \frac{GM}{R^2}$ (1)

Now we want to find the height h above the surface at which the weight of the man is 200 N:

$200 N = W' = mg' = m \frac{GM}{(R+h)^2}$ (2)

If we divide eq.(1) by eq.(2), we get

$\frac{800 N}{200 N}=\frac{W}{W'}=\frac{(R+h)^2}{R^2}$

$4=\frac{(R+h)^2}{R^2}$

By solving the equation, we find:

$4R^2 = (R+h)^2=R^2+2Rh+h^2\\h^2 +2Rh-3R^2 =0$

which has two solutions:

$h=-3R$ --> negative solution, we can ignore it

$h=R$ --> this is our solution

Since the Earth's radius is $R=6.4\cdot 10^6 m$ , the person should be at $h=R=6.4\cdot 10^6 m=6400 km$ above Earth's surface.

5 0

4 years ago

How much power should a braked 1.5t car have to be braked to reduce its speed from 30m / s to 10m / s in 5s?​

How much power should a braked 1.5t car have to be braked to reduce its speed from 30m / s to 10m / s in 5s?