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3 years ago

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Una ardilla corre desarrollando una aceleracion de 3.5 m/s2 la fuerza ejecidas por las patas traseras es de 5 N. Cual sera la ma

sa de la ardilla?

1 answer:

ziro4ka [17]3 years ago

4 0

Answer:

1.43 kg

Explanation:

We can solve the problem by using Newton's second law:

$F=ma$

where

F is the net force on an object

m is its mass

a is its acceleration

For the squirrel in the problem, we have:

$a=3.5 m/s^2$ is the acceleration

F = 5 N is the net force acting on it

Solving for m, we find its mass:

$m=\frac{F}{a}=\frac{5}{3.5}=1.43 kg$

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Calculate the hydrostatic difference in blood pressure between the brain and the foot in a person of height 1.93 m. The density

Slav-nsk [51]

Answer:

Explanation:

Given: Density of blood = 1.03 × 10³ Kg/m³, Height = 1.93 m g = 9.8 m/s²

pressure at the brain is equal to atmospheric pressure. = Hydro-static

pressure(ρ₀)

∴ pressure of the foot = pressure of the brain(ρ₀) + ( density of blood × acceleration due to gravity × height)(ρgh)

Hydro-static pressure = pressure at the feet- pressure at the brain(ρ₀)

Hydro-static pressure (Δp) = (ρgh + ρ₀) - ρ₀ = ρgh

Hydro-static pressure = 1.03 × 10³ × 9.8 × 1.93 = 1.948 × 10⁴ Pa

∴ Hydro-static pressure ≈ 1.95 × 10⁴ Pa

3 0

3 years ago

According to Oxford Dictionaries, a spit take is an act of suddenly spitting out liquid one is drinking in response to something

Tasya [4]

Answer:

The pressure is $p_1 = 4051.4 \ Pa$

Explanation:

From the question we are told that

The gauge pressure at the mouth is $p_1$

The radius of the column is $r_2 = 4 \ mm = 0.004 \ m$

The speed of the liquid outside the body is $v_2 = 3.1 \ m/s$

The area of the column is $A_2$

The area inside the mouth $A_1 = 10 A_2$

Generally according to continuity equation

$v_1 A_1 = v_2 A_2$

=> $v_ 1 = v_2 * \frac{A_2}{A_1}$

=> $v_ 1 = 3.1 * \frac{1}{10}$

=> $v_ 1 = 0.31 \ m/s$

So

$A_1 = 10A_2$

=> $\pi * r_1^2 = 10(\pi * r_2^2)$

=> $r_1 = 10 * r_2$

substituting values

$r_1 = 10 * 0.004$

$r_1 =0.04 \ m$

Now the height of inside the mouth is $h_1 = d = 2r_1 = 2* 0.04 = 0.08\ m$

Now the height of the column is $h_2 = d = 2r_2 = 2* 0.004 = 0.008\ m$

Generally according to Bernoulli's equation

$p_1 = [\frac{1}{2} \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ]$

Now $\rho = 1000 \ kg m^{-3}$ which is the density of water

$p_2$ is the gauge pressure of the atmosphere which is zero

So

$p_1 = [(0.5 * 1000 * (3.1)^2) +(0.008 * 1000 * 9.8) + 0]-$

$[(0.5 * 1000 * 0.31^2) + (0.08*1000 * 9.8)]$

$p_1 = 4051.4 \ Pa$

8 0

4 years ago

How many joules of heat are needed to raise the temperature of 50.0 g of aluminum from 10°C to 110°C, if the specific heat of al

dusya [7]

Answer:

Heat required to raise the temperature of the aluminium is 4750 J

Explanation:

As we know that the heat energy required to raise the temperature of the aluminium is given as

$Q = ms\Delta T$

here we know that

m = 50 g

$\Delta T = 110 - 10$

$\Delta T = 100 ^oC$

so we have

$Q = 50(0.95)(100)$

$Q = 4750 J$

5 0

4 years ago

Whitch two options are forms of kinetic energy?

valentina_108 [34]

Answer:the witch has nothing to do with the problem

Explanation:

7 0

3 years ago

A machinist is required to manufacture a circular metal disk with area 1900 cm2. (a) What radius produces such a disk? (Round yo

alexgriva [62]

Answer:

24.5987 cm

Explanation:

A = 1900 cm^2

Let r be the radius of disc.

The area of disc is given by

A = π r²

Where, π = 31.4

1900 = 3.14 x r²

r² = 605.095

r = 24.5987 cm

6 0

3 years ago