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3 years ago

A human hair has a thickness of about 70 micrometers. How many meters is this?

Physics

2 answers:

Digiron [165]3 years ago

7 0

One meter is equal 1x10^6 micrometers so 70micrometer/10^(6)micrometer = 7x10^(-5) meters or .00007 meters

Aleksandr [31]3 years ago

6 0

This is 7e-5 meters. Hope this helps! Please mark brainliest. :)

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Brainly malfunction?| I'm not sure why but...

Brut [27]

Answer:

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Explanation:

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3 years ago

car was moving in a straight road of length 320 km it covered 240 km with an average velocity 75 km/hr then it ran out of fuel a

Stella [2.4K]

The average velocity of the car for the whole journey is 69.57 km/h.

The given parameters:

Length of the road, L = 320 km
Distance covered = 240 km at 75 km/h
time spent refueling, t₂ = 0.6 hr
Final velocity, = 100 km/hr

The time spent by the before refueling is calculated as follows;

$t = \frac{d}{v} \\\\t_1 = \frac{240}{75} \\\\t_1 = 3.2 \ hours$

The time spent by the car for the remaining journey;

$t_3 = \frac{320 - 240}{100} \\\\t_3 = 0.8 \ hr$

The total time of the journey is calculated as follows;

$t = t_1 + t_2 + t_3\\\\t = 3.2 \ hr \ + \ 0.6 \ hr \ + \ 0.8 \ hr\\\\t = 4.6 \ hours$

The average velocity of the car for the whole journey is calculated as follows;

$v = \frac{total \ distance }{total \ time} \\\\v = \frac{320}{4.6} \\\\v = 69.57 \ km/h$

Learn more about average velocity here: brainly.com/question/6504879

6 0

3 years ago

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 m /

Sidana [21]

Answer:

The maximum height is 2881.2 m.

Explanation:

Given that,

Acceleration = 29.4 m/s²

Time = 7.00 s

We need to calculate the distance

Using equation of motion

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

Put the value into the formula

$s=0+\dfrac{1}{2}\times29.4\times7^2$

$s=720.3\ m$

We need to calculate the velocity

Using formula of velocity

$v=a\times t$

Put the value into the formula

$v=29.4\times7$

$v=205.8\ m/s$

We need to calculate the height

Using formula of height

$H=\dfrac{v^2}{2g}$

Put the value into the formula

$H=\dfrac{(205.8)^2}{2\times9.8}$

$H=2160.9\ m$

We need to calculate the maximum height

Using formula for maximum height

$H'=H+s$

Put the value into the formula

$H'=2160.9+720.3$

$H'=2881.2\ m$

Hence, The maximum height is 2881.2 m.

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3 years ago

CNA Secondary and Preparatory School

ale4655 [162]

Answer:

2:13 2) cm' c6462.

Explanation:

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3 years ago

We are designing a crude propulsion mechanism for a science fair demonstration. One of our team members stands on a skateboardth

Scrat [10]

Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.

initial instant. before the crash

p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

p_f = (m + M) v

the moment is preserved

p₀ = p_f

m v₀ = (m + M) v

v = $\frac{m}{m+M} \ v_o$

for when n balls have collided

v = $\frac{m}{n \ m + M}$ v₀

Now let's analyze the case of the bouncing ball (elastic)

initial instant

p₀ = m v₀

final moment

p_f = m v_{1f} + M v_{2f}

p₀ = p_f

m v₀ = m v_{1f} + M v_{2f}

m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

K₀ = K_f

½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

v₁ = v_{1f} v₂ = v_{2f}

m (v₀² - v₁²) = M v₂²

let's use the identity

(a² - b²) = (a + b) (a-b)

we write our equations

m (v₀ - v₁) = M v₂ (1)

m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

m (v₀ - v₁) = M (v₀ + v₁)

v₀ (m -M) = (m + M) v₁

v₁ = $\frac{m-M}{m + M}$ v₀

we substitute in equation 1 to find v₂

$\frac{M}{m}$ v₂ = v₀ - $\frac{m-M}{m+M}$ v₀

v₂ = $\frac{m}{M} ( 1 - \frac{m-M}{m+M} ) \ v_o$

v₂ = $\frac{m}{M} ( \frac{2M}{m+M} ) \ \ v_o$

v₂ = $\frac{2m}{m +M} \ v_o$

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.

8 0

3 years ago