How can shoe and tire marks be preserved in mud?

Compared to yellow light,orange light has

ArbitrLikvidat [17]

So we wonder how can we compare yellow and orange light. Light is a small art of electromagnetic spectrum which is a continuum of all electromagnetic waves arranged by wavelenght and frequency. So on the EM spectrum orange has greaterr wavelangth than yellow but smaller frequency because wavelength and frequency are connected by the equation: c=f*lambda, where lambda is the wavelength and c is the speed of light and f is the frequency. We see that the higher the frequency the smaller the wavelength and vice versa.

5 0

4 years ago

The manometer shown in fig. 2 contains water and kerosene. with both tubes open to the atmosphere, the free-surface elevations d

ozzi

The solution for this problem is: In the figure, you now know that total length of the kerosene column
So at x – xPatm + Pkg(H0 th) = Pa + Pwgh
Now H0 + h = 20 + 91.1 mm = 111.1 mm
Therefore = Pkg 0.1111 – P2g= h = 56 x 0.111 – 98 / 1000 x 9.81= 0.081 m or 81 mn
Therefore H0 = 111.1 - 81= 30.1 mm

6 0

3 years ago

A centrifuge in a medical laboratory rotates at an

suter [353]

Answer:

- 273.77 rad/s^2

Explanation:

fo = 3800 rev/min = 3800 / 60 rps = 63.33 rps

f = 0

ωo = 2 π fo = 2 x 3.14 x 63.33 = 397.71 rad/s

ω = 2 π f = 0

θ = 46 revolutions = 46 x 2π radian = 288.88 radian

Let α be the angular acceleration of the centrifuge

Use third equation of motion for rotational motion

$\omega^{2}=\omega _{0}^{2}+2\alpha \theta$

$0^{2}=397.71^{2}+2 \times \alpha \times 288.88$

α = - 273.77 rad/s^2

4 0

3 years ago

Nina is working with electricity and shortens the wires through which the current is flowing. What is Nina most likely trying to

wariber [46]

The answer is D i took the test but got it wrong but i got to see the answers

6 0

3 years ago

Read 2 more answers

A basketball player standing up with the hoop launches the ball straight up with an initial velocity of v_o = 3.75 m/s from 2.5

denis23 [38]

Answer:

a) The maximum height the ball will achieve above the launch point is 0.2 m.

b) The minimum velocity with which the ball must be launched is 4.43 m/s or 0.174 in/ms.

Explanation:

a)

For the height reached, we use 3rd equation of motion:

2gh = Vf² - Vo²

Here,

Vo = 3.75 m/s

Vf = 0m/s, since ball stops at the highest point

g = -9.8 m/s² (negative sign for upward motion)

h = maximum height reached by ball

therefore, eqn becomes:

2(-9.8m/s²)(h) = (0 m/s)² - (3.75 m/s²)²

h = 0.2 m

b)

To find out the initial speed to reach the hoop at height of 3.5 m, we again use 3rd eqn. of motion with h= 3.5 m - 2.5m = 1 m (taking launch point as reference), and Vo as unknown:

2(-9.8m/s²)(1 m) = (0 m/s)² - (Vo)²

(Vo)² = 19.6 m²/s²

Vo = √19.6 m²/s²

Vo = 4.43 m/s

Vo = (4.43 m/s)(1 s/1000 ms)(39.37 in/1 m)

Vo = 0.174 in/ms

6 0

3 years ago