Any one have a clue on this one?

A truck traveled 400 meters north in 80 seconds and then it traveled 300 meters east in 70 seconds. The magnitude of the average

Varvara68 [4.7K]

Answer:

3.3m/s

Explanation:

You first get the total time (80 + 70 = 150s).

Then you would find the displacement of the truck. To do that you would do component method (vector addition), so since its a right triangle (North and East), displacement is 400^2 + 300^2 = d^2.

d= 500m.

So now that you have displacement and time, you can find the velocity:

v=d/t

v=500/150

v=3.3

5 0

3 years ago

Maurice pulls on the end of a spring scale. He lets go of the end and observes the spring snap back into place. What force resto

andreev551 [17]

Elastic is the right answer.

3 0

3 years ago

Read 2 more answers

In your own words, explain why we see a rainbow or colorful sunset. Explain the properties and interactions that make this displ

Maksim231197 [3]

Wee see rainbows due to the geometries of the raindrops. when the sun shines behind, rays of light enter the raindrops and this light are refracted. The lights are then reflected from the back of the raindrop and refracted again as it passes the rain drop. Refraction in this sense is the cause for splitting the light into several colors.

7 0

3 years ago

For safety reasons, a worker’s eye travel time in a certain operation must be separated from the manual elements that follow. Th

iogann1982 [59]

Answer:

The answer is 12.67 TMU

Explanation:

Recall that,

worker’s eyes travel distance must be = 20 in.

The perpendicular distance from her eyes to the line of travel is =24 in

What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?

Now,

We solve for the given problem.

Eye travel is = 15.2 * T/D

=15.2 * 20 in/24 in

so,

= 12.67 TMU

Therefore, the MTM -1 of normal time that should be allowed for the eye travel element is = 12.67 TMU

7 0

4 years ago

Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What p

Temka [501]

Answer:

The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

Explanation:

From General gas equation.

PV = nRT............................... Equation 1

Where n = number of moles, V = volume, P = pressure, T = temperature, P = pressure, V = volume.

n = mass/molar mass .................. Equation 2

substituting equation 2 into equation 1.

PV = (mass/molar mass)RT

⇒ Mass/molar mass = PV/RT..................... Equation 3

But mass = Density × Volume

⇒ M = D × V.................... Equation 4

Where D = density, M = mass

Substituting equation 4 into equation 3

DV/molar mass = PV/RT............ Equation 5

Dividing both side of the equation by Volume (V) in Equation 5

D/molar mass = P/RT .............. Equation 6

Cross multiplying equation 6

D × RT = P × molar mass

∴ Molar mass = (D × RT)/P.................. Equation 7

Where D = 0.518 g/L , R = 0.0821 atm dm³/K.mol,

T = 25°C = 25 + 273 = 298 K,

P =721 mmHg = (721/760) atm= 0.949 atm

Substituting these values into equation 7

Molar mass = (0.518 × 0.0821 × 298)/0.949

Molar mass = 13.35 g/mole

The molar mass of the mixture is =13.35 g/mole

Let y be the mole fraction of Helium and 1-y be the mole fraction of oxygen.

∴ 13.35 = 4(y) + 32(1-y)

13.35 = 4y + 32 - 32y

Collecting like terms in the equation,

32y - 4y = 32 - 13.35

28y = 18.65

y = 18.65/28

y =0.666

y = 0.666 × 100 = 66.6%

∴The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

6 0

3 years ago