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4 years ago

A ray diagram is shown. Which statement best describes the diagram?

1 answer:

6 0

The statement that best describes the diagram is:

"The diagram is destined to remain forever
an enigma wrapped in a mystery, since its
willing audience was denied the merest peek."

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If a car can go from 20m/s to 40m/s in 4.0 secs, what would be it’s acceleration?

MariettaO [177]

This is a problem that would be a good test of your understanding rather than your ability to work the formulas. 5m/s² means that the velocity increase each second is 5 m/s. So 4 s of that acceleration would increase the speed (in m/s) from 20 to 40. (Speed increase each second is 5 m/s. We need an increase of 20 m/s.)

Since the acceleration is uniform during those 4 s, we can use the simple average speed of 30 m/s. 30 m/s * 4 s = 120 m.

5 0

4 years ago

What is the acceleration experienced by a car that takes 10s to reach 27m/s from rest

Lerok [7]

|acceleration| = (change in speed) / (time for the change)

Change in the car's speed = (27 - 0) = 27 m/s
Time for the change = 10 sec

|acceleration| = (27 m/s) / (10 s) = 2.7 m/s² .

That's the magnitude of the car's acceleration.
We don't know anything about its direction.

5 0

3 years ago

Consider helium at 350 K and 0.75 m3/kg. Using Eq. 12-3, determine the change in pressure corresponding to an increase of (a) 1

Alex777 [14]

Answer:

a) $(dP)_{v} = 9.692 kPa$

b) $(dP)_{T} = -9.692 kPa$

c) dP = 0 Pa

Explanation:

The specifies equation is :

$dz = (\frac{\delta z}{\delta x}) _{y} dx + (\frac{\delta z}{\delta y}) _{x} dy$

Note that:

$dP = \frac{R}{v} dT - \frac{RT}{v^{2} } dV$

1% increase in temperature at specific volume:

$dT = \frac{0.01}{1} *350\\dT = 3.5 K$

a) Change in pressure of helium at constant volume:

$(dP)_{v} = \frac{R}{v} dT$

R = 2.0769 kJ/kg-K

dT = 3.5 K

v = 0.75 m³/kg

$(dP)_{v} = \frac{2.0769}{0.75} * 3.5\\(dP)_{v} = 9.692 kPa$

dv = (1%/100%) *0.75

dv = 0.0075 m³/kg

Change in pressure of helium at constant temperature:

$(dP)_{T} = \frac{-RT}{v^{2} } dv$

R = 2.0769 kJ/kg-K

T = 350 K

v = 0.75 m³/kg

dv = 0.0075 m³/kg

$(dP)_{T} = \frac{-(2.0769*350)}{0.75^{2} } *0.0075\\(dP)_{T} = -9.692 kPa$

c) The change in pressure of helium :

$dP = (dP)_{v} + (dP)_{T}$

dP = 9.692 - 9.692

dP = 0

5 0

4 years ago

Albert Jr. takes his new skateboard out to the top of the loading ramp. The ramp is 30 m long and sloped at an angle of 15 degre

WINSTONCH [101]

Answer:

vf = 12.343 m/s

Explanation:

Given:

- The Length of the ramp s = 30 m

- The angle of the ramp θ = 15 degrees

- Initial velocity at the top of ramp vi = 0

- Neglect Friction

Find:

How fast is Albert Jr. going at the bottom of the ramp?

Solution:

- We can use conservation of energy for the skateboard at top and bottom of the ramp.

ΔK.E = ΔP.E

Where, ΔK.E : Change in kinetic energy

ΔP.E : Change in gravitational potential energy

0.5*m*(vf^2 - vi^2) = m*g*Δh

Where, m: The mass of the object

Δh: Change in elevation from top to bottom

Δh = s*sin(θ)

- Substitute and simplify:

0.5*(vf^2 - vi^2) = g*s*sin(θ)

vf^2 = 2*g*s*sin(θ)

vf = √(2*g*s*sin(θ)) = √(2*9.81*30*sin(15))

vf = 12.343 m/s

- The velocity at the bottom of ramp would be 12.343 m/s

5 0

3 years ago

g Problem 4. A 2000 kg railway freight car coasts at 4.4 m/s underneath a grain terminal, which dumps grain directly down into t

pogonyaev

Answer: 2933 kg

Explanation:

Given

From the law of conservation of linear momentum, we know that

m(i)v(i) = m(f)v(f), where there is no external force acting on the system

m(i) = initial mass of the freight car system

m(f) = maximum mass of the freight car system

v(i) = initial linear velocity of the system

v(f) = final linear velocity of the system

2000 * 4.4 = m(f) * 3

3m(f) = 8800

m(f) = 8800 / 3

m(f) = 2933.3 kg

Therefore the maximum mass of grain that it can accept is 2933 kg

5 0

4 years ago