When a light beam emerges from water into the air, the average light speed ____

krok68 [10]

Answer:

Faith, to me is almost like hope but for faith you need to believe. Some people associate the word faith with religion. But that's not it. Faith is really just a stronger term than believing someone. Like you can believe in someone but not have faith in that same person. You can have faith in someone but to do that you have to believe and trust them.

Explanation:

6 0

2 years ago

A horizontal force of 25 N is required to push a wagon across a sidewalk at a constant speed.

lions [1.4K]

Answer:

a. The net force acting on the wagon is zero

b. The value of the force of friction is 25 N

c. The effect would be is the frictional force must be increase

Explanation:

Lets explain how to solve the problem

A horizontal force of 25 N is required to push a wagon across a sidewalk

at a constant speed

a.

→ Constant speed means acceleration is zero

→ Force = mass × acceleration

→ Force = mass × 0

→ Force = 0

The net force acting on the wagon is zero

b.

The wagon moves with constant speed, then

→ ∑ force = 0

There are two forces acting on the wagon external force of 25 N,

and the frictional force which is opposite to the direction of motion

→ 25 - Frictional force = 0

→ Frictional force = 25 N

The value of the force of friction is 25 N

c.

The force increased to 30 N

If the wagon still moves with constant speed

Due to Newton's law ∑ force in the direction of motion = mass ×

acceleration, but acceleration is 0 because the wagon moves with

constant speed, then ∑ F = 0

So the frictional force must increase to 30 N

The effect would be is the frictional force must be increase

If the frictional force does not increase the wagon will accelerate

6 0

4 years ago

A sprinter reaches his maximum speed in 2.6 seconds from rest with constant acceleration. He then maintains that speed and finis

Delvig [45]

Answer: maximum speed vmax = 11.42m/s

Explanation:

Given that the sprinter maintained constant acceleration during the first 2.6 seconds.

a = vmax/ta .......1

The distance covered during the acceleration period is;

da = 0.5a(ta)^2 .....2

Substituting equation 1 to 2

da = 0.5(vmax/ta)(ta)^2 = 0.5vmax(ta) .....3

The distance covered during the period of constant speed vmax is;

dv = vmax (tv) ......4

The total distance travelled is

d = da + dv = 100 (Given)

da + dv = 100 ......5

Substituting equation 3 and 4 into 5

0.5vmax(ta) + vmax(tv) = 100

vmax ( 0.5ta +tv) = 100

vmax = 100/(0.5ta + tv) ....6

But,

t = ta + tv

tv = t - ta .......7

Substituting equation 6 into equation 7

vmax = 100/(0.5ta + t - ta)

vmax = 100/(t-0.5ta)

t = 10.06 s

ta = 2.6 s

Substituting the values;

vmax = 100/(10.06 -0.5(2.6))

vmax = 11.42m/s

Note:

ta = acceleration time

tv = constant velocity vmax time

t = overall time

da , dv and d = acceleration, constant velocity and overall distance covered respectively.

8 0

4 years ago

Technician A states that a reading of 250 mv from the O2 indicates a rich exhaust. Technician B states that when the PCM receive

Pachacha [2.7K]

Answer:

Technician B is correct

Explanation:

An oxygen sensor will generate about 1.0 volts when the fuel mixture is rich and there is little unburned oxygen in the exhaust. When the mixture is lean, the sensor's output voltage will drop down to about 0.1 volts.

An o2 sensor cannot accurately measure how rich or how well an exhaust system is.

Therefore, Technician B is correct.

7 0

3 years ago

A 10-cm-long archerfish begins such a leap with its body vertical in the water, using its fins and undulating its body to propel

siniylev [52]

Answer:

Explanation:

Velocity will be equal to the area under the curve of the acceleration-time plot.

v = ½(45)(0.075) = 1.6875

v = 1.7 m/s

h = (0² - 1.6875²) / (2(-9.8)) = 0.145288...

h = 15 cm

3 0

3 years ago

When a light beam emerges from water into the air, the average light speed _____.