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4 years ago

Is it possible for speed to be constant when acceleration is not zero?

Physics

1 answer:

Darina [25.2K]4 years ago

3 0

Answer: Yes, it is possible.

Explanation: In order to explain we can show that circular movement has a centripetal acceletarion and the speed is constant, so the centripetal accelaration changes the direction of the vector velocity in order to keep the circular trajectory.

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You’ve had practice calculating the grams of hydrogen gas, but it is also possible to calculate the amount of oxygen gas produce

alekssr [168]

Answer:41.991ml

Explanation:

Equations: 2 H2O → 4H+ + 4e + O2 OXIDATION

2 H+ + 2e → H2 REDUCTION

Electrolysis is the chemical decomposition of compounds when electricity is made to pass through a molten compound or solution.

from the oxidation reaction:

1moles of oxygen requires 4moles of electrons to be discharged at the product

F=96500C/mol

Quantity of charge Q=It

=60*60*0.201A

Q=723.6C

Mole=Q/(F*mole ratio of electron)

Mole= 723.6/(4*96500)

Mole=((1809)/(965000))

M=0.0018746114

M1/M2=V1/V2

1/0.00187=22.4dm^3/V2

V2=22.4*0.00187

V2=0.04199129534dm^3

41.99129534ml

5 0

3 years ago

A train travels 77 kilometers in 1 hour, and the 66 kilometers in 1 hour. What is the average speed?

Kipish [7]

Average speed = (total distance covered) / (time to cover the distance)

Total distance = (77km + 66km) = 143 kilometers

Time to cover the distance = 2 hours

Average speed = (143 km) / (2 hours) = 71.5 km per hour

6 0

3 years ago

You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov

Marina CMI [18]

Answer:

a. The frequency with which the waves strike the hill is 242.61 Hz

b. The frequency of the reflected sound wave is 254.23 Hz

c. The beat frequency produced by the direct and reflected sound is

11.62 Hz

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

$f_{1} = f[\frac{v_{s} }{v_{s} -v} ]$ ..............................1

where $f_{1}$ is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = $36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}$

v = 16.27 m/s

Substituting into equation 1 we have

$f_{1}$ = $231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})$

$f_{1}$ = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

$f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]$

where $f_{2}$ is the frequency of the reflected sound;

$f_{1}$ is the frequency which the wave strikes the hill = 242.61 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

$f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]$

$f_{2}$ = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound and the reflected sound. This can be obtained as follows;

Δf = $f_{2}$ - $f_{1}$

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

3 0

3 years ago

which of the following laboratory tool would be most appropiate for measuring the approximate volume of a liquid?

SIZIF [17.4K]

Beaker would be most appropriate for measuring the approximate volume of a liquid.

7 0

3 years ago

For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0 cmcm long)

loris [4]

This question is incomplete, the complete question is;

Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point 11.5 cm from their point of intersection, as shown in the figure.

For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0 cmcm long) after reflecting from the first mirror

Answer: angle of incidence is 39.4°

Explanation:

Given that;

two plain mirrors intersect at right angle (90°)

distance d = 11.5 cm

S = 28.0 cm

Now the angle that the reflection ray males with first the mirror equal theta (∅)

tan∅ = (S/2) / d

tan∅ = (28/2) / 11.5

tan∅ = 14 / 11.5

tan∅ = 1.2173

∅ = tan⁻¹ (1.2173)

∅ = 50.6°

so angle of incidence = 90° - ∅

= 90° - 50.6°

= 39.4°

Therefore angle of incidence is 39.4°

5 0

3 years ago