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1 year ago

An object on a number line moved from x = 15 cm to x = 165 cm and then

Physics

2 answers:

mel-nik [20]1 year ago

5 0

Answer:

The average velocity of the object is 0.1cm/s

Explanation:

Given that the object travels from point 15cm to 165cm and back to 25cm within 100 seconds

The average velocity is calculated as thus.

Average Velocity = ∆D/t

Where ∆D represent the displacement.

The displacement is calculated as follows.

∆D = End point - Start Point.

From the question, the end and start point are 25cm and 15cm respectively.

Hence,

∆D = 25cm - 15cm

∆D = 10cm.

t = 100 seconds

So, Average Velocity = 10cm/100s

Average Velocity = 0.1cm/s

Hence, the average velocity of the object is 0.1cm/s

olasank [31]1 year ago

3 0

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

$v_{avg}=\frac{x_{all}}{t}$

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

$v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}$

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

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Mrrafil [7]

Answer: -2.5

Explanation:

1/2(-5)= -2.5

-2.5(1)= -2.5

Got it right in Khan Academy. You’re welcome.

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1 year ago

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

valentina_108 [34]

Answer:

v = 69.82 ms^-1

Explanation:

As we know,

R = vi2 sin2Ꝋ / g

vi2 =R g / sin2 Ꝋ where R is range R = 52m, Ꝋ = 3 Degrees

vi2 = 52 x 9.8 / sin 2(3) = 4875.227

v = 69.82 ms^-1

3 0

2 years ago

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

$P_1$ = Pressure at reservoir = 10 atm

$T_1$ = Temperature at reservoir = 300 K

$P_2$ = Pressure at exit = 1 atm

$T_2$ = Temperature at exit

$R_s$ = Mass-specific gas constant = 287 J/kgK

$\gamma$ = Specific heat ratio = 1.4 for air

For isentropic flow

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K$

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

$\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3$

The density of the flow at the exit is 2.2721 kg/m³

4 0

1 year ago

A 600g toy train completes 10 laps of its circular track in 1 min 20s. If the radius of the track is 1.2 m, Find the centripetal

Lynna [10]

Wow ! This will take more than one step, and we'll need to be careful
not to trip over our shoe laces while we're stepping through the problem.

The centripetal acceleration of any object moving in a circle is

      (speed-squared) / (radius of the circle) .

Notice that we won't need to use the mass of the train.

We know the radius of the track. We don't know the trains speed yet,
but we do have enough information to figure it out. That's what we
need to do first.

Speed = (distance traveled) / (time to travel the distance).

Distance = 10 laps of the track.   Well how far is that ? ? ?

1 lap = circumference of the track = (2π) x (radius) = 2.4π meters

10 laps = 24π meters.

Time = 1 minute 20 seconds = 80 seconds

The trains speed is (distance) / (time)

   = (24π meters) / (80 seconds)

   = 0.3 π meters/second .

NOW ... finally, we're ready to find the centripetal acceleration.

<span> (speed)² / (radius)

   = (0.3π m/s)² / (1.2 meters)

   = (0.09π m²/s²) / (1.2 meters)

   =    (0.09π / 1.2) m/s²

   =    0.236 m/s² . (rounded)

If there's another part of the problem that wants you to find
the centripetal FORCE ...

Well, Force = (mass) · (acceleration) .

We know the mass, and we ( I ) just figured out the acceleration,
so you'll have no trouble calculating the centripetal force. </span>

4 0

1 year ago

A ship 1200m off shore fires a gun. how long after the gun is fired will it be heard on the shore?

ryzh [129]

Answer:

We know that the speed of sound is 343 m/s in air

we are also given the distance of the boat from the shore

From the provided data, we can easily find the time taken by the sound to reach the shore using the second equation of motion

s = ut + 1/2 at²

since the acceleration of sound is 0:

s = ut + 1/2 (0)t²

s = ut <em>(here, u is the speed of sound , s is the distance travelled and t is the time taken)</em>

Replacing the variables in the equation with the values we know

1200 = 343 * t

t = 1200 / 343

t = 3.5 seconds (approx)

Therefore, the sound of the gun will be heard at the shore, 3.5 seconds after being fired

6 0

1 year ago