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2 years ago

The amplitude of the wave in the picture below is... Select one: a. 1 cm b. 1.5 cm c. 2 cm d. 3 cm

Physics

1 answer:

antiseptic1488 [7]2 years ago

8 0

Answer:

1cm

Explanation:

The amplitude of a wave is the distance from the center line (in this case the 1 cm marker on the vertical ruler) to the highest or lowest point. For this image the highest point is 2 cm, 2cm-1cm=1cm. The lowest point is 0cm, 1cm-0cm=1cm.

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Two ball bearings of mass m each moving in opposite directins with equal speed v collide head on with each other.predict the out

posledela

Yo sup??

since the collision is elastic therefore we can say that the two balls will then move in opposite direction.

If ball 1 was moving from east to west then after collision it will move from west to east

and if ball 2 was moving from west to east then it will start moving from east to west.

Hope this helps.

7 0

3 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind.

RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity, $\rm v_{SA} = 9 \ m/sec$

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

The seagull's relative velocity with reference to the ground as;

$\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec$

Air velocity with reference to the ground is;

$\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec$

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

$\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec$

The time the bird takes;

$\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7 \ min \ and \ 42 \ sec$

c)\

The total round-trip time compared to what it would be with no wind. is;

$\rm t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec$

The time for the round trip is;

$\rm t = \frac{12 \times 10^ 3 }{ 9 \ m/sec } \\\\ t = 1333.33 \ sec$

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be 4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

4 0

2 years ago

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P P and volume V V satisfy th

Verdich [7]

Answer:

the volume decreases at the rate of 500cm³ in 1 min

Explanation:

given

v = 1000cm³, p = 80kPa, Δp/t= 40kPa/min

PV=C

vΔp + pΔv = 0

differentiate with respect to time

v(Δp/t) + p(Δv/t) = 0

(1000cm³)(40kPa/min) + 80kPa(Δv/t) = 0

40000 + 80kPa(Δv/t) = 0

Δv/t = -40000/80

= -500cm³/min

the volume decreases at the rate of 500cm³ in 1 min

3 0

3 years ago

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re

spayn [35]

The magnitude of the electric field can be calculated using the equation

$E = \frac{F}{q}$ , where F is the Force acting on the proton and q is the charge of the proton.

We know that the charge of the proton is $q = 1.6 x 10^{-19} C$

We have to calculate the Force first.

We know that F = ma from Newton's 2nd law, where m is the mass of the proton and a is the acceleration.

We know that the mass of the proton $m = (1.67) X 10^{-27} kg$

So it turns out that we have to calculate acceleration before anything else.

In order to calculate the acceleration, we make use of the following data from the question:

Initial Velocity of the proton $V_{i} = 0$

Distance traveled $D = 1.70 cm$ = 0.017 m

Time taken for the travel between the plates $t = (1.48) X 10^{-6} s$

Acceleration a = ?

Using the equation, $D = V_{i}t + \frac{1}{2} at^{2}$ , we get

Knowing that initial velocity is 0, the equation reduces to $D = \frac{1}{2}at^{2}$

Rearranging the equation so as to make a the subject of the formula, we have

$a = \frac{2D}{t^{2} }$

Plugging in the numbers and simplifying gives us a = 1.5 x $10^{10} m/s^{2}$

We can now calculate the Force using F = ma

Plugging in the known values, we get F = 2.5 x $10^{-17} N$

Using this, we can calculate E through the equation $E = \frac{F}{q}$

Plugging the numbers and simplifying gets us E = 156.25 N/C

Thus, the magnitude of the electric field between the plates of the capacitor is 156.25 N/C

B) To calculate the Final Velocity of the proton, we can make use of the equation

$V_{f} = V_{i} + at$

Plugging the numbers in and simplifying gets us $V_{f} = (2.22) * 10^{4} m/s$

5 0

3 years ago

Read 2 more answers

How can you drop two eggs the fewest amount of times, without them breaking?

lord [1]

Answer:

Wrap it in cotton??? like alot of it?

6 0

3 years ago