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3 years ago

Which velocity-time graph matches the position-time graph?

Physics

2 answers:

skelet666 [1.2K]3 years ago

7 0

The answer is Graph C. To explain, this is because as we look at the position vs time graph, we see that after the first second, it was 30 meters from the start. That would mean that it took 1 second to get to 30 meters. That is shown in Graph c

Leona [35]3 years ago

6 0

Explanation:

In this question, a position time graph is shown. We need to find the corresponding velocity time graph.

In position time graph, till 1 second the object is in uniform motion and reaches a distance of 30 meters. Then the object is at till 3 seconds. After 3 seconds, the object again reaches a distance of 40 meters uniformly.

Corresponding velocity-time graph is shown in figure (3). The object is moving initially with a speed of 30 m/s. After 1 second, it comes to rest till 3 seconds then it again moves. So, the correct option is (c). Hence, this is the required solution.

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Twin A makes a one way trip at 0.6c to a star 12 light years away while twin B stays on Earth. Each twin sends the other a signa

hram777 [196]

Answer:

8 signals received by twin A during the trip.

Explanation:

Given that,

Distance = 12 light year

Speed = 0.6 c

Time = 1 year

We need to calculate the time by A

Using formula of time

$T=t\sqrt{\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}}$

Put the value into the formula

$T=1\sqrt{\dfrac{1+0.6}{1-0.6}}$

$T=2\ years$

Similarly,

The expression for distance cover by A

$D=d\sqrt{1-\dfrac{v^2}{c^2}}$

$D=12\sqrt{1-(0.6)^2}$

$D=9.6\ ly$

We need to calculate the time

Using formula of time

$t=\dfrac{D}{v}$

$t=\dfrac{9.6}{0.6}$

$t=16\ years$

We need to calculate the signals received by twin A

Using formula for number of signals

$n=\dfrac{t}{T}$

Put the value into the formula

$n=\dfrac{16}{2}$

$n=8\ signals$

Hence, 8 signals received by twin A during the trip.

7 0

3 years ago

What is the logical oreder for these words : verse, interlude,verse,refrain, introduction,refrain,coda,refrain,verse

Nikolay [14]

Sorry to say but I know that t(e introduction is first and the coda is last

3 0

3 years ago

Find the equilibrant of two 10.0-N forces acting upon a body when the angle between the forces is 90° Solve graphically using a

bazaltina [42]

The equilibrant force of the two given forces is 14.14 N.

<h3 /><h3 /><h3>What is equilibrant force?</h3>

This is a single force that balances other given forces.

The given parameters:

First force, F₁ = 10 N
Second force, F₂ = 10 N
Angle between the forces, θ = 90⁰

The equilibrant force of the two given forces is calculated as follows;

$F = \sqrt{F_1 ^2 + F_2 ^2} \\\\F = \sqrt{(10)^2 + (10)^2} \\\\F = 14.14 \ N$

Thus, the equilibrant force of the two given forces is 14.14 N.

Learn more about equilibrant force here: brainly.com/question/8045102

5 0

3 years ago

If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?

sdas [7]

Data:
$f_{2} = 42 Hz$
n (Wave node)
V (Wave belly)
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
$f_{n} = \frac{nV}{2L}$

Wire 2 → 2º Harmonic → n = 2

$f_{n} = \frac{nV}{2L}$
$f_{2} = \frac{2V}{2L} 
$
$2V = f_{2} *2L$
$V = \frac{ f_{2}*2L }{2}$
$V = \frac{42*2L}{2}$
$V = \frac{84L}{2}$
$V = 42L$

Wire 1 → 1º Harmonic or Fundamental rope → n = 1

$f_{n} = \frac{nV}{2L}$
$f_{1} = \frac{1V}{2L}$
$f_{1} = \frac{V}{2L}$

If, We have:
V = 42L
Soon:
$f_{1} = \frac{V}{2L}$
$f_{1} = \frac{42L}{2L}$
$\boxed{f_{1} = 21 Hz}$

Answer:

<span>The fundamental frequency of the string:
</span>21 Hz

7 0

3 years ago

Read 2 more answers

In this graph, what is the displacement of the particle in the last two seconds?

nikklg [1K]

In this graph, what is the displacement of the particle in the last two seconds?of the particle in the last two seconds?

<span>0.2 meters
2 meters
4 meters
6 meters</span>
In this graph, the displacement of the particle in the last two seconds is 2 meters.

7 0

3 years ago