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Svetllana [295]

3 years ago

13

Which graph shows acceleration? A graph of position (meters) versus time (seconds) has a straight line running from 0 seconds 0

meters upward. A graph of position (meters) versus time (seconds) has a concave line running from 0 seconds 0 meters upward. A graph of position (meters) versus time (seconds) has a straight line running from 0 seconds positive number of meters downward to some later time 0 meters.

2 answers:

anzhelika [568]3 years ago

7 0

Answer:

its the middle one

Explanation:

i did it on edge 2021

charle [14.2K]3 years ago

6 0

It's the one that has a concave line running upward.

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jenny's model train is set up on a circular track. There are six telephone poles evenly spaced around the track. It takes the en

d1i1m1o1n [39]

Answer:

T = 60 s

Explanation:

There are 6 poles on the track which are equally spaced

so the angular separation between the poles is given as

$\theta = \frac{2\pi}{6}$

$\theta = \frac{\pi}{3}$

so the angular speed of the train is given as

$\omega = \frac{\theta}{t}$

$\omega = \frac{\pi}{30} rad/s$

now we have time period of the train given as

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{\frac{\pi}{30}}$

$T = 60 s$

5 0

3 years ago

Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted out

blondinia [14]

Answer:

$\theta = 20.98 degree$

Explanation:

As we know that the speed of the sound is given as

$v = 332 + 0.6 t$

now at t = 273 k = 0 degree

$v = 332 m/s$

so we have

$a sin\theta = N\lambda$

$a sin\theta = N(\frac{v_1}{f})$

now when temperature is changed to 313 K we have

$t = 313 - 273 = 40 degree$

now we have

$v = 332 + (0.6)(40)$

$v_2 = 356 m/s$

$a sin\theta' = N(\frac{v_2}{f})$

now from two equations we have

$\frac{sin19.5}{sin\theta} = \frac{332}{356}$

so we have

$sin\theta = 0.358$

$\theta = 20.98 degree$

7 0

3 years ago

As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road i

Darya [45]

<span>As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road is kinetic friction.

We exert force to move the object from rest and in this case, static friction works. But, when the object comes in motion, then kinetic friction works. Here, since the car is driving without slipping means, kinetic friction acts on it. Its also called sliding or dynamic friction.</span>

5 0

3 years ago

When the Moon orbits Earth, what is the centripetal force?

nata0808 [166]

Answer:

Gravity is the centripetal force when the moon orbits the earth.

5 0

2 years ago

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0

3 years ago