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3 years ago

8

Ben is exercising his pony on a large oval track. The long is trotting at a constant speed of 10 m/s around the track. Which sta

tement can be made about the pony's acceleration

1 answer:

nlexa [21]3 years ago

6 0

There is 0 acceleration because the pony is moving at a constant velocity

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PLEASE HELP (WILL GIVE BRAINLIEST TO BEST ANSWER!!!!!)

Luden [163]

Answer:

1) a radio are uses by astronomy

2) 6 bilion waves

3) expert vertified

Explanation:

1) in contrast to an "ordinary" telescope, which receives visible light, a radio telescope "sees" radio waves emitted by radio sources, typically by means of a large parabolic ("dish") antenna, or arrays of them.and Radio telescopes are also the primary means to track space probes, and are used in the SETI project. so must been

radio are almostly

ceiver by astronomy

2) Radio waves have the longest wavelengths in the electromagnetic spectrum.

3)Expert Verified

Radio telescopes are telescopes that are specially designed for observation of long light wavelengths

CARRY ON ✨

3 0

3 years ago

A mine car (mass=390 kg) rolls at a speed of 0.50 m/s on a horizontal track, as the drawing shows. A 250-kg chunk of coal has a

Nady [450]

Answer:

v=0.60 m/s

Explanation:

Given that

m ₁= 390 kg ,u ₁= 0.5 m/s

m₂ = 250 kg ,u₂ = 0.76 m/s

As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.

Pi = Pf

m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v

Now putting the values in the above equation

390 x 0.5 + 250 x 0.76 = (390 + 250 ) v

$v=\dfrac{390\times 0.5+250\times 0.76}{390+250}\ m/s$

v=0.60 m/s

Therefore the velocity of the system will be 0.6 m/s.

8 0

4 years ago

A cylinder which is in a horizontal position contains an unknown noble gas at 4.63 × 104 Pa and is sealed with a massless piston

AleksandrR [38]

Answer:

The change in internal energy of the system is -17746.78 J

Explanation:

Given that,

Pressure $P=4.63\times10^{4}\ Pa$

Remove heat $\Delta U= -1.95\times10^{4}\ J$

Radius = 0.272 m

Distance d = 0.163 m

We need to calculate the internal energy

Using thermodynamics first equation

$dU=Q-W$ ...(I)

Where, dU = internal energy

Q = heat

W = work done

Put the value of W in equation (I)

$dU=Q-PdV$

Where, W = PdV

Put the value in the equation

$dU=-1.95\times10^{4}-(4.63\times10^{4}\times3.14\times(0.272)^2\times(-0.163))$

$dU=-17746.78\ J$

Hence, The change in internal energy of the system is -17746.78 J

3 0

4 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

7 0

4 years ago

Coherent light of frequency 6.37×1014 Hz passes through two thin slits and falls on a screen 88.0 cm away. You observe that the

IgorC [24]

Answer:

The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency $f= 6.37\times10^{14}\ Hz$

Distance of the screen l = 88.0 cm

Position of the third order y =3.10 cm

We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{c}{f}$

where, c = speed of light

f = frequency

Put the value into the formula

$\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}$

$\lambda=471\ nm$

We need to calculate the distance between the two slits

$m\times \lambda=d\sin\theta$

$d =\dfrac{m\times\lambda}{\sin\theta}$

Where, m = number of fringe

d = distance between the two slits

Here, $\sin\theta =\dfrac{y}{l}$

Put the value into the formula

$d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}$

$d=40.11\times10^{-6}\ m$

$d = 40.11\ \mu m$

Hence, The distance between the two slits is 40.11 μm.

7 0

3 years ago