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faust18 [17]
3 years ago
14

A car is traveling around a horizontal circular track with radius r = 220 m at a constant speed v = 16 m/s as shown. The angle θ

A = 36° above the x axis, and the angle θB = 54° below the x axis. 1)
What is the x component of the car’s acceleration when it is at point A
Physics
1 answer:
iogann1982 [59]3 years ago
5 0
Velocity = distance / time = ( 2 * pi * r ) / t = 20.583 m/s 

<span>x component = sine ( 32 ° ) * 20.583 = 10.91 m/s

 hope this helps :)

</span>
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4. How does the type of medium affect a sound wave?
jek_recluse [69]

Answer:

The type of medium affects a sound wave as sound travels with the help of the vibration in particles.

Explanation:

As different mediums have different amount and size of particles, for example, the speed of sound is faster through solid than liquid as solids have closely packed particles whereas liquids are loosely packed.

5 0
3 years ago
A boy throws a ball straight up with a speed of 21.5 m/s. The ball has a mass of 0.19 kg. How much gravitational potential energ
astra-53 [7]

Answer:

Explanation:

The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation

v_f=v_0+at where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...

0 = 21.5 + (-9.8)t and

-21.5 = -9.8t so

t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)

Now we will use that time to find out the max height of the object in the equation

Δx = v_0t+\frac{1}{2}at^2 and filling in:

Δx = 21.5(2.19)+\frac{1}{2}(-9.8)(2.19)^2 which simplifies down a bit to

Δx = 47.1 - 23.5 so

Δx = 23.6 meters.

Now we can plug that in to the PE equation to find the PE of the object:

PE = (.19)(9.8)(23.6) so

PE = 43.9 J

5 0
3 years ago
A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the partic
olga_2 [115]

Answer:

The speed v of the particle at t=5.00 seconds = 43 m/s

Explanation:

Given :

mass m = 5.00 kg

force f(t) = 6.00t2−4.00t+3.00 N

time t between t=0.00 seconds and t=5.00 seconds

From mathematical expression of Newton's second law;

Force = mass (m) x acceleration (a)

F = ma              

a = \frac{F}{m}      ...... (1)

acceleration (a) = \frac{dv}{dt}   ......(2)

substituting (2) into (1)

Hence, F = \frac{mdv}{dt}

\frac{dv}{dt} = \frac{F}{m}

dv = \frac{F}{m} dt

dv = \frac{1}{m}Fdt

Integrating both sides

\int\limits {} \, dv = \frac{1}{m} \int\limits {F(t)} \, dt

The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;

v = \frac{1}{m} \int\limits^5_0 {F(t)} \, dt     ......(3)

Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):

v = \frac{1}{5} \int\limits^5_0 {(6t^{2} - 4t + 3)} \, dt

v = \frac{1}{5} |\frac{6t^{3} }{3} - \frac{4t^{2} }{2} + 3t |^{5}_{0}

v = \frac{1}{5} |(\frac{6(5)^{3} }{3} - \frac{4(5)^{2} }{2} + 3(5)) - 0|

v = \frac{1}{5} |\frac{6(125)}{3} - \frac{4(25)}{2} + 15 |

v = \frac{1}{5} |\frac{750}{3} - \frac{100}{2} + 15 |

v = \frac{1}{5} | 250 - 50 + 15 |

v = \frac{215}{5}

v = 43 meters per second

The speed v of the particle at t=5.00 seconds = 43 m/s

6 0
3 years ago
A waterwheel built in hamah
andreev551 [17]

☁️ Answer ☁️

annyeonghaseyo!

Your answer is:

"A waterwheel built in Hamah, Syria, has a radius of 20.0 m. If the tangential velocity at the wheel’s edge is 7.85 m/s" -Ggle

Hope it helps.

Have a nice day noona/hyung!~  ̄▽ ̄❤️

4 0
3 years ago
A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A
TiliK225 [7]

Answer:

funkin

Explanation:

8 0
4 years ago
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