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Arturiano [62]
2 years ago
9

PLS HELP

Physics
2 answers:
Amanda [17]2 years ago
7 0

V(final) = V(initial) + a•t

Take the numbers you know, pluggum in, and you should get 3.836 m/s .

Note: This is just the final speed. Without any direction information, there's no way to find the velocity.

Zepler [3.9K]2 years ago
5 0

Answer:

hmm first add 32  plus 53 plus 12 trust me and when you get the answer subtract 25

Explanation:

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A 5.22×104 kg railroad car moves on frictionless horizontal rails until it hits a horizontal spring stopper with a force constan
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To solve this problem we will apply the principles of conservation of energy, for which we have to preserve the initial kinetic energy as elastic potential energy at the end of the movement. If said equality is maintained then we can affirm that,

\text{Initial Energy}=\text{Final Energy}

\frac{1}{2} mv^2=\frac{1}{2} kx^2

Here,

m = mass

k = Spring constant

x = Displacement

v = Velocity

Rearranging to find the velocity,

mv^2 = kx^2

v^2 = \frac{kx^2}{m}

v = \sqrt{\frac{kx^2}{m}}

Our values are,

m = 5.22*10^4kg

k = 4.58*10^5N/m

x = 32cm = 0.32m

Replacing our values we have,

v = \sqrt{\frac{(4.58*10^5)(5.22*10^4)}{0.32}}

v = 2.733*10^5m/s

Therefore the velocity is 2.733*10^5m/s

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3 years ago
A freight train rolls along a track with considerable momentum. If it rolls at the same speed but hastwice as much mass, its mom
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Its Momentum Would Be Doubled

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3 years ago
Ceres, Pluto, and Eris are all round in shape and classified as:_________ A) Leftover planetesimals that formed inside the frost
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Ceres, Pluto, and Eris are classified as DWARF PLANET.

A) Leftover planetesimals inside the frost line are known as ASTEROIDS.

B) METEORITES are the pieces of Asteroids which are fallen on the earth's surface.                                                

C) COMETS are the objects which are visible with long tails.            

D) COMETS are also the leftover planetesimals that are occupied by the jovian planets and are formed in the solar system.                

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3 years ago
Definition of AM and Fm Wave? in your own word
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Answer:

In AM broadcasting, the amplitude of the carrier wave is modulated to encode the original sound. In FM broadcasting, the frequency of the carrier wave is modulated to encode the sound. A radio receiver extracts the original program sound from the modulated radio signal and reproduces the sound in a loudspeaker.

8 0
2 years ago
) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha
vaieri [72.5K]

(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

f= 466 Hz

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s

The angular frequency instead is given by

\omega = 2\pi f

And substituting

f = 466 Hz

We find

\omega = 2\pi (466 Hz)=2926.5 rad/s

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz

While the angular frequency is given by:

\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s

(c) 4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s

The minimum angular frequency of the light wave is

\omega_1 = 2.7\cdot 10^{15}rad/s

so the corresponding frequency is

f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz

and the period is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s

The maximum angular frequency of the light wave is

\omega_2 = 4.7\cdot 10^{15}rad/s

so the corresponding frequency is

f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz

and the period is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s

(d) 2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s

In this case, the frequency is

f=5.0 MHz = 5.0 \cdot 10^6 Hz

So the period in this case is

T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6  Hz}=2.0 \cdot 10^{-7} s

While the angular frequency is given by

\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s

7 0
3 years ago
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