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makvit [3.9K]
3 years ago
5

Which force causes all objects with mass to attract one another?

Physics
1 answer:
Scrat [10]3 years ago
5 0
Gravitational force
You might be interested in
A sinusoidal wave of wavelength 2.00m and amplitude 0.100 m travels on a string with a speed of 1.00 m/s to the right. At t = 0
Shkiper50 [21]
  1. The frequency and angular frequency are 0.500 Hertz and 3.142 rad/s. respectively.
  2. The angular wave number is equal to 3.142 rad/m.
  3. The wave function for this wave is given by: y = Asin(kx - ωt + Φ).
  4. The equation of motion for the left end of the string is given by: y = 0.100sin(3.142x - 3.142t + 0).
  5. The equation of motion for the left end of the string at x = 1.50 m to the right is equal to y = 0.100sin(4.71 - 3.142t + 0).The maximum speed of any point on the string is 0.3142 m/s.

<h3>How to calculate the frequency and angular frequency?</h3>

First of all, we would determine the frequency of this wave by using this formula:

Frequency = wavelength/speed

Frequency = 0.100/2.00

Frequency = 0.500 Hertz.

For the angular frequency, we have:

Angular frequency, ω = 2πf

Angular frequency, ω = 2 × 3.142 × 0.500

Angular frequency, ω = 3.142 rad/s.

<h3>How to determine the angular wave number?</h3>

Angular wave number, k = 2π/∧

Angular wave number, k = (2 × 3.142)/2.00

Angular wave number, k = 3.142 rad/m.

<h3>How to determine the wave function for this wave?</h3>

Mathematically, the wave function for this wave is given by:

y = Asin(kx - ωt + Φ)

For the equation of motion for the left end of the string, we have:

y = 0.100sin(3.142x - 3.142t + 0)

For the equation of motion for the left end of the string at x = 1.50 m to the right, we have:

y = 0.100sin(3.142x - 3.142t + 0)

y = 0.100sin(3.142(1.5) - 3.142t + 0)

y = 0.100sin(4.71 - 3.142t + 0)

<h3>What is the maximum speed of any point on the string?</h3>

Vy = 0.100sin(- 3.142)cos(3.142x - 3.142t)

Vy ≤ 0.3142 m/s (since cosine varies +1 and -1).

Read more on wave function here: brainly.com/question/11181093

#SPJ4

Complete Question:

A sinusoidal wave of wavelength 2.00 m and amplitude 0.100 m travels on a string with a speed of 1.00 m/s to the right.  Initially, the left end of the string is at the origin.  Find:

(a) the frequency and angular frequency,

(b) the angular wave number, and

(c) the wave function for this wave.  

Determine the equation of motion in SI units for

(d) the left end of the string, and

(e) the point on the string at x = 1.50 m to the right of the left end.

(f) What is the maximum speed of any point on the string?

5 0
2 years ago
If a boy rides his bicycle 100 meters in 20 seconds to the end of the street how fast did he ride his bike
Gekata [30.6K]
5 meters per second
5 0
3 years ago
3. A tennis ball is thrown straight up into the air with an initial velocity of 8.9 m/s. -9.81 gravity
otez555 [7]

Answer:

a. time is 0.91s

b. height is 4.04m

Explanation:

  • t = u/g
  • t = 8.9/9.81 = 0.91s
  • the time it takes for the tennis ball to reach its highest point is 0.91 seconds

  • S = u²/2g
  • S = 8.9²/2x9.81
  • s = 79.21/19.62
  • s = 4.04m
5 0
3 years ago
What will be the weight of a box, mass 10Kg on the surface of a hypothetical planet having mass double of Earth and the radius i
vlada-n [284]

Answer:

           W ’= 21.78 kg

Explanation:

The expression for weight is

          W = m g

let's look for the acceleration of gravity with the universal law of gravitation

          F = G m M / r2

          F = m (G M / r2)

without comparing the two equations

          g’= G M / r2

in that case M = 2 Mo and r = 3 ro

where mo and ro are the mass and radius of the earth

         we substitute

         g ’= G 2Mo / (3r₀) 2

         G ’= 2/9 G Mo / r₀²

         g ’= 2/9 g

the weight of the body on this planet is

         W ’= m g’

          W ’= m 2/9 g

let's calculate

          W ’= 2/9 10 9.8

           W ’= 21.78 kg

3 0
3 years ago
g A mass of 2.0 kg traveling at 3.0 m/s along a smooth, horizontal plane hits a relaxed spring. The mass is slowed to zero veloc
dezoksy [38]

By the work-energy theorem, the total work done on the mass by the spring is equal to the change in the mass's kinetic energy:

<em>W</em> = ∆<em>K</em>

and the work done by a spring with constant <em>k</em> as it gets compressed a distance <em>x</em> is -1/2 <em>kx</em> ²; the work it does is negative because the restoring force of the spring points opposite the direction in which it's getting compressed.

So we have

-1/2 <em>k</em> (0.15 m)² = 0 - 1/2 (2.0 kg) (3.0 m/s)²

Solve for <em>k</em> to get <em>k</em> = 800 N/m.

8 0
3 years ago
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