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ollegr [7]
3 years ago
12

Two masses, A and B, are attached to different springs. Mass A vibrates with an amplitude of 8 cm at a frequency of 10 Hz and ma

ss B vibrates with an amplitude of 5 cm at a frequency of 16 Hz. How does the maximum speed of A compare to the maximum speed of B
Physics
1 answer:
Anettt [7]3 years ago
8 0

To solve this problem we will apply the concepts of speed given the simple harmonic movement, for which it defines this speed as

v_{max} = \omega A

Here

\omega =Angular velocity

A = Amplitude

Recall that the angular velocity is equivalent in terms of the frequency at

\omega = 2\pi f

If we replace the value we will have then

v_{max} = 2\pi f A

<em>For mass A </em>

v_{max,A} = 2\pi (10)(0.08) = 5.024m/s

<em>For mass B </em>

v_{max.B} = 2\pi (16)(0.05) = 5.024m/s

Therefore they are equal.

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If your speedometer has an uncertainty of 2.5 km/h at a speed of 92 km/h, what is the percent uncertainty
Elis [28]

Answer:

2.7%

Explanation:

Given:

Uncertainty of the speedometer (u)= 2.5km/h

Speed measured at that uncertainty (v) = 92km/h

Percent uncertainty (p) is given as the ratio of the uncertainty to the speed measured then multiplied by 100%. i.e

p = \frac{u}{v} * 100%

p = \frac{2.5}{92} * 100%

p = 2.7%

Therefore, the percent uncertainty is 2.7%

6 0
2 years ago
21. If the Sun's rays were at 45° to a vertical pillar, how would
diamong [38]

Answer:

Let the height of the pole AB = x m. ∴ Length of shadow OB ol the pole AB = x m. Let the angle of elevation be ө, i.e. Hence, the angle of elevation of the Sun's altitude is 45°.

Explanation:

4 0
3 years ago
on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker
DIA [1.3K]
<u>Momentum</u> 
- a vector quantity; has both magnitude and direction
- has the same direction as object's velocity
- can be represented by components x & y.

Find linebacker momentum given m₁ = 120kg, v₁ = 8.6 m/s north
P₁ = m₁v₁
P₁ = (120)(8.6)
[ P₁ = 1032 kg·m/s ] = y-component, linebacker momentum

Find halfback momentum given m₂ = 75kg, v₂ = 7.4 m/s east
P₂ = m₂v₂
P₂ = (75)(7.4)
[ P₂ = 555 kg·m/s ] = x-component, halfback momentum

Find total momentum using x and y components.
P = √(P₁)² + (P₂)²
P = √(1032)² + (555)²
[[ P = 1171.77 kg·m/s ]] = magnitude 

!! Finally, to find the magnitude of velocity, take the divide magnitude of momentum by the total mass of the players.
P = mv
P = (m₁ + m₂)v
1171.77 = (120 + 75)v      <em>[solve for v]</em>
<em />v = 1171.77/195
v = 6.0091 ≈ 6.0 m/s

If asked to find direction, take inverse tan of x and y components.
tanθ = (y/x)
θ = tan⁻¹(1032/555)
[ θ = 61.73° north of east. ]

The magnitude of the velocity at which the two players move together immediately after the collision is approximately 6.0 m/s.
6 0
3 years ago
for the same type of atom with 8 protons, 9 neutrons, and 10 electrons, what type of atom or ion is it​
Ilia_Sergeevich [38]

Answer:

O²⁻

Explanation:

Number of protons  = 8

Number of neutrons  = 9

Number of electrons  = 10

What type of atom or ion is it = ?

Solution:

Protons are the positively charged particle in an atom

Neutrons do not carry any charges

Electrons are negatively charged particles

 For this atom, the number of protons helps to identify what specie it is; so this is an oxygen atom.

Now,

     Charge  = Number of protons  - Number of electrons

      Charge  =  8  - 10  = -2

The charge on the atom is -2 and so it is an oxygen ion with -2 charge

 The ion is O²⁻

8 0
2 years ago
What is the period of a wave that has a frequency of 300 hz?
Butoxors [25]

Answer:

T = 0.003 s

(Period is written as T)

Explanation:

Period = time it takes for one wave to pass (measured in seconds)

frequency = number of cycles that occur in 1 second

(measured in Hz / hertz / 1 second)

Period : T

frequency : f

So, if we know that the frequency of a wave is 300 Hz, we can find the period of the wave from the relation between frequency and period

T =  \frac{1}{f}    f = \frac{1}{T}

to find the period (T) of this wave, we need to plug in the frequency (f) of 300

T = \frac{1}{300}

T = 0.00333333333

So, the period of a wave that has a frequency of 300 Hz is 0.003 s

[the period/T of this wave is 0.003 s]

3 0
1 year ago
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