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Arturiano [62]
3 years ago
12

To determine a waves frequency you must know the??

Physics
2 answers:
saul85 [17]3 years ago
8 0

Answer:

To determine a waves frequency you must know the distance it travels!

Explanation:

Frequency=speed. Something must move for you to determine how fast it  moves!

lutik1710 [3]3 years ago
7 0

Answer: I think, the number of oscillations in a given period of time.

Explanation: Well I guess because in a period time is known as the rate of occurrence of the wave. Hope this helps!

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Can someone please help me. idk what to do​
Alex_Xolod [135]

Answer:

-4.5 m/s

Explanation:

Even without doing calculations, we know that a projectile has the same speed coming down as it did going up at the same height.  Since it went up at 4.5 m/s, it returns down at -4.5 m/s.

To prove it with math, use the given information:

v₀ = 4.5 m/s

a = -9.8 m/s²

Δy = 0 m

Find: v

v² = v₀² + 2aΔy

v² = (4.5 m/s)² + 2 (-9.8 m/s²) (0 m)

v = -4.5 m/s

3 0
3 years ago
20 POINTSSS!!!!!!!!
Soloha48 [4]
Answer: A
Hope this help you!!
7 0
3 years ago
Which of the following is the best reason for using a scanning electron microscope?
torisob [31]

Answer is C: Ability to see three-dimensional images of the surfaces of object

Explanation:

To enable the technician see fractures and broken particles in a better resolution as the SEM sees the peaks and valley of the structure.

8 0
3 years ago
A 1300 kg steel beam is supported by two ropes. (Figure
Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

8 0
2 years ago
A positive test charge of 8.5 × 10 negative 7 Columbus experiences a force of 4.1 × 10 negative 1 N calculate the electric field
Sophie [7]

Explanation:

direction of electric field is same as that of force experienced by the test charge

8 0
3 years ago
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