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3 years ago

can someone please help me A wave’s velocity is 120 m/sec with a frequency of 6 Hz. What is its wavelength?

Physics

2 answers:

Olegator [25]3 years ago

8 0

Answer:

by using formula,

wavelength= velocity/frequency

= 120/6

= 20 meter

Ans: 20 meter

RSB [31]3 years ago

3 0

Answer:

wavelength= velocity

= 120/6= 20 meter

Explanation:

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4. Brandon throws a tennis ball vertically upward. The ball returns to the point of release after 4.0 s. What is the

vekshin1

Answer:

-39.2m/s

Explanation:

Using the equation of motion;

v = u + at

Since the ball is thrown upward, the acceleration due to gravity acting on it will be negative, hence a = -g

v = u - gt

Since g = 9.8m/s²

t = 4.0s

u = 0m/s

v = 0 + (-9.8)(4)

v = -39.2m/s

Hence the speed of the ball before release is -39.2m/s

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3 years ago

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Aloiza [94]

The answer is A. Photosynthesis
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5 0

3 years ago

Read 2 more answers

Can anyone help me? (physics)

Masja [62]

Answer:

The initial velocity of the golf is 15.7 m/s.

The direction of the golf is 57°.

Explanation:

The following data were obtained from the question:

Time of flight (T) = 2.7 secs

Range (R) = 23 m

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =.?

Direction (θ) =.?

T = 2U Sine θ /g

2.7 = 2 × U × Sine θ /9.8

Cross multiply

2.7 × 9.8 = 2 × U × Sine θ

26.46 = 2 × U × Sine θ

Divide both side by 2 × Sine θ

U = 26.46 /2 Sine θ

U = 13.23 / Sine θ ... (1)

R = U² Sine 2θ /g

23 = U² Sine 2θ / 9.8

U = 13.23 / Sine θ

23 = (13.23/ Sine θ)² Sine 2θ / 9.8

23 = (175.0329 / Sine² θ) × Sine 2θ / 9.8

23 = 17.8605/Sine² θ × Sine 2θ

Recall:

Sine 2θ = 2SineθCosθ

23 = 17.8605/ Sine² θ × 2SineθCosθ

23 = 17.8605/ Sine θ × 2Cosθ

23 = 35.721 Cos θ /Sine θ

Cross multiply

23 × Sine θ = 35.721 Cos θ

Divide both side by 23

Sine θ = 35.721 Cos θ /23

Sine θ = 1.5531 × Cos θ

Divide both side by Cos θ

Sine θ /Cos θ = 1.5531

Recall:

Sine θ /Cos θ = Tan θ

Sine θ /Cos θ = 1.5531

Tan θ = 1.5531

Take the inverse of Tan

θ = Tan¯¹ (1.5531)

θ = 57°

Therefore, the direction of the golf is 57°

Thus, the initial velocity can be obtained as follow:

U = 13.23 / Sine θ

θ = 57°

U = 13.23 / Sine 57

U = 13.23/0.8387

U = 15.7 m/s

Therefore, the initial velocity of the golf is 15.7 m/s

8 0

3 years ago

What is the person’s displacement for the trip?<br><br> what is a vector quantity? give an example.

borishaifa [10]

Answer:

+ 3.0 m

Explanation:

displacement is shortest distance from fixed point O in particular direction . in diagram shortest distance at end from O is 3 m and it is right of O so +. HENCE +3.0m

4 0

3 years ago

Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction

Norma-Jean [14]

Answer:

The ball would hit the floor approximately $0.55\; \rm s$ after leaving the table.

The ball would travel approximately $5.5\; \rm m$ horizontally after leaving the table.

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Let $\Delta h$ denote the change to the height of the ball. Let $t$ denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let $v_0(\text{vertical})$ denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible: $\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t$ .

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was $v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}$ .

The height of the table was $1.5\; \rm m$ . Therefore, after hitting the floor, the ball would be $1.5\; \rm m \!$ below where it was before leaving the table. Hence, $\Delta h = -1.5\;\rm m$ .

The equation becomes:

$\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}$ .

Solve for $t$ :

$\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55$ .

In other words, it would take approximately $0.55\; \rm s$ for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at $v(\text{horizontal}) =10\; \rm m \cdot s^{-1}$ ) until the ball hits the floor.

The ball was in the air for approximately $t = 0.55\; \rm s$ and would have travelled approximately $v(\text{horizontal})\cdot t \approx 5.5\;\rm m$ horizontally during the flight.

4 0

3 years ago