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Nookie1986 [14]
3 years ago
5

ILL GIVE BRAINLYEST

Physics
1 answer:
lawyer [7]3 years ago
5 0

Answer:

3rd picture straight line going up right

Explanation:

3rd picture

You might be interested in
A simple pendulum has length of 820mm. Calculate the frequency (g = 9.8 ms -2)<br>​
Vadim26 [7]

Answer:

\huge\boxed{\sf f=0.55 \ Hz}

Explanation:

<u>Given Data:</u>

Length = l = 820 mm = 0.82 m

Acceleration due to gravity = g = 9.8 ms⁻²

<u>Required:</u>

Frequency = f = ?

<u>Formula:</u>

\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} }

<u>Solution:</u>

\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}

7 0
2 years ago
A circular dartboard has a radius of 2 meters and a red circle in the center. Assume you hit the target at a random point. For w
Sati [7]

Answer:

1.549 m

Explanation:

Given:

The radius of the circular board, r = 2 m

The probability of hitting the red is given as 0.6

Now, this probability of hitting the red can be conclude as

0.6 = (Area of red)/ (Total area of the board)

Total area of the board = πr² = π × 2²

let the radius of the red area be R

thus, area of red circle, = πR²

on substituting the value of the area, we have

0.6 = (πR²)/ (π × 2²)

or

R² = 2.4

or

R = 1.549 m

Thus, the radius of the red circle is 1.549 m

3 0
2 years ago
If you're studying the science of sound, you're studying____?
timurjin [86]
The study of sound is called sonics and the study of sound waves are acoustics
3 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
2 years ago
Un joven pelotero llamado Saúl en su primer año de ser firmado en grandes ligas, batea varias veces de cuadrangular, el segundo
Brilliant_brown [7]

Answer:

135-15=120

120÷3=40

40+40+(10)+40+(5)=135

8 0
2 years ago
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