Sara goes to her apartment on the fifth floor by elevator from

If two variables have a non-related relationship and one of the variables is changed, how will the other variable change?

PolarNik [594]

If one of the variables is changed, that tells nothing about what happens to the other one, or IF anything happens, or when, or how long it lasts. Because they are UN-RELATED. You just said so yourself.

None of the choices says this.

6 0

3 years ago

A third baseman makes a throw to first base 40.5 m away. The ball leaves his hand with a speed of 30.0 m/s at a height of 1.4 m

Ugo [173]

Answer:

When the ball goes to first base it will be 4.23 m high.

Explanation:

Horizontal velocity = 30 cos17.3 = 28.64 m/s

Horizontal displacement = 40.5 m

Time

$t=\frac{40.5}{28.64}=1.41s$

Time to reach the goal posts 40.5 m away = 1.41 seconds

Vertical velocity = 30 sin17.3 = 8.92 m/s

Time to reach the goal posts 40.5 m away = 1.41 seconds

Acceleration = -9.81m/s²

Substituting in s = ut + 0.5at²

s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m

Height of throw = 1.4 m

Height traveled by ball = 2.83 m

Total height = 2.83 + 1.4 = 4.23 m

When the ball goes to first base it will be 4.23 m high.

8 0

3 years ago

A string is tied between two posts separated by 2.4 m. When the string is driven by an oscillator at frequency 567 Hz, 5 points

Alex787 [66]

Explanation:

The given data is as follows.

Length (l) = 2.4 m

Frequency (f) = 567 Hz

Formula to calculate the speed of a transverse wave is as follows.

f = $\frac{5}{2l} \times v$

Putting the gicven values into the above formula as follows.

f = $\frac{5}{2l} \times v$

567 Hz = $\frac{5}{2 \times 2.4 m} \times v$

v = 544.32 m/s

Thus, we can conclude that the speed (in m/s) of a transverse wave on this string is 544.32 m/s.

5 0

3 years ago

Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the sa

tester [92]

Answer:

Part a)

$t = 2.83 s$

Part b)

Ball thrown downwards = $v_f = 23.8 m/s$

Ball thrown upwards = $v_f = 23.8 m/s$

Part c)

$d = 22.24 m$

Explanation:

Part a)

Since both the balls are projected with same speed in opposite directions

So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection

Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

$\Delta y = 0 = v_y t + \frac{1}{2}at^2$

$0 = 13.9 t - \frac{1}{2}(9.81) t^2$

$t = 2.83 s$

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

$v_f^2 - v_i^2 = 2 a \Delta y$