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3 years ago

I need help with this too. (im not good at science or math)

Physics

2 answers:

dedylja [7]3 years ago

7 0

I believe 1 m/s.....

mr_godi [17]3 years ago

6 0

Answer:

i belive 1 m/s

Explanation:

dividing displacement from time it should be 1 cuz 5/5 is 1

please tell me if right!

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2. In the pendulum lab, the period (swing time) was O a. the independent variable O b.graphed on the vertical-axis O c.graphed o

Nadya [2.5K]

B is appropriate answer

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3 years ago

A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T. If the pendulum is take

Juli2301 [7.4K]

To solve this problem we will use the definition of the period in a simple pendulum, which warns that it is dependent on its length and gravity as follows:

$T =2\pi \sqrt{\frac{L}{g}}$

Here,

L = Length

g = Acceleration due to gravity

We can realize that $2 \pi$ is a constant so it is proportional to the square root of its length over its gravity,

$T \propto \sqrt{\frac{L}{g}}$

Since the body is in constant free fall, that is, a point where gravity tends to be zero:

$g \rightarrow 0 \Rightarrow T \rightarrow \infty$

The value of the period will tend to infinity. This indicates that the pendulum will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.

5 0

2 years ago

A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of

sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height $\Delta h$ depends on the square of the time:

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ,

where

$\Delta h$ is the change in the ball's height.
$g$ is the acceleration due to gravity.
$t$ is the time for which the ball is in the air.
$v_0$ is the initial vertical velocity of the ball.

The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. $\Delta h = -0.950\;\text{m}$ .
Gravity pulls objects toward the earth, so $g$ is also negative. $g \approx -9.81\;\text{m}\cdot\text{s}^{-2}$ near the surface of the earth.
Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, $v_0 = 0$ .

Solve for $t$ .

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ;

$\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2}$ ;

$\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)}$ ;

$t \approx 0.440315\;\text{s}$ .

What's the initial horizontal velocity of the ball?

Horizontal displacement of the ball: $\Delta x = 0.352\;\text{m}$ ;
Time taken: $\Delta t = 0.440315\;\text{s}$

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

$\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}$ .

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0

3 years ago

Which type of wave would not be classified as a transverse wave?

Jlenok [28]

Longitud wave something like that.

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3 years ago

The first model of the atom was developed through

Ilya [14]

It was developed through Democritus who was a greek philosopher.

Hope this helps

7 0

2 years ago

Read 2 more answers