Ask question

Ask question

All categories

3 years ago

8

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

n the horizontal-axis O d. the variable you controlled

1 answer:

Nadya [2.5K]3 years ago

7 0

B is appropriate answer

You might be interested in

Sound waves produced by a source pass a point five times every second. Which of the following choices correctly describes the pe

Ugo [173]

The period is 1/5 second, and the frequency is 5 Hz.

Explanation:

Let's start by reviewing some definitions about waves:

- The period of a wave is the time taken for the wave to make one complete cycle. It is indicated with T and it is measured in seconds (s)

- The frequency of a wave is the number of complete cycles made by the wave in one second. It is indicated with f and it is measured in Hertz (Hz). It is equal to the reciprocal of the period:

$f=\frac{1}{T}$

where f is the frequency and T is the period.

In this problem, we have a wave that passes a given point five time per second. This means that the number of oscillations per second is five, and so its frequency is:

$f=\frac{5 cycles}{1 sec}=5 Hz$

It follows than the period of the wave is:

$T=\frac{1}{f}=\frac{1}{5}s$

Therefore, the correct statement is

The period is 1/5 second, and the frequency is 5 Hz.

Learn more about waves, frequency and period:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

3 0

3 years ago

NEED ANSWERS ASAP !!!

morpeh [17]

Answer:

I think its B

Explanation:

because "This means that when you rubbed the plastic comb along your hair, your hair resisted the movement of the comb and slowed it down. The friction between two surfaces can cause electrons to be transferred from one surface to the other."

3 0

3 years ago

Read 2 more answers

A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-dire

oee [108]

Answer:

a) v, v

b) 2mv^2

c) Elastic collion

Explanation:

(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction

Here x, y represent direction.They are not variable. 1 and 2 represent before and after.

2vm=v1xm+v2xm, we find v1x=v.

From momentum conservation in y-direction

0 =v1ym+v2ym, we findv1y=v.

(b) By energy conservation principle

Before: K=1/2m(2v)^2=2mv^2.

After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2

(c) The collision is elastic

6 0

3 years ago

A ball of mass m, attached to the end of a horizontal cord, is rotated in a circle of radius r on a frictionless horizontal surf

kirza4 [7]

Answer: $v=\sqrt{\frac{FL}{m}}$

Explanation:

Given

Ball of mass m

maximum Bearable Tension in string is F

Let length of the cord be L m and moving at a speed of v m/s

Here Tension will Provide Centripetal Force

T=Centripetal Force

$F=T=\frac{mv^2}{L}$

$v=\sqrt{\frac{FL}{m}}$

8 0

3 years ago

A spring with spring constant of 34 N/m is stretched 0.12 m from its equilibrium position. How much work must be done to stretch

Nesterboy [21]

Answer:0.253Joules

Explanation:

First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.

F = ke where;

F is the force

k is spring constant = 34N/m

e is the extension = 0.12m

F = 34× 0.12 = 4.08N

To get work done,

Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.

Work done = Force × Distance

Since F = 4.08m, distance = 0.062m

Work done = 4.08 × 0.062

Work done = 0.253Joules

Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules

8 0

3 years ago