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Oliga [24]
2 years ago
10

A simple harmonic oscillator has an amplitude of 3. 50 cm and a maximum speed of 26. 0 cm/s. What is its speed when the displace

ment is 1. 75 cm?.
Physics
1 answer:
Marianna [84]2 years ago
3 0

Answer:

1.) A simple harmonic oscillator has an amplitude of 3.50 cm and a maximum speed of 26.0 cm/s. What is its speed when the displacement is 1.75 cm? 2.) Both pendulum A and B are 3.0 m long. The period of A is T. Pendulum A is twice as heavy as pendulum B. What is the period of B? 3.) The time for one cycle of a periodic process is called the _ ? 4.) In simple harmonic motion, the acceleration is proportional to? 5.) The position of a mass that is oscillating on a spring is given by x= (18.3 cm) cos [(2.35 s-1)t]. What is the frequency of this motion?

Explanation:

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What an object is placed 8 mm from a concave spherical mirror a clear image can be projected on the screen 16 mm in front of me
alexgriva [62]

Concept: The magnification of spherical mirror can be defined by two ways.

(i) In terms of the height of the object and image.

The magnification of the spherical mirror is defined as the ratio of the height of the image'h_{i}' to the height of the object 'h_{o}'. It is denoted by letter 'm'.

Mathematically, it can be written as

m= \frac{h_{i}}{h_{o}}   ------------(1)

(ii) In terms of the object's and image's distances.

The magnification of the spherical mirror is defined as the negative ratio of the image distance'd_{i}' to the object distance 'd_{o}'.

Mathematically, it can be written as

m= - \frac{d_{i}}{d_{o}}   ------------(2)

Now, from equation (1) and (2) we have,

m = \frac{h_{i}}{h_{o}}   = -  \frac{d_{i}}{d_{o}}  -----------(3)

Given: Spherical Concave Mirror,

We will consider positive sign for object's and image's distance because both are in front of the mirror.

Object distance (d_{o}) = + 8 mm.

Image distance (d_{i}) = + 16 mm

Object's height (h_{o}) = + 4 mm

Image's height (h_{i}) =?

Now, apply equation (3)

\frac{h_{i}}{h_{o}}   = - \frac{d_{i}}{d_{o}}

Or,   \frac{h_{i}}{4 mm}   = - \frac{+16 mm}{+8 mm}

Or, hi = - 8 mm

Here; negative sign means, the image will be inverted.

The image's height will be 8 mm.

4 0
3 years ago
An athlete is working out in the weight room. He steadily holds 50 kilograms above his head for 10 seconds. Which statement is t
postnew [5]

Answer:

A. The athlete isn’t doing any work because he doesn’t move the weight.

Explanation:

We must remember the definition of work, which says that work is equal to the product of mass by the distance displaced. In this case, the athlete only does work when he lifts the weight from the ground to the point where he holds the weight suspended.

So when he's holding the weight, he doesn't do any work.

3 0
3 years ago
The charge of a single electron
vlabodo [156]
B. Electrons are negative so one single 1 would be negative 1
7 0
3 years ago
A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth
Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

\tan\theta=\dfrac{8}{3}

\thta=\tan^{-1}\dfrac{8}{3}

\theta=69.44^{\circ}

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

mg=2T\cos\theta

Put the value into the formula

3\times9.8=2T\times\cos69.44

T=\dfrac{3\times9.8}{2\times\cos69.44}

T=41.85\ N  

Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.

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3 years ago
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(I will give brainliest whoever helps me !!)
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C. Forces have mass and take up space
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