NEED ANSWER FAST WILL GIVE BRAINLIEST ANSWER IF CORRECT.

Is the maximum displacement of a point on a vibrating body or wave.

Georgia [21]

The correct answer is amplitude

7 0

3 years ago

An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.

Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

$u -v= d$

$u=71+26=97\ cm$

We need to calculate the focal length

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}$

$\dfrac{1}{f}=-\dfrac{71}{2522}$

$f=-35.52\ cm$

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value in to the formula

$\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}$

$\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}$

$\dfrac{1}{v}=-\dfrac{124}{2755}$

$v=-22.21\ cm$

We need to calculate the magnification

Using formula of magnification

$m=\dfrac{v}{u}$

$m=\dfrac{22.21}{95}$

$m=0.233$

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

4 0

3 years ago

Gasoline is an example of an accelerant. True False

Kay [80]

Answer:

true

Explanation:

7 0

2 years ago

Read 2 more answers

You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 76.2 kg hop on board for a ride

marin [14]

a) The spring constant is 12,103 N/m

b) The mass of the trailer 2,678 kg

c) The frequency of oscillation is 0.478 Hz

d) The time taken for 10 oscillations is 20.9 s

Explanation:

a)

When the two children jumps on board of the trailer, the two springs compresses by a certain amount

$\Delta x = 6.17 cm = 0.0617 m$

Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:

$2mg = k'\Delta x$ (1)

where

m = 76.2 kg is the mass of each children

$g=9.8 m/s^2$ is the acceleration of gravity

$k'$ is the equivalent spring constant of the 2-spring system

For two springs in parallel each with constant k,

$k'=k+k=2k$

Substituting into (1) and solving for k, we find:

$2mg=2k\Delta x\\k=\frac{mg}{\Delta x}=\frac{(76.2)(9.8)}{0.0617}=12,103 N/m$

b)

The period of the oscillating system is given by

$T=2\pi \sqrt{\frac{m}{k'}}$

where

And for the system in the problem, we know that

T = 2.09 s is the period of oscillation

m is the mass of the trailer

$k'=2k=2(12,103)=24,206 N/m$ is the equivalent spring constant of the system

Solving the equation for m, we find the mass of the trailer:

$m=(\frac{T}{2\pi})^2 k'=(\frac{2.09}{2\pi})^2 (24,206)=2,678 kg$

c)

The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:

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

where

f is the frequency

T is the period

In this problem, we have

T = 2.09 s is the period

Therefore, the frequency of oscillation is

$f=\frac{1}{2.09}=0.478 Hz$

d)

The period of the system is

T = 2.09 s

And this time is the time it takes for the trailer to complete one oscillation.

In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.

Therefore, the time needed for 10 oscillations is:

$t=10T=10(2.09)=20.9 s$

#LearnwithBrainly

7 0

3 years ago

A ball of radius r rolls on the inside of a track of radius R. If the ball starts from rest at the vertical edge of the track, f

meriva

Answer:

$v=\sqrt{\dfrac{10g(R-r)}{7}}$

Explanation:

Given that

Radius of track = R

Radius of ball = r

The ball can be treated as solid sphere, so

The moment of inertia of ball

$I=\dfrac{2}{5}mr^2$

When the ball reach at the lowest position then it will have both angular and linear speed.

Condition for rolling without slipping v= ωr

Form energy conservation

$mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

v= ωr

$I=\dfrac{2}{5}mr^2$

$mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2$

$mg(R-r)=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mv^2$

$2mg(R-r)=mv^2+\dfrac{2}{5}mv^2$

$2g(R-r)=\dfrac{7}{5}v^2$

$v=\sqrt{\dfrac{10g(R-r)}{7}}$

3 0

3 years ago