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saul85 [17]
3 years ago
6

_______________have a negative charge and are located on the outside of the nucleus.

Physics
1 answer:
GuDViN [60]3 years ago
6 0
May 5th is Adachi day change your profile picture

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A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the mer
son4ous [18]

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, $\omega = \frac{1}{8}$  rev/sec

                             $=\frac{2 \pi \times 7.5}{8}$  rad/sec

                              = 5.89 rad/sec

Therefore, force required,

$F=m.\omega^2.r$

   $$=1640 \times (5.89)^2 \times 7.5  

   = 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

   = 427126.9 x 7.5

   = 3,203,451.75 J

6 0
2 years ago
If a dog is mass iS 14.3 kg what is it’s weight on earth
BigorU [14]

Answer:

Weight of the dog on surface of earth is 140.14 Newton.

Given:

mass of the dog = 14.3 kg

To find:

Weight of the dog = ?

Formula used:

Weight of the dog is given by,

W = mg

Where, W = weight of the dog

m = mass of the dog

g = acceleration due to gravity

Solution:

Weight of the dog is given by,

W = mg

Where, W = weight of the dog

m = mass of the dog = 14.3 kg

g = acceleration due to gravity

W = 14.3 × 9.8

W = 140.14 Newton

Weight of the dog on surface of earth is 140.14 Newton.


8 0
3 years ago
Two cars collide and come to a complete stop. where did all of their energy go?
MA_775_DIABLO [31]
Most of the energy will be absorbed by the materials that make up the cars, causing them to deform. The energy will also be converted into sound energy, causing a loud bang upon collision. Also, some energy will be converted to thermal energy, which will cause the cars to heat up slightly.
3 0
3 years ago
2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate
adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

  • v is image distance
  • u is object distance, u is 10 cm
  • f is focal length, f is 5 cm

{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

{ \tt{f + fM = v}} \\  { \tt{f(1 +M) = v }} \\ { \tt{1 +M =  \frac{v}{f}  }} \\  \\ { \boxed{ \mathfrak{formular :  } \: { \tt{ M =  \frac{v}{f}  - 1 }}}}

  • v is image distance, v is 10 cm
  • f is focal length, f is 5 cm
  • M is magnification.

{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

  • Image is magnified
  • Image is erect or upright
  • Image is inverted
  • Image distance is identical to object distance.
4 0
2 years ago
A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i
Fofino [41]

Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

(c) 11.518 m/s

Solution:

As per the question:

The eqn of the displacement is given by:

y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]          (1)

n = 4

Now,

We know the standard eqn is given by:

y = AsinKxcos\omega t           (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = 14.4 m^{- 1}

\omega = 166\ rad/s

where

A = Amplitude

K = Propagation constant

\omega = angular velocity

Now, to calculate the string's wavelength,

(a) K = \frac{2\pi}{\lambda}

where

K = propagation vector

\lambda = \frac{2\pi}{K}

\lambda = \frac{2\pi}{14.4} = 0.436\ m

(b) The length of the string is given by:

l = \frac{n\lambda}{2}

l = \frac{4\times 0.436}{2} = 0.872\ m

(c)  Now, we first find the frequency of the wave:

\omega = 2\pi f

f = \frac{\omega}{2\pi}

f = \frac{2\pi}{166} = 26.42\ Hz

Now,

Speed of the wave is given by:

v = f\lambda

v = 26.419\times 0.436 = 11.518\ m/s

4 0
3 years ago
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