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Zielflug [23.3K]
3 years ago
10

In liquids, the particles vibrate and change position, but the particles in liquids do not have more motion than the particles i

n gases. true or false
Physics
1 answer:
dexar [7]3 years ago
5 0
The right answer is true because particles are allowed to freely move in gases but in liquids the particles can not leave the other particles, they may move freely but with other particles but gas particles can be found individually roaming around
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We know that the magnitudes of the negative charge on the electron and the positive charge on the proton are equal. Suppose, how
TEA [102]

Answer:3,45 x 10^9 N

Explanation: We have considered the total charge for each coin , this is the total atoms x 29 electrons for cooper and multiplier by electron charge, the total charge for each coin is 0,464 C

Finally we use the Coulomb law,

F=k Q/ (r)^2

6 0
3 years ago
A mouse ran 100 centimeters in
olga_2 [115]

Answer:

his speed is 5cm a second

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3 years ago
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An oil slick on water is 120 nm thick and illuminated by white light incident perpendicular to its surface. What color does the
gregori [183]

Answer:

\lambda = 672 nm

so this is nearly red colour light

Explanation:

As we know that the interference of light from reflected light then the path difference is given as

\Delta x = 2\mu t + \frac{\lambda}{2}

now we know that for constructive interference of light the path difference is given as

\Delta x = n\lambda

so we will have

2\mu t + \frac{\lambda}{2} = N\lambda

so we will have

4\mu t = \lambda

\lambda = 2(1.40)(120nm)

\lambda = 672 nm

so this is nearly red colour light

8 0
3 years ago
A 75.0kg bicyclist (including the bicycle) is pedaling to the right, causing her speed to increase at a rate of 2.20m/s^2, despi
malfutka [58]

1) 4 forces

2) 165 N

3) 225 N

Explanation:

1)

There are in total 4 forces acting on the bicylist:

- The gravitational force on the byciclist, acting vertically downward, of magnitude mg, where m is the mass of the bicyclist and g is the acceleration due to gravity

- The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force

- The force of push F, acting horizontally forward, given by the push exerted by the bicylist on the pedals

- The air drag, R, of magnitude R = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist

2)

The magnitude of the net force on the bicyclist can be calculated by considering separately the two directions.

- Along the vertical direction, we have the gravitational force (downward) and the normal force (upward); these two forces are equal in magnitude, since the acceleration of the bicyclist along this direction is zero, therefore the net force in this direction is zero.

- Along the horizontal direction, the two forces (forward force of push and air drag) are balanced, since the acceleration is non-zero, so we can use Newton's second law of motion to find the net force on the bicylist:

F_{net}=ma

where

F_{net} is the net force

m = 75.0 kg is the mass of the bicyclist

a=2.20 m/s^2 is its acceleration

Solving, we find the net force:

F_{net}=(75.0)(2.20)=165 N

3)

In this part, we basically want to find the forward force of push, F.

We can rewrite the net force acting on the bicyclist as

F_{net}=F-R

where:

F is the forward force of push

R is the air drag

We know that:

F_{net}=165 N is the net force on the bicyclist

R = 60.0 N is the magnitude of the air drag

Therefore, by re-arranging the equation, we can find the force generated by the bicylicst by pedaling:

F=F_{net}+R=165+60=225 N

6 0
3 years ago
A diverging lens of focal length 18.0m is used to view a shark that is 90.0m away from the lens. If the image formed is 1.0m lon
Reil [10]

Answer:

i. + 22.5 m ii. 4.0 m

Explanation:

i. Image distance

Using the lens formula

1/u + 1/v = 1/f where f = focal length = + 18.0 m, u = object distance = distance of shark away from lens = + 90.0 m and v = image distance from lens = unknown

So, we find v

1/v = 1/f - 1/u

= 1/+18 - 1/+90

= (5 - 1)/90

= 4/90

v = 90/4

= + 22.5 m

So the image is real and formed 22.5 m away on the other side of the lens.

ii Length of Shark

Using the magnification formula, m = image height/object height = image distance/object distance. image height = 1.0 m where object height = length of shark.

m = image distance/object distance

= v/u

= +22.5/+90

= 0.25

0.25 = image height/object height

So,

object height = image height/0.25

= 1.0 m/0.25

= 4.0 m

So, the length of the shark is 4.0 m

7 0
3 years ago
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