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galben [10]
3 years ago
6

To form a solution_____

Physics
1 answer:
Kipish [7]3 years ago
5 0
A. One substance must dissolve in another
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Which describes the relationship between photon energy and the color of light?
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Photon energy is directly proportional to the frequency of electromagnetic radiation.
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So the photon energy increases as you scan the chart of visible colors
moving from the red end of the rainbow to the blue end.
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A man runs at an average speed of 5.0m/s how long will it take him to run 5.2km on a perfectly straight line
Anni [7]

The time taken is 1040 s.

<h3>What is speed?</h3>

The term speed refers to the rate at which the distance changes per unit time. This is why we define speed as the ratio of the distance to time for a body that is moving along a straight line.

Now;

We must first convert the distance to meters;

distance = 5.2km  or 5200m

Speed = distance/time

time = distance/speed

time =  5200m/5 m/s

time = 1040 s

Learn more about speed:brainly.com/question/28224010

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8 0
1 year ago
A 22kg Accelerates at a rate of 2.3 m/s. What is the magnitude of the net force acting on the bike?
Tcecarenko [31]

magnitude of the net force = mass x acceleraton

                                             = 22 x 2.3

                                             =50.6 N

7 0
3 years ago
A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se
zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}

Also,

\vec{E} = \frac{2K\lambda }{R}                  (1)

where

K = electrostatic constant = \frac{1}{4\pi \epsilon_{o} R}

R = Distance

\lambda = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

\vec{E} = 1800\ N/C     (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C

\vec{E'} = 1800\ N/C     (towards)

Now, the total field at the origin is the sum of both the fields:

\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

7 0
3 years ago
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