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Leto [7]
3 years ago
5

X-rays are used to take pictures of your bones. UV rays just burn your skin. Why do you think x-rays can go through skin and UV

rays cannot?
(Please help, And try to be as detailed as possible.)​
Physics
1 answer:
Elena L [17]3 years ago
6 0

Answer:

yes x rays are like light rays the difference is that they v

can oass through stuffs

skin and fats doesn't block more of the x rays

Explanation:

why uv cannot go through it us dangerous causing skin disease and cancer,skin itchness and redness and ut is cumulative

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This picture represents the electric field diagram between two particles with static charges. Do the two particles have the same
dexar [7]

Answers:

No, They will attract each other, B, and neither direction

Explanation:

Since the two already presented particles in the diagram represent both opposing charges due to the direction of the arrows (the arrows facing away from the particle shows a positive charge and the particles facing towards the particle show a negative charge), not only because of this but as the arrows between the particles show an attracting magnetic field, then it can be concluded that the particles will attract to each other and if another particle was introduced into the diagram of a positive charge, then it would attract to the negatively charged particle. If you have any questions or need further explanation, please comment below. E2021, have a great day.

7 0
3 years ago
A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500
defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

M = \frac{\mu_0 N_1 N_2A_1}{l_1}

Where,

\mu =permeability of free space

N_1 = Number of turns in solenoid 1

N_2 = Number of turns in solenoid 2

A_1= Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

\mu_0 = 4\pi *10^{-7}H/m

N_1 = 500

N_2 = 40

A = 7.5*10^{-4}m^2

l = 0.5m

Substituting,

M = \frac{\mu_0 N_1 N_2A_2}{l_1}

M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}

M = 3.77*10^{-4}H

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0
4 years ago
When drawing ray diagrams involving thin lenses, how many rays (at a minimum) are needed show the image distance and magnificati
Rufina [12.5K]

Answer:

GGG he fggfggfufbvg I Rd cbh

6 0
4 years ago
An object is launched with an initial velocity of 50.0 m/s at a launch angle of 36.9∘ above the horizontal. part a determine x-v
ankoles [38]
Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal

Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s

The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s

The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.

Answer:

  t, s    x, m
------  --------
     0   0
     1   39.98
     2   79.79
     3   112.68
     4   159.58
     5   199.47
     6   239.37

5 0
3 years ago
Read 2 more answers
Describe a situation where you can be traveling at a low speed but have an extremely high velocity
Reil [10]

Answer:

Cruising at 35,000 feet in an airliner, straight toward the east,

at 500 miles per hour

Explanation:

3 0
3 years ago
Read 2 more answers
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