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DENIUS [597]
3 years ago
10

If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular path were reduced,

what would happen?
Physics
2 answers:
Tanya [424]3 years ago
8 0

For PF STUDENTS:  The car would begin to move in the direction it was headed in a straight line.

saw5 [17]3 years ago
3 0
As the question states the reduction of the centripetal force is a reduction of the frictional force. By reducing these forces (they are the same force) the tires might not follow the circular path, this is the car could slip because the force would be insufficient to keep the circular path.
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If you wanted to detect x rays coming from the sun, where would you place the detector? Why?
Damm [24]
You would have to place your sensor above earth's atmosphere because it blocks out nearly all x-rays. this is why we have the Chandra observatory

hope this helps
4 0
3 years ago
Read 2 more answers
In the steady state 1.2 ✕ 1018 electrons per second enter bulb 1. There are 6.3 ✕ 1028 mobile electrons per cubic meter in tungs
bekas [8.4K]

Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

V=\frac{I}{nAq}

Where,

V= Drift Velocity

I= Flow of current

n= number of electrons

q = charge of electron

A = cross-section area.

For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is

\frac{I}{q} = 1.2*10^{18}

A= 1.3*10^{-8}m^2

n=6.3*10^{28} e/m^3

\omicron{O} = 1.2*10^{-4}(m/s)(N/c) Mobility

We can find the drift velocity replacing,

V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}

V= 1.465*10^-3m/s

The electric field is given by,

E= \frac{V}{\omicron{O}}

E=\frac{1.465*10^-3}{1.2*10^{-4}}

E=12.2V/m

7 0
3 years ago
A stone is thrown at an angle of 30 degrees above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. A
natta225 [31]

v = initial velocity of launch of the stone = 12 m/s

θ = angle of the velocity from the horizontal = 30

Consider the motion of the stone along the vertical direction taking upward direction as positive and down direction as negative.

v₀ = initial velocity along vertical direction = v Sinθ = 12 Sin30 = 6 m/s

a = acceleration of the stone = - 9.8 m/s²

t = time of travel = 4.8 s

Y = vertical displacement of stone = vertical height of the cliff = ?

using the kinematics equation

Y = v₀ t + (0.5) a t²

inserting the values

Y = 6 (4.8) + (0.5) (- 9.8) (4.8)²

Y = - 84.1 m

hence the height of the cliff comes out to be 84.1 m

5 0
2 years ago
If a substance is found to be reactive flammable soluble and explosive what observation is also a physical property
Ainat [17]

Answer:

  • <u><em>soluble</em></u>

Explanation:

Chemical properties only manifest when a chemical reaction occurs. Being reactive, flammable and explosive are chemical properties, because they involve chemical reactions: the substances are changed; the chemical bonds of some substances, called reactants, are broken, and the chemical bonds are created, forming other substances, called products.

Solubility is a<em> physical property</em> because during dissolution no new substances are formed. You can prove it when the solvent evaporates leaving behind the same original substance.

The the observation that the substance is <em>soluble</em> is describing a <em>physical property.</em>

3 0
3 years ago
Which telescope would be better viewing a faint, distant star? Why?
grin007 [14]
Reflecting telescope. Reflecting telescopes tend to have larger objective (due to the use of mirrors, mirrors are a lot cheaper than lenses) and have the ability to collect more light, while refracting telescopes are limited to objective lenses with smaller diameters due to their structural limitations (chromatic abbreviation, for example). Therefore, reflecting telescopes should be better at viewing faint distant stars
4 0
3 years ago
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