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TiliK225 [7]
3 years ago
15

A child threw a stone straight down off a high bridge. The initial velocity of the stone was 15.0 m/s (include units and proper

significant figures). How fast was the stone moving the instant before it hit the river 3.00 sec later?
Physics
1 answer:
Andre45 [30]3 years ago
8 0

Answer:

The speed of the stone before it hit the river 3.00 sec later. Let v is the velocity at that instant is 45 m/s.

Explanation:

Given that, a child threw a stone straight down off a high bridge.

Initial velocity of the stone, u = 15 m/s

We need to find the speed of the stone before it hit the river 3.00 sec later. Let v is the velocity at that instant. When it come down, it is moving under the action of gravity. Using equation of motion as :

v=u+gt\\\\v=15+10\times 3\\\\v=45\ m/s

So, the speed of the stone before it hit the river 3.00 sec later. Let v is the velocity at that instant is 45 m/s.

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A charged particle having mass 6.64 x 10-27 kg (that of a helium atom) moving at 8.70 x 105 m/s perpendicular to a 1.30-T magnet
Fiesta28 [93]

Answer:

the charge of the particle is 2.47 x 10⁻¹⁹ C

Explanation:

Given;

mass of the particle, m = 6.64 x 10⁻²⁷ kg

velocity of the particle, v = 8.7 x 10⁵ m/s

strength of the magnetic field, B = 1.3 T

radius of the circle, r = 18 mm = 1.8 x 10⁻³ m

The magnetic force experienced by the charge is calculated as;

F = ma = qvB

where;

q is the charge of the particle

a is the acceleration of the charge in the circular path

a = \frac{v^2}{r} \\\\ma = qvB\\\\q = \frac{ma}{vB} \\\\q = \frac{mv^2}{rvB} = \frac{mv}{rB} \\\\q = \frac{(6.64\times 10^{-27} ) \times (8.7\times 10^5)}{(1.8\times 10^{-2}) \times (1.3)} \\\\q = 2.47 \ \times 10^{-19} \ C

Therefore, the charge of the particle is 2.47 x 10⁻¹⁹ C

6 0
3 years ago
If your chunk of gold weighed 1 N in which case would you have the largest mass of gold?
kotykmax [81]
Ah ha !  Very interesting question.
Thought-provoking, even.

You have something that weighs 1 Newton, and you want to know 
the situation in which the object would have the greatest mass.

          Weight = (mass) x (local gravity)

          Mass  =  (weight) / (local gravity)

          Mass  =  (1 Newton) / (local gravity)

"Local gravity" is the denominator of the fraction, so the fraction
has its greatest value when 'local gravity' is smallest.  This is the
clue that gives it away.

If somebody offers you 1 chunk of gold that weighs 1 Newton,
you say to him:

   "Fine !  Great !  Golly gee, that's sure generous of you.  
But before you start weighing the chunk to give me, I want you
to take your gold and your scale to Pluto, and weigh my chunk
there.   And if you don't mind, be quick about it."

The local acceleration of gravity on Pluto is  0.62 m/s² ,
but on Earth, it's 9.81 m/s.

So if he weighs 1 Newton of gold for you on Pluto, its mass will be
1.613 kilograms, and it'll weigh 15.82 Newtons here on Earth. 

That's almost 3.6 pounds of gold, worth over $57,000 !


It would be even better if you could convince him to weigh it on
Halley's Comet, or on any asteroid.  Wherever he's willing to go
that has the smallest gravity.  That's the place where the largest
mass weighs 1 Newton.

3 0
3 years ago
How is resonance used in musical instruments?
9966 [12]

Answer:

The answer is B: to Amplify the sound

7 0
3 years ago
Read 2 more answers
Suppose the original segment of wire is stretched to 10 times its original length. How much charge must be added to the wire to
Debora [2.8K]

Here we want to study how the linear charge density changes as we change the measures of our body.

We will find that we need to add 9*Q of charge to keep the linear charge density unchanged.

<em>I will take two assumptions:</em>

The charge is homogeneous, so the density is constant all along the wire.

As we work with a linear charge density we work in one dimension, so the wire "has no radius"

Originally, the wire has a charge Q and a length L.

The linear charge density will be given by:

λ = Q/L

Now the length of the wire is stretched to 10 times the original length, so we have:

L' = 10*L

We want to find the value of Q' such that λ' (the <u>linear density of the stretched wire</u>) is still equal to λ.

Then we will have:

λ' = Q'/L' = Q'/(10*L) = λ = Q/L

Q'/(10*L) = Q/L

Q'/10 = Q

Q' = 10*Q

So the new <u>charge must be 10 times the original charge</u>, this means that we need to add 9*Q of charge to keep the linear charge density unchanged.

If you want to learn more, you can read:

brainly.com/question/14514975

6 0
2 years ago
What is the definition of permitted orbits ?
givi [52]
.The path of a celestial body or an artificial satellite as it revolves around another body due to their mutual gravitational <span>attraction.</span>
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