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Ber [7]
3 years ago
7

A ball is thrown with a velocity of 35 meters per second at an angle of 30° above the horizontal. which quantity has a magnitude

of zero when the ball is at the highest point in its trajectory?
Physics
1 answer:
enot [183]3 years ago
5 0
The quantity that has a magnitude of zero when the ball is at the highest point in its trajectory is the vertical velocity.

In fact, the motion of the ball consists of two separate motions:
- the horizontal motion, on the x-axis, which is a uniform motion with constant velocity v_x=v_0 cos 30^{\circ}, where v_0=35 m/s
- the vertical motion, on the y-axis, which is a uniformly accelerated motion with constant acceleration g=9.81 m/s^2 directed downwards, and with initial velocity v_y=v_= sin 30^{\circ}. Due to the presence of the acceleration g on the vertical direction (pointing in the opposite direction of the initial vertical velocity), the vertical velocity of the ball decreases as it goes higher, up to a point where it becomes zero and it reverses its direction: when the vertical velocity becomes zero, the ball has reached its maximum height. 
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Which of the following is not an example of kinetic energy being converted to potential energy?
KengaRu [80]

The list of choices you provided with your question
is utterly devoid of any such examples.

6 0
4 years ago
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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
4 years ago
What is elasticity in polymer generally related to?
Andrej [43]
The elasticity of a polymer is primarily due to the structure of the molecule and the cross-linking between strands. Hydrogen bonding is a contributor to the shape of the molecule, but not a major player in terms of elasticity. We would have to answer "false".
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6 0
3 years ago
Read 2 more answers
A student makes the following statement
Aliun [14]

Answer:

The current is not used up. The electrons flow through the entire circuit and this travel is the current. They flow until they are not charged anymore. That is also why the circuit must be closed or else electrons would escape not just light it up for a second then go out.

Explanation:

8 0
4 years ago
A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied horizontally. Therefore, the work done is

W=(230)(4.0)(cos 0^{\circ})=920 J

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

W=(1300)(4.0)(cos 0^{\circ})=5200 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
3 years ago
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