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vagabundo [1.1K]
3 years ago
14

Two people are sitting on wheeled chairs 1 metre apart. Person A is holding a 2kg ball. Person A throws the ball to Person. What

happens to person A? Explain.
a.) their chair stays still
b.) chair rolls backwards
c.) chair rolls forwards
Physics
1 answer:
zhuklara [117]3 years ago
4 0

Answer:

B - Person A's chair rolls backwards.

Explanation:

This can be seen simply through the Conservation of Momentum. When the ball is thrown to be, the ball's momentum(+m) is transfererd to B, so B's chair moves forward(think about it - it makes sense, doesn't it?), meaning A's chair moves backwards.

So, B.

EXTRA TIP: Try drawing a diagram to help you!  

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What is the uncertainty of the position of the bacterium? express your answer with the appropriate units?
lbvjy [14]
For two un-related quantities, the Heisenberg uncertainty equations holds: the prduct of the two uncertainty quantities is greater than \hbar/2
Example of unrelated quantities are position and momentum, energy and time. 
Thus
\Delta x*\Delta p  \ \textgreater \ \hbar/2
Knowing the speed of the bacteria the uncertainty in its position is
\Delta x \ \textgreater \ \hbar/(2 \Delta p) =\hbar/(2mv)
4 0
3 years ago
Ten students stand in a circle and are told to make a transverse wave. What best describes the motion of the students? Each stud
Bond [772]

Lifting hands and the down by one student at a time best describe the presentation of the transverse wave by students. Option D is correct.

<h3>What is a Transverse wave?</h3>
  • The wave in which the oscillation of particles is is perpendicular to the direction of energy transfer.

  • The students can make a transverse wave by raising their hands up and then down, one student at a time.

  • The raised hand represents the oscillation of particles while the sequence of the raising hand represents the direction of energy transfer.

Therefore, lifting hands and the down by one student at a time best describe the presentation of the transverse wave by students.

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3 0
3 years ago
Read 2 more answers
A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and c
Nady [450]

(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

I= \Delta p = m \Delta v = m (v-u)

where

m is the mass of the object

\Delta v is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s

(b) 155 N

The impulse can also be rewritten as

I=F \Delta t

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

\Delta t is the duration of the collision

In this situation, we have

\Delta t = 0.06 s

So we can re-arrange the equation to find the magnitude of the average force:

F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N

6 0
3 years ago
2. A 2000 kg car with speed 12.0 m/s hits a tree. The tree does not move or
krek1111 [17]

a) The work done by the tree is -1.44\cdot 10^5 J

b) The amount of force applied is 2880 N

Explanation:

a)

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

W=K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

where

W is the work done on the car

m is the mass of the car

u is its initial speed

v is its final speed

For the car in this problem, we have:

m = 2000 kg

u = 12.0 m/s

v = 0 (since the car comes to a stop, after the crash)

Therefore, the work done by the tree on the car is:

W=0-\frac{1}{2}(2000)(12.0)^2=-1.44\cdot 10^5 J

The work is negative because it is done in the direction opposite to the direction of motion of the car.

b)

The work done by the tree on the car can also be rewritten as

W=Fd

where

F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

W=-1.44\cdot 10^5 J is the work done

d=50.0 cm = 0.50 m is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

Learn more about work:

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#LearnwithBrainly

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3 years ago
A solid ball with a mass of 4.25 kg and a
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7.20 units long yea
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