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Tresset [83]
3 years ago
15

A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 6 m, y = 8.5 m, and has veloci

ty ~vo = (9 m/s) ˆı + (−2.5 m/s) ˆ . The acceleration is given by ~a = (4.5 m/s 2 ) ˆı + (3 m/s 2 ) ˆ . What is the x component of velocity after 3.5 s? Answer in units of m/s.
Physics
1 answer:
andreev551 [17]3 years ago
8 0

Answer:

24.75 m/s

Explanation:

Since it's only asking for the component we can use v_final=v_initial+at which then gives v_final=9+3.5*4.5=24.75 m/s

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What is the force felt by the electrons and the nuclei in the rod when the external field described in the problem introduction
fomenos

Complete Question

The complete question is shown on the uploaded image

Answer:

The nuclei experience a force that will move it to the right of the conductor rod while the electrons experience a force that will move it to the left side of the conductor rod.

Explanation:

The force that act on the charges(both the positive and the negative charge ) is Mathematically expressed as

                      F = qE   => E = F/q

where F is the force, q is the charge and E is the electric field

we can see that the force is directly proportional to the electric field,so an increase in the electric field would increase the force.

we can also see that the electric field is given as force per unit charge and generally the direction of this field is taken to be the direction of the force it would exert on a positive test charge

Now from the question we are being told that the external electric field is the direction of the positive x axis

Hence this field would drive the positive charge i.e the nuclei to the right.

In order to further explain let consider this

Generally the electric field is always radially outward  originating from a positive point charge and radially in toward a negative point charge.

what  this means for this question, is that the positive point charge is on the left side of the electric field while the negative point charge is at the right side of the field.

According to Coulomb's law which states that unlike term attract while like terms repel, it means that  the electron would move to the left of the conductor rod   while the nuclei would move to the right side of the conductor rod.

7 0
3 years ago
PLEASE TRY TO ANSWER AS MANY QUESTIONS AS YOU CAN !
suter [353]
Good luck with solving this
3 0
3 years ago
Read 2 more answers
Calculate the work done if a boy lifts a bag of cement 500N to the floor of a lorry 2.5m above the ground
eduard

Answer:

W = 1250 J = 1.25 KJ

Explanation:

The work done by the boy is due to the change in the position of the cement vertically. Hence, the work done in this case will be equal to the potential energy of the cement:

W = P.E\\W = mgh

where,

W = Work done = ?

mg = W = weight of cement = 500 N

h = height covered = 2.5 m

Therefore,

W = (500\ N)(2.5\ m)

<u>W = 1250 J = 1.25 KJ</u>

4 0
2 years ago
Juan is receiving bills from the hospital even though his insurance company said his procedure would be covered in full. What st
BaLLatris [955]

==> Jot down notes before and after making each call, and avoid calling during business hours when people are busy.

==> Save all emails and bills, and follow up to make sure people did what they said they would do.

5 0
3 years ago
A worker pushed a 33 kg block 6.1 m along a level floor at constant speed with a force directed 23° below the horizontal. if the
jenyasd209 [6]
The work done occurs only in the direction the block was moved - horizontally. Work is given by:

W = F(h) * d

Where F(h) is the force applied in that direction (horizontal) and d is the distance in that direction. In this case, F(h) is the horizontal component of the applied force, F(app). However, the question doesn't give us F(app), so we need to find it some other way.

Since the block is moving at a constant speed, we know the horizontal forces must be balanced so that the net force is 0. This means that F(h) must be exactly balanced by the friction force, f. We can express F(h) as a function of F(app):

F(h) = F(app)cos(23)

Friction is a little trickier - since the block is being PUSHED into the ground a bit by the vertical component of the applied force, F(v), the normal force, N, is actually a bit more than mg:

N = mg + F(v) = mg + F(app)sin(23)

Now we can get down to business and solve for F(app) - as mentioned above:

F(h) = f
F(h) = uN
F(h) = u * (mg + F(v))
F(app)cos(23) = 0.20 * (33 * 9.8 + F(app)sin(23))
F(app) = 76.8

Now that we have F(app), we can find the exact value of F(h):

F(h) = F(app)cos(23)
F(h) = 76.8cos(23)
F(h) = 70.7

And now that we have F(h), we can find W:
W = F(h) * d
W = 70.7 * 6.1
W = 431.3

Therefore, the work done by the worker's force is 431.3 J. This also represents the increase in thermal energy of the block-floor system.
3 0
3 years ago
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