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4 years ago

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WILL MARK AS BRAINLIEST...,.. A constant force vector f =2i cap+3j cap-5k cap acts on a particle and displaces it from (1,2,-3)

m to (2,5,-1) m. find the work done by the force .

1 answer:

Montano1993 [528]4 years ago

6 0

Explanation:

Given that,

Force, $F=2i+3j-5k$

The particle displaces from (1,2,-3) m to (2,5,-1) m.

We need to find the work done by the force. Work done by the force is given by :

W = Fd

It is equal to the dot product of force and displacement.

Displacement from (1,2,-3) m to (2,5,-1) m is (2-1, 5-2, -1-(-3)) or (1, 3, 2) m

Work done,

$W=F{\cdot} d\\\\W=(2i+3j-5k){\cdot} (i+3j+2k)$

We know that, i.i=j.j=k.k=1

So,

$W=1\ J$

So, the work done by the force is 1 J.

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3 years ago

The change of an atom from the excited state to the ground state always requires _______

lawyer [7]

Answer: emission of electromagnetic radiation

Explanation:

When an atom is in its <u>ground state</u>, its electrons fill the lower energy orbitals completely before they begin to occupy higher energy orbitals.

On the other hand, when an atom is <u>excited (</u>it has left its ground state, in which each electron occupies its place in its orbit, around the nucleus), some electron jumps out of the orbit it occupied in its fundamental state to an outer orbit, further away from the nucleus and then return to the ground state, emitting in the form of electromagnetic radiation (light which may be visible or not) the energy received.

So, when an excited electron passes from an outer orbit to the ground state, it produces electromagnetic radiation of a specific wavelength that depends on the amount of energy the electron releases in the process.

Therefore:

<h3>The change of an atom from the excited state to the ground state always requires <u>the emission of electromagnetic radiation.</u></h3>

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3 years ago

A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surfa

nataly862011 [7]

Answer:

a) $f=0\ N$

b) $f=6\ N$

c) $F=16\ N$

d) $a=2.45\ m.s^{-2}$

Explanation:

Given:

weight of the box on the horizontal surface, $w=40\ N$

coefficient of static friction between the surface and the box, $\mu_s=0.2$

coefficient of kinetic friction between the surface and the box, $\mu_k=0.2$

a)

When no horizontal force acts on the box then according to the Newton's first law of motion there will be no any force of friction acting on the body but just a vertical component is balanced by the normal reaction.

b)

Now force on the box, $F=6\ N$

So there we have the maximum force of static friction as:

$f_s=\mu_s.N$

here:

N = normal force equal to the weight of the body

$f_s=0.4\times 40$

$f_s=16\ N$

Now the magnitude of the static frictional force is equal to the applied force on the box. So,

$f=6\ N$

c)

Since we have the maximum static frictional force between the two surfaces as:

$f_s=16\ N$

So, the applied force must be equal to this limiting value.
<u>So the applied force must be:</u>

<u /> $F=16\ N$ <u />

d)

Now when the box has started its motion then the minimum intensity of the force to keep the box moving is equals to the kinetic frictional force:

$F_k=\mu_k.N$

$F_k=0.2\times 40\\F_k=8\ N$

e)

The value of friction force:

Since the box is moving, so the maximum friction is the kinetic friction:

$F_k=8\ N$

The applied force is :

$F=18\ N$

<u>So the acceleration will be due to :</u>

$\Delta F=F-F_k$

$\Delta F=10\ N$

now we know that:

$a=\frac{\Delta F}{m}$

$a=10\div\frac{40}{9.8}$

$a=2.45\ m.s^{-2}$

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3 years ago

a basketball is shot at 14 m/s at a 65 degree angle. what is the magnitude only (no direction) of the velocity of the ball 2.16

Vinvika [58]

Answer:

10.4 m/s

Explanation:

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3 years ago

A block of mass 5 kg slides down an inclined plane that has an angle of 10. If the inclined plane has no friction and the block

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The kinetic energy of the block when it reaches the bottom is 39.2 J.

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Apply the principle of conservation of energy.

K.E(bottom) = P.E(top)

P.E(top) = mgh

where;

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g is acceleration due to gravity
h is the vertical height of fall

P.E(top) = 5 x 9.8 x 0.8

P.E(top) = K.E(bottom) = 39.2 J

Learn more about kinetic energy here: brainly.com/question/25959744

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2 years ago