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2 years ago

Shareen performs a skit to model a method of charging. In the skit, a painter shakes her hand and gets paint on her.

Physics

2 answers:

dimaraw [331]2 years ago

8 0

Answer:

B

Explanation:

It models conduction because the painter represents a charged object and the paint represents electrons that are transferred through contact.

Veseljchak [2.6K]2 years ago

5 0

Answer:

It models conduction because the painter represents a charged object and the paint represents electrons that are transferred through contact.

Explanation:

Conduction phenomenon of charging is the process of charging in which two bodies are made in contact with each other so that charges are transferred due to potential difference of two bodies.

here we know that when hands are shake then it will have paint on it. so here due to hand shake the hands are in contact with charge particles and due to contact the electrons are transferred to the hand.

Now here we need to assume that charge of paint must be opposite that of the charge on the hand because only due to opposite charge attraction the paint must be transferred to the hand

SO here correct answer will be

It models conduction because the painter represents a charged object and the paint represents electrons that are transferred through contact.

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An 80 kg skateboarder moving at 3 m/s pushes off with her back foot to move faster. If her velocity increases to 5 m/s, what is

Arturiano [62]

1) The kinetic energy of an object is given by:
$K= \frac{1}{2}mv^2$
where m is the object's mass and v its speed.

By using this equation, we find the initial kinetic energy of the skateboarder:
$K_i= \frac{1}{2}(80 kg)(3 m/s)^2=360 J$
and the final kinetic energy as well:
$K_f= \frac{1}{2}(80 kg)(5 m/s)^2=1000 J$

So, her change in kinetic energy is
$\Delta K=K_f-K_i=1000 J-360 J=640 J$

2) The work-energy theorem states that the work done to increase the speed of an object is equal to the variation of kinetic energy of the object:
$W=\Delta K$
Therefore, the work done by the skateboarder is
$W=\Delta K=640 J$

7 0

2 years ago

Read 2 more answers

A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu

melomori [17]

Answer:

please read the answer below

Explanation:

The angular momentum is given by

$|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)$

By taking into account the angles between the vectors r and v in each case we obtain:

v=(2,0)

r=(0,1)

angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

r=(0,-1)

angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

r=(1,0)

angle = 0°

r and v are parallel

L = 0kgm/s

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

r=(1,1)

angle = 45°

$L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}$

r=(-1,1)

angle = 45°

the same as e):

L = 5kgm/s

r=(-1,-1)

angle = 135°

$L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}$

r=(1,-1)

angle = 135°

the same as g):

L = 5kgm/s

hope this helps!!

4 0

2 years ago

Calculate the binding energy e of the boron nucleus 11 5b (1ev=1.602×10−19j). express your answer in millions of electron volts

nignag [31]

Depends on the precision you're working to. proton mass ~ 1.00728 amu neutron mass ~ 1.00866 amu electron mass ~ electron mass = 0.000549 amu Binding mass is: mass of constituents - mass of atom Eg for nitrogen: (7*1.00728)-(7*1.00866)-(7*0.000549) -14.003074 = 0.11235amu Binding energy is: E=mc^2 where c is the speed of light. Nuclear physics is usually done in MeV[1] where 1 amu is about 931.5MeV/c^2. So: 0.11235 * 931.5 = 104.6MeV Binding energy per nucleon is total energy divided by number of nucleons. 104.6/14 = 7.47MeV This is probably about right; it sounds like the right size! Do the same thing for D/E/F and recheck using your numbers & you shouldn't go far wrong :) 1 - have you done this? MeV is Mega electron Volts, where one electronVolt (or eV) is the change in potential energy by moving one electron up a 1 volt potential. ie energy = charge * potential, so 1eV is about 1.6x10^-19J (the same number as the charge of an electron but in Joules). It's a measure of energy, but by E=mc^2 you can swap between energy and mass using the c^2 factor. Most nuclear physicists report mass in units of MeV/c^2 - so you know that its rest mass energy is that number in MeV.

7 0

2 years ago

The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0. Exp

olganol [36]

Answer:

Explanation:

We are given that

Initial velocity=u=0

We have to find the magnitude of p of the momentum of the particle at time t.

Let mass of particle=m

Applied force=F

Acceleration, $a=\frac{F}{m}$

Final velocity , $v=a+ut$

Substitute the values

$v=0+\frac{F}{m}t=\frac{F}{m}t$

We know that

Momentum, p=mv

Using the formula

$p=m\times \frac{F}{m}t=Ft$

6 0

2 years ago

An green hoop with mass mh = 2.8 kg and radius rh = 0.13 m hangs from a string that goes over a blue solid disk pulley with mass

Otrada [13]

The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table.
So by applying the theorem of parallel axes, the moment of inertia of the sphere is computed by:I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere.
I = 7/5*mR^2 M = 7/5*m
Therefore, linear acceleration is computed by:F/m = 27.468 / (2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2

7 0

2 years ago