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3 years ago

1 poir

Physics

1 answer:

djverab [1.8K]3 years ago

4 0

Answer:

D) 735 J(oules)

Explanation:

Work is defined as force * distance

Force is defined as mass * acceleration

Given a mass of 15 kg and a gravitational acceleration of 9.8 m/s² since the box is being lifted up, the force being applied to the box is 15 kg * 9.8 m/s² = 147 N

Since the distance is 5 meters, the work done is 147 N * 5 m = 735 N/m = 735 J, making D the correct answer.

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An electron is isolated from an atom and exists in vacuum. A group of scientists collectively state that they can remove part of

lesya692 [45]

D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values

Explanation:

The electric charge of an object is a property of the object that is related to the ability of the object to experience/exert an electric force: if the object is electrically charge, then it is attracted or repelled by other electrically charged object.

The electric charge of an object depends on the amount of charged particles it has on it. In particular, the fundamental particles that carry electric charge are:

Protons: they carry electric charge of +e
Electrons: they carry electric charge of -e

Where "e" is the fundamental charge ( $e=1.6\cdot 10^{-19}C$ ). Therefore, one proton carry a charge of +e and one electron carry a charge of -e.

An electron is a fundamental particle: this means that it cannot be divided into smaller particles. This also means that it is not possible to remove part of the charge of the electron: in fact, it is said that electric charge exists only as discrete values, being a multiple of $e$ . Therefore, the correct statement is

D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values

Learn more about particles:

brainly.com/question/2757829

#LearnwithBrainly

5 0

3 years ago

A sock stuck to the inside of the clothes dryer spins around the drum once every 2.0 s at a distance of 0.50 m from the center o

Rashid [163]

a) 1.57 m/s

The sock spins once every 2.0 seconds, so its period is

T = 2.0 s

Therefore, the angular velocity of the sock is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{2.0}=3.14 rad/s$

The linear speed of the sock is given by

$v=\omega r$

where

$\omega$ is the angular velocity

r = 0.50 m is the radius of the circular path of the sock

Substituting, we find:

$v=(3.14)(0.50)=1.57 m/s$

B) Faster

In this case, the drum is twice as wide, so the new radius of the circular path of the sock is twice the previous one:

$r' = 2r = 1.00 m$

At the same time, the drum spins at the same frequency as before, therefore the angular frequency as not changed:

$\omega' = \omega = 3.14 rad/s$

Therefore, the new linear speed would be:

$v'=\omega' r' = \omega (2r)$

And substituting,

$v'=(3.14)(1.00)=3.14 rad/s = 2v$

So, we see that the linear speed has doubled.

8 0

3 years ago

Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105

Dovator [93]

It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
All you need is F=(K*Q1*Q2)/r^2 
Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you 
(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C

3 0

3 years ago

Read 2 more answers

Which substance will cool down the slowest during the night, and why?

Studentka2010 [4]

Cool down the Slowest

B. Seawater, because it heats up slower and gives away heat slower than sand.

Read the question wrong the first time

5 0

3 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

Conservation of linear momentum:

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

Conservation of kinetic energy:

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

3 years ago