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3 years ago

Positively charged particle A.proton B.neutron C.electron

2 answers:

swat323 years ago

5 0

A positively charged particle is a proton.

Negatively charged particles are electrons.

Particles with no charge are neutrons.

o-na [289]3 years ago

5 0

A positively charged particle is:

A. (Proton)

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Calculate the index of refraction for a medium in which the speed of light is 2.1x 108 m/s. The speed of light in vacuum is 3x10

strojnjashka [21]

Answer:

n = 1.42

Explanation:

The refractive index for a medium is given by the ratio of the speed of light in vacuum to the speed of light in a medium.

$n=\dfrac{c}{v}\\\\n=\dfrac{3\times 10^8}{2.1\times 10^8}\\\\n = 1.42$

So, the refractive index of the medium is 1.42.

5 0

3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

rjkz [21]

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$

and

F cos θ = f

when substituting one into the other we get:

$F cos \theta=\mu_{s}(F sin \theta +mg)$

which can be solved for the static friction coefficient so we get:

$\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

which is the answer to part b.

3 0

3 years ago

Read 2 more answers

A car is traveling at 24.0 m/s when the driver suddenly applies the brakes, causing the car to slow down with constant accelerat

Elan Coil [88]

Answer:

The answer to your question is : vf = 15.18 m/s

Explanation:

Data

vo = 24 m/s

d = 120 m

vf = ? when d = 60.0 m

Formula

vf² = vo² + 2ad

For d =100m

a = (vf² - vo²) / 2d

a = (0 -24²) / 2(100)

a = -576/200

a = 2.88 m/s²

Now, when d = 60

vf² = (24)² - 2(2.88)(60)

vf² = 576 - 345.6

vf² = 230.4

vf = 15.18 m/s

4 0

3 years ago

Read 2 more answers

A bullet travels at 850 m/s. how long will it take a bullet to go 1 km?

Dmitrij [34]

The speed of bullet =
850 m/s

Distance given = 1 km = 1000m

S = D/t
t • S = D/t • t
St = D
St/S = D/S
t = D/S
t = 1000m/850m/s
t = 1.176 s

It will take the bullet 1.176 or about 1.18 seconds to go 1 km.

8 0

3 years ago

If the distance between us and a star is doubled, with everything else remaining the same, the luminosity Group of answer choice

Savatey [412]

Answer:

remains the same, but the apparent brightness is decreased by a factor of four.

Explanation:

A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.

It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).

The luminosity of a star refers to the total amount of light radiated by the star per second and it is measured in watts (w).

The apparent brightness of a star is a measure of the rate at which radiated energy from a star reaches an observer on Earth per square meter per second.

The apparent brightness of a star is measured in watts per square meter.

If the distance between us (humans) and a star is doubled, with everything else remaining the same, the luminosity remains the same, but the apparent brightness is decreased by a factor of four (4).

Some of the examples of stars are;

- Canopus.

- Sun (closest to the Earth)

- Betelgeuse.

- Antares.

- Vega.

8 0

3 years ago