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Makovka662 [10]
3 years ago
13

The beginning development of a star is marked by a supernova explosion, with the gases present in the nebula being forced

Physics
1 answer:
GenaCL600 [577]3 years ago
4 0

The beginning development of a star is marked by a supernova explosion, with the gases present in the nebula being forced to scatter. As the star shrinks, radiation of the surface increases and create pressure on the outside shell to push it away and forming a planetary nebula or white dwarf.

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You are travelling by skateboard at 3.0 m/s and start to accelerate. If you
vovangra [49]

Answer: 8m/s

Explanation:

Vs= 3 m/s

Vf=?

a=0.5m/s²

t=10s

-----------

a=Vf-Vs/t

at=Vf-Vs

0.5*10s=Vf-Vs

5m/s=Vf-3m/s

5m/s+3m/s=Vf

Vf=8m/s

6 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
4 years ago
Read 2 more answers
Which type of molecule is added to this genetic sequece might create a insertion mutation? A-G-A-T-G-C-A-G-A-G-T-T-A-C-G-G
Jet001 [13]

D. Adding another base pair will re-arrange your DNA sequences and cause an insertion mutation. This will make your codons group of differently, and possibly give you a BAD mutation. However, sometimes the codons still make the same proteins as its supposed to, the mutation will NOT affect you.


Example:


THE BIG FAT CAT ATE THE RAT


Now, if i were to add a letter (X) to this and make the letters group up in three aka the codons:


THE XBI GFA TCA TAT ETH ERA T


As you can you can see, adding a base pair in a DNA insertion will usually have a negative effect, specifically a insertion mutation.



3 0
4 years ago
A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel
dybincka [34]

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

W=12,279.75J

The workdone by the gas on the bullet is 12,279.75J

3 0
4 years ago
When the mass of the cylinder increased by a factor of 3, from 1.0 kg to 3.0 kg, what happened to the cylinder’s gravitational p
const2013 [10]

Answer: Fourth option. It increased by a factor of 3.

Solution:

m1=1.0 kg

Cylinder's gravitational potential energy: Ep=m*g*h

Ep1=(1.0 kg)*g*h

Ep1=g*h

m2=3.0 kg

Ep2=(3.0 kg)*g*h

Ep2=3*g*h

Replacing g*h by Ep1 in the equation above:

Ep2=3*Ep1

Then, the cylinder's gravitational potential energy increased by a factor of 3.

3 0
4 years ago
Read 2 more answers
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