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Nuetrik [128]
3 years ago
9

NEED HELP ASAP!! I AM MARKING BRAINLIEST!!

Physics
1 answer:
iogann1982 [59]3 years ago
8 0

So the air evaporate s and mix the temptures and marks air power to the same theme as the other the two degrees  lower and lower until they are the same

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What is the average velocity of a dragster that starts from rest at the starting line and finishes a race at 2600 m/s in 12 seco
xxMikexx [17]
Using the equation v(avg)=distance/time 
and the equation v=v(original)+a(t)
solve for acceleration
2600=0+a(12)
a=216.66666 m/s^2

Then, you use the equation
v^2=v(original)+2a*(change in x)
2600^2=2(216.666666)*change in x
6760000/2/216.666666 = 15600 meters which is the length of the race

Then using v(avg)=x/t
15600/12= 1300 m/s
7 0
3 years ago
What is the rhyme scheme in Emily Dickinson’s poem, “Hope is the thing with feathers”?
kaheart [24]
Answer is B. ABAB. Hope it helped you, and have a great day.
-Charlie
5 0
3 years ago
Not in book
umka2103 [35]

Answer:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
  • second charge, q_2=3\times 10^{-3}\ C
  • position of first charge, x_1=-2\ m
  • position of second charge, x_2=-1\ m

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

E_1=E_2

  • since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}

3(r^2+1+2r)=5r^2

2r^2-6r-3=0

r=3.4365 \&\ r=-0.4365

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

x=-1+3.4365=2.4365\ m

and

x=-1-0.4365=-1.4365\ m

6 0
3 years ago
How can a 1kg ball have more kinetic energy than a 100kg ball? Explain both using words and by providing a numerical example
MariettaO [177]

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.

<u>Explanation</u>:

We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the (velocity)^2 and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:

\text { Kinetic Energy }=\frac{1}{2} m v^{2}

Better understood from numerical example as given:

If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?

This can be solved as follows:

\text { Kinetic Energy of } \mathrm{A}=\frac{1}{2} 50 \times 5^{2}=625 \mathrm{J}

\text { Kinetic Energy bf } \mathrm{B}=\frac{1}{2} 100 \times 2.5^{2}=312.5 \mathrm{J}

It shows that man A will have more K.E.

Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.

4 0
3 years ago
The specific heat of a certain type of metal is 0.128 J/(g⋅∘C).0.128 J/(g⋅∘C). What is the final temperature if 305 J305 J of he
Makovka662 [10]

Answer:

45.3°C

Explanation:

Heat gained = mass × specific heat × increase in temperature

q = mC (T − T₀)

Given C = 0.128 J/g/°C, m = 94.0 g, q = 305 J, and T₀ = 20.0°C:

305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)

T = 45.3°C

6 0
3 years ago
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