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malfutka [58]
2 years ago
10

Now switch to “Position vs. Time #2” on the Choose a graph menu. Once again, experiment until you are able to reproduce the grap

h.
What did the caterpillar have to do during the horizontal segments of the graph?
Physics
2 answers:
seraphim [82]2 years ago
6 0

Answer:

The caterpillar had to move across the horizontal segments. I hope this helps :)

Explanation:

Alex17521 [72]2 years ago
3 0
The caterpillar had to had the horizontal segments of the graphs because
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true or false Both the large loose rocks and the small loose rocks used to be part of earth's solid rock layer
salantis [7]
Hello Micu212006 


Question: <span> Both the large loose rocks and the small loose rocks used to be part of earth's solid rock layer
</span><span>
Answer: True


Hope This Helps!
-Chris </span>
8 0
3 years ago
A typical meteor that hits the earth's upper atmosphere has a mass of only 2.5 g, about the same as a penny, but it is moving at
attashe74 [19]

Answer:

Answer:u=66.67 m/s

Explanation:

Given

mass of meteor m=2.5 gm\approx 2.5\times 10^{-3} kg

velocity of meteor v=40km/s \approx 40000 m/s

Kinetic Energy of Meteor

K.E.=\frac{mv^2}{2}

K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}

K.E.=2\times 10^6 J

Kinetic Energy of Car

=\frac{1}{2}\times Mu^2

=\frac{1}{2}\times 900\times u^2

\frac{1}{2}\times 900\times u^2=2\times 10^6  

900\times u^2=4\times 10^6

u^2=\frac{4}{9}\times 10^4

u=\frac{2}{3}\times 10^2

u=66.67 m/s

8 0
2 years ago
What are two ways an engineer can build a car in order for it to accelerate faster
Ket [755]

Explanation:

Take F=ma

a = F/m

For a higher, F higher or m lower

Means higher horse power for engine or lower mass for the car

4 0
2 years ago
Please help, it's grade 7 science. :)
xeze [42]

Answer:

The air in the soccer ball in cold weather will decrease slightly in size and it becomes flat. The air in the soccer ball in hot weather will seem flat because the low preasure leads to lower bounce in the ball.

The metal door frame in cold weather contracts and the wood contracts more in the winter. The metal door frame in hot weather thermal blowing can occur on the outer surface of the metal door frame. Hopefully that is what you were looking for have a good day.

4 0
2 years ago
Read 2 more answers
To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where
BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

\int E.dA = \frac{q}{\epcilon_0}

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

q = \int \rho dV

q = \rho * \pi r^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r}{2 \epcilon_0}

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

q = \int \rho dV

q = \rho * \pi r_0^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r_0^2}{2 \epcilon_0 r}


7 0
3 years ago
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