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dalvyx [7]
3 years ago
7

Which phrase describes the paleo-Indians of prehistoric Arkansas?

Physics
2 answers:
sergejj [24]3 years ago
5 0

question (1)           D. hunted with stone-tipped spears


question (2)          A. developed widespread agriculture

Vlad [161]3 years ago
4 0

The first one is D hunted with stone-tipped spears

The second one is A

Hope this helps

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Three carts of masses 4.0 kg, 10kg, and 3.0 kg move on a frictionless track with speeds of v1 = 5.0m/s, v2=3.0m/s, and v3=-3.6 m
gogolik [260]

2.24 m/s is the calculated velocity.

Initial velocity (u) squared plus two times the acceleration (a) times the displacement equals final velocity (v) squared (s). Final velocity (v) is equal to the square root of initial velocity (u) squared plus two times the acceleration (a) times displacement when v is the variable being solved for (s).

The cart's masses and speeds are known.

M1 = 4.00 kg, M2 = 10.0 kg, M3 = 3.00 kg, etc.

v1 = 5.00 m/s = 5.00 m/s, v2 = 3.00 m/s = 3.00 m/s, v3 = -4.00 m/s = 4.00 m/s, and m1v1+m2v2+m3v3 = (m1+m2+m3) v=d frac m 1v 1+m 2v 2+m 3v 3, where (m1+m2 + m3) is the product of (v1 v 1+m2v2+m3v3).

"m 1+m 2+m 3" is equivalent to "m 1+m2+m3/m1v 1+m2v2 +m3v3"

the three carts' final velocities are calculated as follows: v=d frac

{4.00kg\sdot5.00m/s+10.0kg\sdot3.00m/s-3.00kg\sdot4.00m/s} 4.24m/s = 4.50kg+10.0kg+3.00kg vs. 4.50kg+10.0kg+3.00kg

5.00m/s/4.00kg/5.00m/s+10.0m/s/3.00m/s/4.00m/s =2.24m/s.

2.24 m/s is the calculated final velocity.

Learn more about velocity here-

brainly.com/question/18084516

#SPJ9

7 0
2 years ago
2. How can people know that they've learned?
photoshop1234 [79]
Answer: People know that they jap e learned by their mistakes or by someone ya being them something
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3 years ago
A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C
mario62 [17]

Answer:

3.62 m  and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}

\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}

\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}

x = - 1.4 m

6 0
3 years ago
Which two layers make up the Earth's upper mantle?
Bad White [126]

Answer: The lithosphere and the asthenosphere

4 0
3 years ago
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About what fraction of earth's freshwater is in the form of ice
Bingel [31]
68% So 68/100 of freshwater is found in ice
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3 years ago
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