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uranmaximum [27]
3 years ago
12

What processes were involved in the formation of the model? How is the model similar to the real sedimentary sample? How is the

model different from the real sedimentary sample?

Physics
1 answer:
Stels [109]3 years ago
8 0

Answer:

1. sediment layering and compacting on top of each other and solidifying

2. sediment layering and compacting (not as much) on top of each other and solidifying, just not as much

Explanation:

hope this helps! :))

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A car accelerates from 30 m/s to 50 m/s in 2 seconds. Calculate the cars acceleration
masha68 [24]

Answer:

applying 1st eq of motion vf=vi+at we have to find a=vf-vi/t here a=50-30/2=10 so we got a=10m/s²

5 0
3 years ago
Please<br>help<br>me<br>with this problem​
uranmaximum [27]

Answer:

speeding up

Explanation:

because its speeding up, theres going to be more newtons in the back

i really hope this is right, tell me if so

3 0
2 years ago
A box is sliding up an incline that makes an angle of 14.0° with respect to the horizontal. the coefficient of kinetic friction
mezya [45]
A box is sliding up an incline that makes an angle of 14.0° with respect to the horizontal. the coefficient of kinetic friction between the box and the surface of the incline is 0.180. the initial speed of the box at the bottom of the incline is 2.20 m/s. how far does the box travel along the incline before coming to rest?
6 0
3 years ago
What are two examples of common units for each of the above measurements
kap26 [50]

Density: g/mL, kg/cubic meter  

Volume: L, teaspoon  

Mass: g, MeV/sq. C

3 0
3 years ago
A satellite that goes around the earth once every 24 hours (86,400 s) is called a geosynchronous satellite. If a geosynchronous
Olegator [25]

Answer:

42244138.951 m

Explanation:

G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²

r = Radius of orbit from center of earth

M = Mass of Earth = 5.98 × 10²⁴ kg

m = Mass of Satellite

The satellite revolves around the Earth at a constant speed

Speed = Distance / Time

The distance is the perimeter of the orbit

v=\frac{2\pi \times r}{24\times 3600}

The Centripetal force of the satellite is balanced by the universal gravitational force

m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m

The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m

6 0
3 years ago
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