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Andreyy89
4 years ago
10

Which of the following most completely describes an object’s motion?

Physics
2 answers:
Ronch [10]4 years ago
7 0
Position, speed, direction, and acceleration
Katena32 [7]4 years ago
5 0
Position, velocity, acceleration, (j-e-r-k (its the third derivative of position wrt time), snap, crackle, pop)... don't forget the last ones :D
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Can someone help me find which one is a sedimentary rock
SashulF [63]
Byevyebsndkneknddnudd
7 0
3 years ago
Brandon buys a new seadoo he goes 12 km north from the beach he jumps wakes for 6 km to the east the chases a boat 10 km north w
d1i1m1o1n [39]

Answer:

Distance covered 28 km

displacement is 22.8 km North-East

Explanation:

Distance shows how far apart objects or points are from each other. The distance he covered is the sum of all the distance travelled. Therefore:

Distance covered = 12 km + 6 km + 10 km = 28 km

Displacement is a vector quantity (has direction). It is the overall change in position.

The total distance traveled north = 12 km + 10 km = 22 km

The distance traveled east = 6 km

The displacement (d) is:

d² = 22² + 6² = 484 + 36

d² = 520

d = √520 = 22.8 km

Therefore the displacement is 22.8 km North-East

3 0
3 years ago
Major storms get their energy from WATER VAPOR in the air. Which step of the water cycle would give energy to a storm
IrinaK [193]
Evaporation is the process of water being evaporated and rising. This water vapor condenses at a certain point and mixes with aerosols to create a cloud.
7 0
3 years ago
An object has an acceleration of 6 m/s/s. If the net force
Bond [772]

Answer:

F = ma \\ a =  \frac{F}{m} = 6  \\ when \: force \: is \: trippled \: and \: mass \: is \: halved \\ a =  \frac{3F}{ \frac{1}{2} m}  =   \frac{3}{2}  \times 6 \\ a =  \frac{(3 \times 6)}{2}  \\ a =  \frac{18}{2}  \\ a = 9 \: m {s}^{ - 2}

6 0
3 years ago
A 0.500 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 40.0 pC charge on it
vladimir1956 [14]

We can find the potential using the equation

V = kQ/r

Where V is the potential, k corresponds to the Coulomb constant

k = 9. 10^9 N N. m^2/C^2 ,Q is the charge and r is the radius of the sphere (r = 2.5.10^-3 m)

Substituting in the equation we obtain

V = (9. 10^9 N N. m^2/C^2)*(40.10^-12 C)/(2.5.10^-3 m)  =

V =  144 V (Potential near its surface)

4 0
4 years ago
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