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Romashka-Z-Leto [24]
3 years ago
8

Which of the following best defines

Physics
1 answer:
marta [7]3 years ago
6 0

Answer:

I'd say D

Explanation:

because not all weather happens within the atmosphere, and most weather depends on region (lile if your near the equator or not)

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What information do you need to describe an object's location
stira [4]
longitude and latitude<span />
4 0
3 years ago
A 3.5 x 10-6 C charge is located 0.28 m from a 2.8 x 10-6 C charge. What is the magnitude of the force being exerted on the smal
Margarita [4]
The electrostatic force between two charges is given by Coulomb's law:
F=k_e  \frac{q_1 q_2}{r^2}
where
ke is the Coulomb's constant
q1 is the first charge
q2 is the second charge
r is the separation between the two charges

By substituting the data of the problem into the equation, we can find the magnitude of the force between the two charges:
F=(8.99 \cdot 10^{9} Nm^2C^{-2} ) \frac{(3.5 \cdot 10^{-6} N/C)(2.8 \cdot 10^{-6}N/C)}{(0.28m)^2}=1.2 N
7 0
3 years ago
A square sheet of rubber has sides that are 20 cm long. What is the area of the square of rubber in cm squared?
Alinara [238K]

Answer:

400cm^2

Explanation:

sides are 20cm long Area for a square is a squared

since all the lides are of equal length you can just choose one side.

20squared is 400

20 x 20 = 400cm squared

Hope this helps :)

6 0
3 years ago
Which statement best describes the importance of joints in the roadways of Bridges ?
Sergeeva-Olga [200]

Answer:

c. Joints allow the roadway to expand and contract as cars put force on the bridge

Explanation:

The reasons why joints are allowed on roadway is to accommodate the contraction and expansion of the road as cars put force on them.

  • Most materials used in making roadways are susceptible to expansion and contraction.
  • When a measure of force is applied their length either increases or decreases depending on the type of force.
  • To accommodate these changes, joints are placed in roadways
3 0
3 years ago
Answer it pls!!!!!!!!!!!
Archy [21]

Answer:

Fractional error = 0.17

Percent error = 17%

F = 112 ± 19 N

Explanation:

Plug in the values to find the force:

F = (3.5 kg) (20 m/s)² / (12.5 m) = 112 N

Find the fractional error:

ΔF/F = Δm/m + 2Δv/v + Δr/r

ΔF/F = 0.1/3.5 + 2(1/20) + 0.5/12.5

ΔF/F = 0.17

Multiply by 100% to find the percent error:

ΔF/F × 100% = 17%

Solve for the absolute error:

ΔF = 0.17 × 112 N = 19 N

Therefore, the force is:

F = 112 ± 19 N

8 0
3 years ago
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