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KATRIN_1 [288]
3 years ago
9

Which is located farther west the central valley or the coast ranges?

Physics
1 answer:
astra-53 [7]3 years ago
8 0
The question is concerned with the regions found within California, which are the Coastal Region, Mountain Region, Central Valley Region, and the Desert Region.
The Coastal Region is located furthest to the west out of all of these regions. The Coastal Region is where the California meets the Pacific Ocean, and it has a somewhat moderate and constant climate throughout the year due to its location near the ocean.
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What is the displacement of the particle in the time interval 7 seconds to 8 seconds? A. 0 meters B. 1.5 meters C. 3 meters D. 7
Dmitrij [34]

Answer: B. 1.5 meters

Explanation:

4 0
2 years ago
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Solutions with ions that react with acids or bases to lessen their effects are fubrefs
Softa [21]

Answer:

Buffers

Explanation:

A buffer solution is a solution containing weak acids and their salts or weak bases and their salts.

A buffer solution is an equilibrium system that resists changes in pH or pOH when a small amount of an acid or base is added hence it is a solution of fairly constant pH value.

4 0
3 years ago
What is the relationship between air pressure and air temperature?
LenKa [72]
The relationship between the two is that air temperature changes the air pressure. For example, as the air warms up the molecules in the air become more active and they use up more individual space even though there is the same<span> number of molecules. This causes an </span>increase<span> in the air pressure.</span>
8 0
3 years ago
A 1.0 kg rock is thrown straight upward with an initial speed of 8.0 m/s. What is its speed
Ronch [10]

Answer:5.7m/s

Explanation:

Mass=1kg

Initial velocity=u=8m/s

height=h=1.6m

Final velocity =v

Acceleration due to gravity=g=9.8m/s^2

v^2=u^2-2xgxh

v^2=8^2-2x9.8x1.6

v^2=8x8-2x9.8x1.6

v^2=64-31.36

v^2=32.64

Take the square root of both sides

√(v^2)=√(32.64)

v=5.7

Speed at the height of 1.6m is 5.7m/s

8 0
3 years ago
A charge of 50 µC is placed on the y axis at y = 3.0 cm and a 77-µC charge is placed on the x axis at x = 4.0 cm. If both charge
zheka24 [161]

Answer:

The acceleration of an electron is 1.2\times10^{20}\ m/s^2

Explanation:

Given that,

One Charge = 50 μC

Distance on y axis = 3.0 cm

Second charge = 77 μC

Distance on x axis = 4.0 cm

We need to calculate the force on electron due to q₁

Using formula of force

F_{1}=\dfrac{kq_{1}q}{r^2}

Here, q = charge of electron

Put the value into the formula

F_{1}=\dfrac{9\times10^{9}\times50\times10^{-6}\times1.6\times10^{-19}}{(3\times10^{-2})^2}

F_{1}=8\times10^{-11}j\ \ N

We need to calculate the force on electron due to q₂

Using formula of force

F_{2}=\dfrac{kq_{2}q}{r^2}

Here, q = charge of electron

Put the value into the formula

F_{2}=\dfrac{9\times10^{9}\times77\times10^{-6}\times1.6\times10^{-19}}{(4\times10^{-2})^2}

F_{2}=6.93\times10^{-11}i\ \ N

We need to calculate the net force

Using formula of net force

F=F_{1}+F_{2}

Put the value into the formula

F=8\times10^{-11}j+6.93\times10^{-11}i

The magnitude of the net force

F=\sqrt{(8\times10^{-11})^2+(6.93\times10^{-11})^2}

F=1.058\times10^{-10}\ N

We need to calculate the acceleration of an electron

Using formula of force

F = ma

a=\dfrac{F}{m}

Put the value into the formula

a=\dfrac{1.058\times10^{-10}}{9.1\times10^{-31}}

a=1.2\times10^{20}\ m/s^2

Hence, The acceleration of an electron is 1.2\times10^{20}\ m/s^2

3 0
3 years ago
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