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2 years ago

For a storm to be a blizzard, the wind must be at least

1 answer:

7 0

For a storm to be a blizzard, the wind must be at least 35 miles per hour. This is just one criteria for considering a storm to be a blizzard. The wind speed of 35 miles per hour should reduce the visibility to less than 400 meters. The last criteria is that the storm must continue for a time frame of at least 3 hours.

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CAN SOMEONE DO THIS TO ME ITS OK IF YOUR DRAWING IS NOT THAT COOL OR GOOD AND DONT GET A PHOTO FROM SOCIAL MEDIA

Svet_ta [14]

bc i cant draw good sorry but hope it helps!

and can i have brainliest? ight plz can i have brainliest?

ill give another picture of the volley ball thing

is that better?

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3 years ago

In this photograph, a soccer player is about to kick the ball. Use the situation to explain that when two objects interact, the

Makovka662 [10]

hey you look nice (pic).

According to Newton’s first law, if no force is applied to a ball, it will continue moving at the same speed and direction as it did before. When we put the ball on the grass it stays in its place, namely it stays in zero motion since no force is applied to it. However, after we kick the ball, it will continue moving in the direction we kicked it. Its speed will drop gradually, due to friction (a force applied on the ball in the opposite direction to its motion), but the direction of its motion will remain the same.

According to Newton’s second law, a force applied to an object changes that object’s acceleration – namely, the rate at which the speed of the object changes. When we kick the ball, the force we apply to it causes it to accelerate from a speed of 0 to a speed of dozens of kilometers per hour. When the ball is released from the foot, it begins to decelerate (negative acceleration) due to the force of friction that is exerted upon it (as we observed in the previous example). If we were to kick a ball in outer space, where there is no friction, it would accelerate during the kick, and then continue moving at a constant speed in the direction that we kicked at, until it hits some other object or another force is applied to it.

8 0

3 years ago

A 5.00-V battery charges the parallel plates in a capacitor, with a plate area of 865 mm2 and an air-filled separation of 3.00 m

Westkost [7]

Answer:

W = 3.21x10⁻¹¹ J

Explanation:

The work required to separate the plates can be calculated using the following equation:

$W = U_{2} - U_{1} = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2})$

<u>Where</u>:

U₂: is the final stored energy

U₁: is the initial stored energy

C₂: is the final capacitance

C₁: is the initial capacitance

V₁: is the initial potential difference = 5.00 V

V₂: is the final potential difference

The initial and final capacitance is:

$C_{1} = \epsilon_{0}*\frac{A}{d_{1}}$

<u>Where</u>:

ε₀: is the vacuum permittivity = 8.85x10⁻¹² C²/(N*m²)

d: is the initial distance = 3.00 mm = 3.00x10⁻³ m

A: is the plate area = 865 mm² = 8.65x10⁻⁴ m²

$C_{1} = \epsilon_{0}*\frac{A}{d_{1}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 \cdot 10^{-3} m} = 2.55 \cdot 10^{-12} F$

Similarly, C₂ is:

$C_{2} = \epsilon_{0}*\frac{A}{d_{2}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 + 3.00 \cdot 10^{-3} m} = 1.28 \cdot 10^{-12} F$

Now, V₂ can be calculated by finding the initial charge (q₁):

$q_{1} = C_{1}V_{1} = 2.55 \cdot 10^{-12} F*5.00 V = 1.28 \cdot 10^{-11} C$

Since, q₁ is equal to q₂, V₂ is:

$V_{2} = \frac{q_{2}}{C_{2}} = \frac{1.28 \cdot 10^{-11} C}{1.28 \cdot 10^{-12} F} = 10 V$

Finally, we can find the work:

$W = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2}) = \frac{1}{2}(1.28 \cdot 10^{-12} F*(10 V)^{2} - 2.55 \cdot 10^{-12} F(5.00 V)^{2}) = 3.21 \cdot 10^{-11} J$