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3 years ago

Lightning is an example of what kind of electricity

Physics

2 answers:

n200080 [17]3 years ago

8 0

Answer:

The answer is A. Static discharge

Explanation:

ki77a [65]3 years ago

7 0

the answer would be DISCHARGE

I ʝυʂƚ ƚσσƙ ƚԋҽ TEST

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A hover craft also known as a air cushion vehicle glides on a cushion of air allowing it to travel with equal is land or water.

dimaraw [331]

Answer: M = 1797.75 kg

Explanation:

given parameters are;

speed V = 26.7 M/S.

momentum P = 4.8×10^4 KGM/S.

What was the mass of the V. A-3?

Momentum P is the product of mass and velocity. That is, P = MV

Substitute V and P into the formula

4.8×10^4 = 26.7 × M

Make M the subject of formula

M = 4.8×10^4/ 26.7

M = 1797.75 kg

Therefore, the mass of the V. A-3 was 1797.75 kg

4 0

3 years ago

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LuckyWell [14K]

1.Landslide 2. Delta 3. Moving water 4. Erosion 5. Abrasion and Deflation

6. Winds 8. Sediment it can erode

Sorry, don't know 7.

7 0

4 years ago

Vector A with arrow, which is directed along an x axis, is to be added to vector B with arrow, which has a magnitude of 5.5 m. T

ohaa [14]

Answer:

Magnitude of vector A = 0.904

Explanation:

Vector A , which is directed along an x axis, that is

$\vec{A}=x_A\hat{i}$

Vector B , which has a magnitude of 5.5 m

$\vec{B}=x_B\hat{i}+y_B\hat{j}$

$\sqrt{x_{B}^{2}+y_{B}^{2}}=5.5\\\\x_{B}^{2}+y_{B}^{2}=30.25$

The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A $\vec{A}+\vec{B}=6x_A\hat{j}\\\\x_A\hat{i}+x_B\hat{i}+y_B\hat{j}=6x_A\hat{j}$

Comparing we will get

$x_A=-x_B\\\\y_B=6x_A$

Substituting in $x_{B}^{2}+y_{B}^{2}=30.25$

$\left (-x_{A} \right )^{2}+\left (6x_{A} \right )^{2}=30.25\\\\37x_{A}^2=30.25\\\\x_{A}=0.904$

So we have

$\vec{A}=0.904\hat{i}$

Magnitude of vector A = 0.904

8 0

3 years ago

A 4.0-m-diameter playground merry-go-round, with a moment of inertia of

HACTEHA [7]

Answer:

$7.1$ ms⁻¹

Explanation:

$d$ = diameter of merry-go-round = 4 m

$r$ = radius of merry-go-round = $\frac{d}{2}$ = $\frac{4}{2}$ = 2 m

$I$ = moment of inertia = 500 kgm²

$w_{i}$ = angular velocity of merry-go-round before ryan jumps = 2.0 rad/s

$w_{f}$ = angular velocity of merry-go-round after ryan jumps = 0 rad/s

$v$ = velocity of ryan before jumping onto the merry-go-round

$m$ = mass of ryan = 70 kg

Using conservation of angular momentum

$Iw_{i} - m v r = (I + mr^{2})w_{f}$

$(500)(2.0) - (70) v (2) = (I + mr^{2})(0)$

$1000 = 140 v$

$v = 7.1$ ms⁻¹

5 0

3 years ago

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sesenic [268]

Answer:

Explanation:

4 0

3 years ago