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2 years ago

List the names of two tectonic plates that are NOT part of the ring of fire

Physics

1 answer:

Ann [662]2 years ago

5 0

Nolur acil lütfen yalvarırım yalvarırım

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What will happen to plant height if the amount of available light is reduced due to global dimming?

SpyIntel [72]

The plants would wilt and decrease in height due to the lack of sunlight. I hope I helped!

7 0

3 years ago

How much work did the movers do (horizontally) pushing a 41.0-kg crate 10.6 m across a rough floor without acceleration, if the

VLD [36.1K]

Answer:

The required work done is $2555.448~J$

Explanation:

Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if ' $\mu_{k}$ ' be the coefficient of friction, then

$f = \mu_{k} \times N = \mu_{k} \times Mg$

where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.

Since the application of force by the movers does not create any acceleration to the block, we can write

$F = f = \mu_{k} \times M \times g = 0.6 \times 41~Kg~ \times 9.8~m~s^{-2} = 241.08~N$

So the work done (W) in moving the crate by a distance s = 10.6 m is

$W = F \times s = 241.08~N \times 10.6~m = 2555.448 J$

5 0

3 years ago

A 0.85 N force exists between a 7.1 * 10 ^ - 6 * C charge 5.4 m away. What is the magnitude of the second charge ? Please show w

katovenus [111]

Answer:

Explanation:

Force between charge is given by the following expression

F = k Q₁ Q₂ / R² , k = 9 x 10⁹ , Q₁ and Q₂ are charges , R is distance between charges .

Putting the given values ,

.85 = 9 x 10⁹ x 7.1 x 10⁻⁶ x Q₂ / 5.4²

Q₂ = .85 x 5.4² / (9 x 10⁹ x 7.1 x 10⁻⁶ )

= .38788 x 10⁻³ C .

= 387.88 x 10⁻⁶ C .

4 0

3 years ago

Two electric charges are moved so that they are twice as far apart as they had originally been. Is the force they experience fro

Natasha_Volkova [10]

Answer:

No.

Explanation:

The force that two particle experience is inversely proportional to the sqare of the distance, this is:

$F \ \alpha \ \frac{1}{D^{2}}$ for a distance D

If we move them so that D is doubled:

$\frac{1}{2^{2}.D^{2} }$ = $\frac{1}{4} \eq \frac{1}{.D^{2} } \eq$

Then the force they experience is one fourth of the original.

5 0

3 years ago

Match each media portrayal with the type of behavior the media portrayal is

dsp73

The hero attending a funeral is safe behavior while the hero driving fast is riskier behavior

4 0

2 years ago