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3 years ago

Please help. 8th grade science

Physics

1 answer:

Aneli [31]3 years ago

3 0

Answer:

false

Explanation:

It doesn't the copper wire wouldn't even be pulled by the magnet at all and the electricity would stay inside of the the force of the copper wire

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A tennis ball is served horizontally from 2.4m above the ground at net is 12m away and point 0.9 high will be ball clear the net

Schach [20]

Explanation:

Let us first calculate long does it take to go 12m at 30m/s( assumed speed)

12/30 = 0.4 seconds

horizontal distance the ball drop in that time

H= (0)(0.4)+1/2(-9.8)(0.4)2

H= -0.78m

negative sign shows that the height of the ball at the net from the top.

Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m

As 1.62m>0.9m so the ball will clear the net.

H_1= V0y t’ + ½ g t’^2

-2.4= (0)t’ + ½ (-9.8) t’^2

t’= 0.69s

X’=V0x t’

X’=(30)(0.96)

X’= 20.7m

3 0

3 years ago

The temperature on the moon is

Nuetrik [128]

B.

It can go from very hot to very cold, it depends on the area of the moon and where the sunlight hits.

4 0

3 years ago

Read 2 more answers

What is the order of the events for the water cycle on a typical warm day?

Rudik [331]

B precipitation,condensation,precipitation

6 0

2 years ago

A ball is dropped from rest at a point 12 m above the ground into a smooth, frictionless chute. The ball exits the chute 2 m abo

Nonamiya [84]

Answer:

29,7 m

Explanation:

We need to devide the problem in two parts:

A) Energy

B) MRUV

<u>Energy:</u>

Since no friction between pint (1) and (2), then the energy conservatets:

Energy = constant ----> Ek(cinética) + Ep(potencial) = constant

Ek1 + Ep1 = Ek2 + Ep2

Ek1 = 0 ; because V1 is zero (the ball is "dropped")

Ep1 = m*g*H1

Ep2= m*g*H2

Then:

Ek2 = m*g*(H1-H2)

By definition of cinetic energy:

m*(V2)²/2 = m*g*(H1-H2) ---> $V2 = \sqrt{(2*g*(H1-H2)}$

Replaced values: V2 = 14,0 m/s

The decomposition of the velocity (V2), gives a for the horizontal component:

V2x = V2*cos(α)

Then the traveled distance is:

X = V2*cos(α)*t.... but what time?

The time what takes the ball hit the ground.

Since: Y3 - Y2 = V2*t + (1/2)*(-g)*t²

In the vertical axis:

Y3 = 0 ; Y2 = H2 = 2 m

Reeplacing:

-2 = 14*t + (1/2)*(-9,81)*t²

solving the ecuation, the only positive solution is:

t = 2,99 sec ≈ 3 sec

Then, for the distance:

X = V2*cos(α)*t = (14 m/s)*(cos45°)*(3sec) ≈ 29,7 m

6 0

3 years ago

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A Imagine you derive the following expression by analyzing the physics of a particular system: v2 + 2az. The problem requires so

Nuetrik [128]

Answer:

z = 93.2 m

Explanation:

We can appreciate that this expression is equivalent to the linear motion equation with constant acceleration

v² = v₀² + 2 a d

If we make a term-to-term comparison with the expression obtained, they are equivalent

u² = v² + 2 a z

From here we can clear the position

2 a z = u² –v²

z = (u² –v²) / 2 a

Let's calculate

For the speed to reduce the acceleration must be negative

z = (0 - 21.8²) / 2(- 2.55)

z = 93.2 m

7 0

3 years ago