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3 years ago

A ball is thrown horizontally from a 20-m-high building with a speed of 5.0m/s

1 answer:

3 0

1) Horizontal velocity, Vx

Vx is constant and equal to 5.0 m/s

The graph Vx(t) = 5.0 m/s is a horizontal line (parallel to the horizontal t-axis). that intercepts the vertical-axis at 5.0 and runs from t = 0 to t = 2.

2) Vertical velocity, Vy

Vy = gt = 10t

d = at^2 / 2 => tmax = √(2d / a) = √(2*20m/10m/s^2) = 2 s

Graph Vy is an inclined line with slope 10 m/s^2, that runs from t =0 to t = 2 s, and passes through the points (0,0), (1,10), and (2,20).

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The Rite of Spring was composed by

7nadin3 [17]

The 'Rite of Spring' was composed by Igor Stravinsky.

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3 years ago

A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners ar

finlep [7]

Answer:

B) 1.0 × 10^5 V

Explanation:

Electric Potential Due To Point Charges

The electric potential produced from a point charge Q at a distance r from the charge is

$\displaystyle V=k\frac{Q}{r}$

The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.

We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is

$d=\sqrt2 a$

where a is the length of the side.

The distance from any corner to the center is half the diagonal, thus

$\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}$

$\displaystyle r=\frac{1}{\sqrt{2}}=0.707\ m$

The total potential is

$V_t=V_1+V_2+V_3+V_4$

Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of $\pm 3\mu\ C$ . Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.

$\displaystyle V_1=V_2=k\frac{Q}{r}=9\times 10^9 \frac{4\times 10^{-6}}{0.707}$

$V_1=V_2=50912\ V$

The total potential is

$V_t=50912\ V+50912\ V=1\times 10^5\ V$

$\boxed{V_t=1\times 10^5\ V}$

6 0

3 years ago

A ship 1200m off shore fires a gun. how long after the gun is fired will it be heard on the shore?

ryzh [129]

Answer:

We know that the speed of sound is 343 m/s in air

we are also given the distance of the boat from the shore

From the provided data, we can easily find the time taken by the sound to reach the shore using the second equation of motion

s = ut + 1/2 at²

since the acceleration of sound is 0:

s = ut + 1/2 (0)t²

s = ut (here, u is the speed of sound , s is the distance travelled and t is the time taken)

Replacing the variables in the equation with the values we know

1200 = 343 * t

t = 1200 / 343

t = 3.5 seconds (approx)

Therefore, the sound of the gun will be heard at the shore, 3.5 seconds after being fired

6 0

3 years ago

When you throw a ball, the work you do to accelerate it equals the kinetic energy the ball gains. if you do twice as much work w

nata0808 [166]

It doesn't because when u threw it the first time, u notice that the ball eventually came to a stop because of the force that was acting upon it. Although when u throw it harder it will start out faster than the first time u threw it because u put more kinetic energy onto the ball. But the same thing happens with this ball that happened to the second ball, they both have a type of force acting upon them.

8 0

3 years ago

Select the correct answer.

BaLLatris [955]

It is due to the excess stress.

6 0

3 years ago