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3 years ago

Quando ocorre um eclipse total da lua?

1 answer:

3 0

Um eclipse solar ocorre quando a lua fica entre a Terra e o sol, e a lua projeta uma sombra sobre a Terra. Um eclipse solar só pode ocorrer na fase da lua nova, quando a lua passa diretamente entre o sol e a Terra e suas sombras caem sobre a superfície da Terra.

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With time running out in a game, Rachel runs towards the basket at a speed of 2.5 meters per second and from half-court, launche

finlep [7]

Answer:

Rachel(2.5,0)

ball(6.5,4.7)

b.R=10.15m/s, 27.57deg

Explanation:

The reference angle of Rachel is 0 $0^{0} while the ball is at 36^{0}$

resolving rachel's speed to the horizontal, we have

Ux=2.5cos0

Ux=2.5m/s

resolving rachel's speed to the vertical we have,

Uy=2.5sin0

Uy=0

for the ball

resolving the speed to its horizontal component

Ux=8cos36

Ux=6.5m/s

Uy=8sin36

Uy=4.7m/s

Rachel(2.5,0)

ball(6.5,4.7)

To get the resultant of their speed

Add the horizontal speed of rachel to that of the ball to get the total horizontal speed

Add the vertical speed of rachel and the ball to get the total vertical speed component

TUx=2.5+6.5=9M/S

TUy=0+4.7=4.7m/s

R= $\sqrt{(TUx^2+TUy^2}$

R= $\sqrt{(9^2+4.7^2}$

R= $\sqrt{(103)}$

R=10.15m/s

the direction

tan $\alpha$ =TUy/TUx

tan $\alpha$ =4.7/9

$\alpha$ =tan^-1(0.522)

$\alpha$ =27.57deg

4 0

3 years ago

I need friends so if you want to be my friend let me know

nydimaria [60]

Answer:

sure if I can figure out how to follow you

Explanation:

recenty joined

3 0

2 years ago

A barrel 1 m tall and 60 cm in diameter is filled to the top with water. What is the pressure it exerts on the floor beneath it?

allsm [11]

Answer:

The pressure on the ground is about 9779.5 Pascal.

The pressure can be reduced by distributing the weight over a larger area using, for example, a thin plate with an area larger than the circular area of the barrel's bottom side. See more details further below.

Explanation:

Start with the formula for pressure

(pressure P) = (Force F) / (Area A)

In order to determine the pressure the barrel exerts on the floor area, we need the calculate the its weight first

$F_g = m \cdot g$

where m is the mass of the barrel and g the gravitational acceleration. We can estimate this mass using the volume of a cylinder with radius 30 cm and height 1m, the density of the water, and the assumption that the container mass is negligible:

$V = h\pi r^2=1m \cdot \pi\cdot 0.3^2 m^2\approx 0.283m^3$

The density of water is 997 kg/m^3, so the mass of the barrel is:

$m = V\cdot \rho = 0.283 m^3 \cdot 997 \frac{kg}{m^3}= 282.151kg$

and so the weight is

$F_g = 282.151kg\cdot 9.8\frac{m}{s^2}=2765.08N$

and so the pressure is

$P = \frac{F}{A} = \frac{F}{\pi r^2}= \frac{2765.08N}{\pi \cdot 0.3^2 m^2}\approx 9779.5 Pa$

This answers the first part of the question.

The second part of the question asks for ways to reduce the above pressure without changing the amount of water. Since the pressure is directly proportional to the weight (determined by the water) and indirectly proportional to the area, changing the area offers itself here. Specifically, we could insert a thin plate (of negligible additional weight) to spread the weight of the barrel over a larger area. Alternatively, the barrel could be reshaped (if this is allowed) into one with a larger diameter (and smaller height), which would achieve a reduction of the pressure.

7 0

3 years ago

which will have the greatest amplitude and pitch when shaken, a jar half fill with water and 2 marble or a jar with no water and

masha68 [24]

Answer:

a jar with no water and 2 marbles

Explanation:

When a jar is half fill with water and 2 marbles, the amplitude pitch of the sound produced is lower because the water has caused the glass to vibrate slower. When there is no water and two marbles in the jar, the amplitude and pitch is higher because the jar is vibrating faster.

8 0

4 years ago

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$

$7.05 + 30.5 = \frac{0.266}{r}$

$r = 7.08 \times 10^{-3} m$

so distance moved by the particle is given as

$d = r_1 - r_2$

$d = 0.0377 - 0.00708$

$d = 0.0306 m$

6 0

3 years ago