Using the periodic table, choose the more reactive nonmetal. Si or N

Why should the safety net that is used in circuses under trapezoids be little tightened?

bixtya [17]

So that there isn't too much force restricting the structure of the circus and cause disaster

8 0

3 years ago

Our milky way galaxy is 100000 lyly in diameter. a spaceship crossing the galaxy measures the galaxy's diameter to be a mere 1.

Sidana [21]

The speed of the spaceship relative to the galaxy is 0.99999995c.

A light-year measures distance rather than time (as the name might imply). A light-year is a distance a light beam travels in one year on Earth, which is roughly 6 trillion miles (9.7 trillion kilometers). One light-year equals 5,878,625,370,000 miles. Light moves at a speed of 670,616,629 mph (1,079,252,849 km/h) in a vacuum.We multiply this speed by the number of hours in a year to calculate the distance of a light-year (8,766).

The Milky way galaxy is 100,000 light years in diameter.

The galaxy's diameter is a mere 1. 0 ly.

We know that ;

$L = L_0 \sqrt{1-\frac{v^2}{c^2} }$

L = 1 light year

L₀ = 100,000 light year

$1 = 100,000 \sqrt{1-\frac{v^2}{c^2} }$

$1 = 100,000 \sqrt{1-\frac{v^2}{(3*10^8)^2} }$

$\frac{1}{100,000} = \sqrt{1-\frac{v^2}{c^2} }$

$v = 0.999999995$ $c$

Therefore, the speed of the spaceship relative to the galaxy is 0.99999995c.

Learn more about a light year here:

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5 0

1 year ago

For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo

pickupchik [31]

The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.

In that particular situation, you can prove it like this: 

initial velocity is Vo 
launch angle is α 

initial vertical velocity is 
Vv = Vo×sin(α) 

horizontal velocity is 
Vh = Vo×cos(α) 

total time in the air is the the time it needs to fall back to a height of 0 m, so 
d = v×t + a×t²/2 
where 
d = distance = 0 m 
v = initial vertical velocity = Vv = Vo×sin(α) 
t = time = ? 
a = acceleration by gravity = g (= -9.8 m/s²) 
so 
0 = Vo×sin(α)×t + g×t²/2 
0 = (Vo×sin(α) + g×t/2)×t 
t = 0 (obviously, the projectile is at height 0 m at time = 0s) 
or 
Vo×sin(α) + g×t/2 = 0 
t = -2×Vo×sin(α)/g 

Now look at the horizontal range. 
r = v × t 
where 
r = horizontal range = ? 
v = horizontal velocity = Vh = Vo×cos(α) 
t = time = -2×Vo×sin(α)/g 
so 
r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) 
r = -(Vo)²×sin(2α)/g 

To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. 

dr/dα = d[-(Vo)²×sin(2α)/g] / dα 
dr/dα = -(Vo)²/g × d[sin(2α)] / dα 
dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα 
dr/dα = -2 × (Vo)² × cos(2α) / g 

Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when 
cos(2α) = 0 
2α = 90° 
α = 45°

4 0

3 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch

Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: $-3.62\cdot 10^7 N/C$

c) Electric field 8.00 cm outside the paint layer: $-7.27\cdot 10^7 N/C$

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius $r$ as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

$q=0$

So we have

$\int E dS=0$

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

$r=R=0.065 m$

The area of the surface is

$A=4\pi R^2$

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

$E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}$

The charge included within the sphere in this case is the charge on the paint layer, therefore

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

So, the electric field is:

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C$

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

$r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m$

Gauss theorem this time becomes

$E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

And the charge included within the sphere is again the charge on the paint layer,

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C$

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5 0

3 years ago

Mercury has a density of 13.6 g/cm , whereas the density of water is 1.00 g/cm. The pressure of the atmosphere is high enough to

kenny6666 [7]

10.3

Explanation:

Step 1:

The pressure exerted by any liquid column of height, h density d is given by the formula P = h * d * g

Step 2:

It is given that one atmosphere pressure pushes up 76.0 cm of mercury, we need to calculate the level of water that will be pushed by the same pressure.

Step 3:

Since the pressure pushing up mercury and water is the same

$h_{mercury}$ * $d_{mercury}$ * g = $h_{water}$ * $d_{water}$ * g