The hottest climates on Earth are located near the Equator because this region

You see a total eclipse of the sun. Three Saros cycles later, you find yourself at the same location.

alukav5142 [94]

Answer:

The correct answer is B. You would not have to travel far to see another total eclipse .

Explanation:

it is said that the way to predict or know beforehand if there is a lunar or solar eclipse is just to know the what is called the saros which are periods of exactly 223 synodic months and the eclipses are repeated or occurred in the same location every 3 Saros cycles.

5 0

3 years ago

a water bomber flying with a horizontal speed of 85m/s at a height of 3000m drops a load on a fire below. How far in front of th

Andreyy89

Answer:

2081.65 m

Explanation:

We'll begin by calculating the time taken for the load to get to the target. This can be obtained as follow:

Height (h) = 3000 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

3000 = ½ × 10 × t²

3000 = 5 × t²

Divide both side by 5

t² = 3000 / 5

t² = 600

Take the square root of both side

t = √600

t = 24.49 s

Finally, we shall determine the distance from the target at which the load should be released. This can be obtained as follow:

Horizontal velocity (u) = 85 m/s

Time (t) = 24.49 s

Horizontal distance (s) =?

s = ut

s = 85 × 24.49

s = 2081.65 m

Thus, the load should be released from 2081.65 m.

3 0

3 years ago

A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of rho0 is placed in a container of water. Initially

ivanzaharov [21]

Answer:

a) s,f,r b) r c) f

Explanation:

To determine what happens with the sphere we use Newton's second law with the Archimedes principle that states that the thrust (B) on a body is equal to the weight of the liquid dislodged

For the sphere to be in equilibrium the sum of forces is zero

B - W = 0

B = W = mg

Now let's use the concept of density for the body and water

Solid sphere

ρ = m / V

V = 4/3 π r³

m = ρ₀ (4/3 π r³)

W = ρ₀ (4/3 π r³) g

Water (a)

ρ = mₐ / Vₐ

mₐ = ρ Vₐ

B = ρ Vₐ g

Let's replace and simplify

ρ Vₐ g = ρ₀ (4/3 π r³) g

ρ Vf = ρ₀ (4/3 π r³) (1)

For the initial condition with rho, mo and ro the height of the water is H, let's analyze each case

a) We have the same mass, but less radius, as density is mass over volume density increases

r <ro V <V₀ ⇒ ρ₁> ρ₀

When analyzing the equation (1) on the right side, this case is the most complicated because I can make the relationship between the density of the sphere and its volume change even when the mass is constant

Assume the three possibilities

- The product of (ρ₁ V) that does not matter in that case the left side does not change and the mark remains the same (s)

- The product (ρ₁ V) increases the left side must increase so the mark goes up (r)

- The product (ρ₁ V) decreases the left side should go down, so the low mark (f)

b) sphere the same radius, but the density increases.

In this case the right side of the equation (1) increases, therefore the left side must increase so that the volume must increase and consequently increase the height (r)

c) you have the same radius, but the mass decreases

r = r₀ V = V₀ m <m₀ ρ₁ <ρ₀

The right side of the equation decreases, because the density decreases, the left side must decrease, for this the volume must decrease, lowering the height (f)

8 0

3 years ago

A simple experiment to measure the speed of sound doesn't involve a stopwatch. You can fill up along tube with water and put a t

Serhud [2]

Answer:

Explanation:

In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.

To find the fundamental mode you use:

$f_n=\frac{nv_s}{4L}$

n: mode of the sound

vs: sound speed

L: length of the column of air in the tube.

A) The fundamental mode id obtained for n=1:

$f_1=\frac{v_s}{4L}$

B) For the 3rd harmonic you have:

$f_3=\frac{3v_s}{4L}$

C) For the 2nd harmonic:

$f_2=\frac{2v_s}{4L}$

7 0

3 years ago

The velocity of a ball changes from ‹ 9, −6, 0 › m/s to ‹ 8.96, −6.12, 0 › m/s in 0.02 s, due to the gravitational attraction of

Alenkasestr [34]

Answer:

a) $a=(-2,-6,0)m/s^2$ , with a magnitude of $6.3m/s^2$

b) $\frac{\Delta p}{\Delta t}=(-0.24,-0.72,0)Kgm/s^2$ , with a magnitude of $0.76Kgm/s^2$

c) $F=(-0.24,-0.72,0)N$ , with a magnitude of $0.76N$

Explanation:

We have:

$v_{ix}=9m/s, v_{iy}=-6m/s, v_{iz}=0m/s\\v_{fx}=8.96m/s, v_{fy}=-6.12m/s, v_{fz}=0m/s\\t=0.02s, m=0.12Kg$

We can calculate each component of the acceleration using its definition $a=\frac{\Delta v}{\Delta t}$

$a_x=\frac{v_{fx}-v_{ix}}{t} = \frac{(8.96m/s)-(9m/s)}{0.02s} =-2m/s^2\\a_y=\frac{v_{fy}-v_{iy}}{t} = \frac{(-6.12m/s)-(-6m/s)}{0.02s} =-6m/s^2\\a_y=\frac{v_{fz}-v_{iz}}{t} = \frac{(0m/s)-(0m/s)}{0.02s} =0m/s^2\\$

The rate of change of momentum of the ball is $\frac{\Delta p}{\Delta t} = \frac{\Delta mv}{\Delta t} = \frac{m\Delta v}{\Delta t} = ma$

So for each coordinate:

$\frac{\Delta p_x}{\Delta t}=-0.24Kgm/s^2\\\frac{\Delta p_y}{\Delta t}=-0.72Kgm/s^2\\\frac{\Delta p_z}{\Delta t}=0Kgm/s^2$

And these are equal to the components of the net force since F=ma.

If magnitudes is what is asked:

$a=\sqrt{a_x+a_y+a_z} =6.3m/s^2\\F=ma=\frac{\Delta p}{\Delta t}=0.76N$

3 0

3 years ago