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Montano1993 [528]
3 years ago
13

PLS HELP ME I WILL GIVE YOU BRAINLIEST

Physics
2 answers:
ElenaW [278]3 years ago
8 0
3. Millions of years ago, the Earth was not divided into 7 continents. It was only a very large land mass, the scientists named that one land mass “Pangaea”
4. If the tectonic plates are disturbed, they can cause mass earthquakes and tsunamis.
5. The ring of fire is a massive collection of volcanoes under the pacific ocean. They all form the shape of a ring, hence the name.
SORRY, I COULDNT ANSWER THE FIRST TWO QUESTIONS.
Maksim231197 [3]3 years ago
5 0
Isn’t that for earth science though
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If the speed of the wave is 12 m/s and the frequency is 2.3 Hz, what is the <br> wavelength?
-BARSIC- [3]

Answer: 5.2

Explanation: 12/ 2.3

3 0
2 years ago
PLEASE ANSWER ILL GIVE YOU BRAIN!!!!
solniwko [45]

Answer:

I think the first two one is 70 the second on is 14

I hope this helps if I’m wrong I’m sorry

Explanation:

3 0
3 years ago
(Superposition, quadratic formuła) The Earth (mass M) is at origin, and the Moon (mass m ) is located at a distance d. (a) Using
jekas [21]

Answer:

a)    Fa = G m2 [M / r² - m / (d-r)²]  

b) r2 = 31 10⁶ m

Explanation:

The equation of the law of universal gravitation is

            F = G m1 m1/ r²

The value of the gravitation constant is 6.67 10-11 N m²/kg². This force is always attractive.

Let's calculate the value of that force on the spacecraft, add the strength

Earth's force to the ship

            F1 = G M m2 / r²

The moon force ship

            F2 = G m m2 / (d-r)²

Total force is

            Fa = F1 — F2

            Fa = G M M2 / r² - G m m2 / (d-r)²

            Fa = G m2 [M / r² - m / (d-r)²]

This is the force on the spaceship

b)  Let's look for the point where the force is zero, for this we can see that the value of the bracket must be zero

           Fa = 0

           [M / r² - m / (d-r)²] = 0

            M / r² = m / (d-r)²

           (d-r)² = m/M   r²

           d² -2rd + r² - m/M   r² = 0

           r² [m/M - 1] + r 2d - d² = 0

This is a second degree equation for r, we solve the to find the results.

           r = {-2d ±√[4d² - 4 [m/M -1] (-d²)]} / (2 [m/M-1])

           r = {-2d ± √ [4d² (1 + (m/M-1)]} / 2(m/M-1)

           r = {-2d ± 2d √(m/M)}  / (2(m/M-1))

           r = 2d {-1 ± √(m/M)} / 2(m/M-1)

           r = d [-1 ± √(m/M)] /  (m/M-1)

To find the explicit value we substitute the values ​​that we can find in tables

          m = 7.36 1022 kg

          M = 5.98 1024 kg

          d = 380000 km (1000m / 1 km) = 380 10⁶ m

          r = 380 10⁶ [-1 ±√(7.36 10²² / 5.98 10²⁴)] /(7.36 10²² / 5.98 / 10²⁴ -1)

          r = 380 10⁶ [-1 - √ (1.23 10²)] / (123-1)

          r = 380 106 [-1 ± 11] / 122

 

          r1 = 380 10⁶ 10/122 = 380 10⁶ 0.08197

          r1 =  31 10⁶ m

          r2 = 380 106 [-12/122] = 380 10 6 0.09836

          r2 = -37 10⁶ m

The correct distance is the positive r2 = 31 10⁶ m

c) let's use Newton's second law, to find the acceleration in the spacecraft

 

          F = m a

          a = Fa / m2 = G m2 [M / r² - m / (d-r)²] / m2

          a = G [M/r² - m/(d-r)²]

Since we have acceleration, we can use the definition of kinematics

           a = dv / dt = dv / dr dr / dt = dv / dr v

           v dv = a dr

           v dV = G [M /r² - m /(d-r)²] dr

We integrate

            ½ (V² - Vo²) = G [M (-1 /r) -m (1 / (d-r)

We evaluate between the initial point where we can assume that the initial velocity is zero for the Xo position and the final point with velocity v at the points

            V² = 2G [M (1 / Xo - 1 /X) - m (1 / (d-X) - 1 /(d-xo)]

            V² = [-M /X -m /(d-X)] 2G + constant

We now use the definition of speed

            v = dx / dt

            dx = V dt

We substitute, perform the integral and simplify, if we can make the constant zero

             dx = √([-M / X -m / (d-X)] 2G) dt

             dx / √([-M / X -m / (d-X)] 2G = dt

8 0
3 years ago
what is the true answer in this question?Making a schedule on the things that need to be accomplished yes or no​
musickatia [10]

Answer:

yes

Explanation:

this is true don't believe other people whether right or wrong

5 0
3 years ago
Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds.
s344n2d4d5 [400]

This question is incomplete, the complete question is;

A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = -16t² + 8√t, where t is the time in seconds and s(t) is the distance in feet from the top of the building.

(a) Write a formula for the average velocity of the ball for t near 4.

(b) Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds

(c) What is your estimate for the instantaneous velocity of the ball at t = 4

Answer:

a)

Average velocity, (Vavg)  of the ball for t near 4.

Vavg = [s(4) - s(0)] / (4 - 0)

Where s(4) = -16 × 4² + 8 × √4= - 240 m

s(0) = -16 × 0 + 8 * 0 = 0

b)

duration = 1 sec

Vavg = [s(5) - s(4)] / (5 - 4)

s(5) = -16 × 52 + 8 × √5 = - 382 m

s(4) = -16 × 42 + 8  √4 = - 240 m

Vavg = (-382 - (-240)) / (5 - 4)

Vavg = - 142.1 m/s

duration = 0.5 sec

Vavg = [s(4.5) - s(4)] / (4.5 - 4)

s(4.5) = -16 × 4.52 + 8 × √4.5 = - 307 m

s(4) = -16 × 42 + 8 × √4 = - 240 m

Vavg = (-307 - (-240)) / (4.5 - 4)

Vavg= - 134.1 m/s

duration = 0.05 sec

Vavg = [s(4.05) - s(4)] / (4.05 - 4)

s(4.05) = -16 × 4.052 + 8 × √4.05 = - 246 m

s(4) = -16 × 42 + 8 × √4 = - 240 m

Vavg = (-246 - (-240)) / (4.05 - 4)

Vavg= - 126.8 m/s

c)

Instantaneous velocity, v = ds/dt

= - 16 × 2 × t + 8 ×× (0.5 / √t )

= - 32 × t + 4/√t

ds/dt at t = 4 is,

v = - 32 × 4 + 4 / √4

= - 126 m/s

5 0
3 years ago
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