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Kazeer [188]
3 years ago
14

Volcanoes tend to erupt at places where

Physics
2 answers:
tigry1 [53]3 years ago
4 0

Answer:

it is 3

Explanation:

because the crack will be open for the magma to come out

yulyashka [42]3 years ago
3 0
The answer is 1.
Most volcanoes that erupt are located along the circum-pacific.
There is convergent plate boundary at the circus pacific.
Convergent plate boundary is an area on Earth where two or more lithospheric plates collide.
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A 438kg car is accelerating east at 2.55m/s^2. What is the total force acting east on the car
lisabon 2012 [21]

Answer:

<h2>1116.9 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 438 × 2.55

We have the final answer as

<h3>1116.9 N</h3>

Hope this helps you

5 0
3 years ago
a 0.40kg soccer ball approaches a player horizontally with a velocity of 18m/s north. the player strikes the ball and causes it
denis23 [38]
Impulse = Ft = (m)(delta v)
delta v = change in velocity = velocity final - velocity initial.
= -22m/s - +18m/s = -40m/s.
mdeltav = (0.40kg)(-40m/s) = -16kgm/s or -16Ns.

4 0
4 years ago
If a proton and an electron are released when they are 7.00×10−10 m apart (typical atomic distances), find the initial accelerat
Ugo [173]

Answer:

The acceleration of the proton is 2.823 x 10¹⁷ m/s²

The acceleration of the electron is 5.175 x 10²⁰ m/s²

Explanation:

Given;

distance between the electron and proton, r = 7 x 10⁻¹⁰ m

mass of proton, m_p = 1.67 x 10⁻²⁷ kg

mass of electron, m_e = 9.11 x 10⁻³¹ kg

The attractive force between the two charges is given by Coulomb's law;

F = \frac{k(q_p)(q_e)}{r^2}

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/c²

F = \frac{k(q_p)(q_e)}{r^2} \\\\F = \frac{(9*10^9)(1.602*10^{-19})(1.602*10^{-19})}{(7*10^{-10})^2} \\\\F = 4.714 *10^{-10} \ N

Acceleration of proton is given by;

F = ma

F = m_pa_p\\\\a_p = \frac{F}{m_p}\\\\a_p = \frac{4.714*10^{-10}}{1.67*10^{-27}}\\\\a_p = 2.823 *10^{17} \ m/s^2

Acceleration of the electron is given by;

F = m_ea_e\\\\a_e = \frac{F}{m_e}\\\\a_e = \frac{4.714*10^{-10}}{9.11*10^{-31}}\\\\a_e = 5.175 *10^{20} \ m/s^2

7 0
3 years ago
A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observ
siniylev [52]

Answer:

m=1.53kg    

Explanation:

To solve this problem we use the Hooke's Law:

F=k*\Delta x     (1)

F is the Force needed to expand or compress the spring by a distance Δx.

The spring stretches 0.2cm per Newton, in other words:

1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm  

The force applied is due to the weight

F=mg

We replace in (1):

mg=k*\Delta x  

We solve the equation for m:

m=k*\Delta x/g=5*3/9.81=1.53kg    

7 0
3 years ago
How are mechanical waves and electromagnetic waves alike?
miss Akunina [59]

I got this answer from the internet

6 0
3 years ago
Read 2 more answers
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