A seismic wave is energy released as the result of rock movement along a fault.<br> T/F

Calculate the pressure when a force of 200N acts on an area of 0.05cm2<br> .

deff fn [24]

FORCE ACTING ON THE BODY , F = 200N

AREA ON WHICH THE FORCE IS ACTING , A = 0.05 CM²

= 0.05*(10⁻²)

= 0.05*10⁻² M²

TO CONVERT CM TO M * BY 10⁻²

PRESSURE = FORCE / AREA

PRESSURE = 200/0.05*10⁻⁴

= 200*10⁴/0.05

= 200*10⁴*10²/5

= 20*10⁵/5

= 4*10⁵ N/M².

HOPE THIS HELPS.

4 0

2 years ago

An unstable atomic nucleus of mass 1.58 10-26 kg initially at rest disintegrates into three particles. One of the particles, of

lara31 [8.8K]

Answer:

a The velocity of the third particle is $v_3 = (-15.629*10^{6}i -14.45*10^{6}j)\ m/s$

b The total kinetic energy increase $E = 6.50 *10^{-13} J$

Explanation:

To solve this question we are going to be using the concept of conservation of momentum which is mathematically represented as

$m_1v_1 +m_2v_2 +m_3v_3 = 0$

From the question

Mass of unstable atomic nucleus $m_0= 1.58 *10^{-26} kg$

Mass of x-axis particle $m_1 = 8.44* 10^{-27} kg$

Velocity of x-axis particle $v_ 1 = 4.00* 10^6 m/s$

Mass of y-axis particle $m_2 =5.20* 10^{-27} kg$

Velocity of y-axis particle $v_2 = 6.00 *10^6 m/s$

Since the initial mass is known we can obtain the mass of the third particle

Using the conservation of mass mathematical expression

$m_3 = m_0 - (m_1 +m_2)$

$m_3 = 1.58 *10^{-26} - ( 8.44* 10^{-27} + 5.20* 10^-27)$

$= 2.16 *10 ^{-27}kg$

Using the mathematical expression above for conservation of momentum

$(8.44 * 10^{-27})( 4.00* 10^6i) + (5.20 *10^{-27})(6.00* 10^6j) +(2.16*0^{-27})v_3 =0$

$v_3 = (-15.629*10^{6}i -14.45*10^{6}j)\ m/s$

The above is the vector solution now to obtain the value in numeric form we evaluate the magnitude of the vector

$Mag = \sqrt{(15.629*10^6)^2 +(14.45*10^6)^2}$

$= 21.285*10^6 \ m/s$

To obtain the total increase in kinetic energy which is mathematically represented as

$E = \frac{1}{2}m_1v_1 + \frac{1}{2}m_2v_2 + \frac{1}{2} m_3 v_3$

$E =\frac{1}{2} [(5.20*10^{-27})(6.0*0^{6})^2 + (8.44*10^{-27})(4.00*10^6)^2+\\\\(2.16*10^{-27})(21.285*10^6)^2]$

$= 6.50 *10^{-13} J$

5 0

3 years ago

Why doesn't the skater ever regain his potential energy?

lina2011 [118]

Answer:

potential energy is stored energy...basically energy gathered from staying still so because the skater is in motion he has lost the potential energy and turned it into kinetic energy

Explanation:

7 0

3 years ago

Identify the scenario below as to whether they increase, decrease, or not change the frequency in which a string vibrates and th

telo118 [61]

Answer:1. Increase 2. Decrease 3. Decrease

Explanation:

4 0

3 years ago

If two identical objects are dropped one after 1 s delay with respect to another then in the absence of the air resistance *

zloy xaker [14]

Answer:

A.No. Assuming no other factors (such as air resistance) the first object will have a velocity of 32 feet/second when the other is dropped. Since they will both have the same acceleration, the first distance between them will increase by 32 feet per second.

Explanation:

5 0

3 years ago

A seismic wave is energy released as the result of rock movement along a fault. T/F