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3 years ago

What is the minimum total energy released when

Physics

2 answers:

Karo-lina-s [1.5K]3 years ago

7 0

Answer:

(1) $1.64 \times 10^{-13} J$

Explanation:

Energy released in this process is given by

$Q = \Delta m c^2$

here we have

$\Delta m = 9.1 \times 10^{-31} kg$

since here mass of two electrons converted into energy so we have

$Q = 2(9.1 /times 10^{-31})(3\times 10^8)^2$

$Q = 1.64 \times 10^{-13} J$

so here energy released is the energy of rest mass energy due to two charges i.e. electrons and positrons

Ganezh [65]3 years ago

6 0

What is the minimum total energy released when an electron and its antiparticle (positron) annihilate each other?
The minimum total energy released when an electron and its antiparticle (positron) annihilate each other is 2.73 × 10^–22 J. The answer is number 4.

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Suppose a car is traveling at +25.0 m/s, and the driver sees a traffic light turn red. After 0.340 s has elapsed (the reaction t

scoundrel [369]

First, we will get the distance traveled before the driver applied the brakes.
distance = velocity * time
distance = 25*0.34 = 8.5 m

Now, we will calculated the distance that the car traveled after the driver applied the brakes. To do this, we will use the equation of motion:
vf^2 = vi^2 + 2*a*d where:
vf = zero, vi = 25 m/s and a = -7 m/s^2
Note: The negative sign is only to show deceleration
d = 1/2*(625) /(7) = 44.6428 m

The total stopping distance = 8.5 + 44.6428 = 53.1428 m

3 0

3 years ago

You hold glider AA of mass 0.125 kgkg and glider BB of mass 0.375 kgkg at rest on an air track with a compressed spring of negli

andrew11 [14]

Answer:

The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction

Explanation:

Inelastic Collision

Given data

mass of glider A m1= 0.125kg

initial velocity u1=0

final velocity v1= 0.600 m/s

mass of glider B m2= 0.375kg

initial velocity u2=0

final velocity v2=?

We know that the expression for the conservation of momentum is given as

m1u1+m2u2=m1v1+m2v2

since u1=u2=u=0m/s

u(m1+m2)=m1v1+m2v2

substituting we have

0(0.125+0.0375)=0.125*0.6+0.375*v2

0=0.075+0.375v2

0.375v2=-0.075

v2=-0.075/0.375

v2=-0.2m/s

The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction

3 0

3 years ago

Directions: Select ALL the correct answers.

nordsb [41]

Definitely D. The brakes on a bike rub against the wheel. Not sure about the others.

5 0

3 years ago

A(n) 10.1 g bullet is fired into a(n) 2.41 kg ballistic pendulum and becomes embedded in it. The acceleration of gravity is 9.8

fomenos

Answer:

$v = 186.90\,\frac{m}{s}$

Explanation:

The motion of ballistic pendulum is modelled by the appropriate use of the Principle of Energy Conservation:

$\frac{1}{2}\cdot (m_{p}+m_{b})\cdot v^{2} = (m_{p}+m_{b})\cdot g \cdot h$

The final velocity of the system formed by the ballistic pendulum and the bullet is:

$v = \sqrt{2\cdot g\cdot h}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.031\,m)}$

$v\approx 0.78\,\frac{m}{s}$

Initial velocity of the bullet can be calculated from the expression derived of the Principle of Momentum:

$(0.0101\,kg)\cdot v = (2.41\,kg + 0.0101\,kg)\cdot (0.78\,\frac{m}{s} )$

$v = 186.90\,\frac{m}{s}$

5 0

3 years ago

A 4.00 kg block is suspended from a spring with k 500 N/m. A 50.0 g bullet is fired into the block from directly below with a sp

Yanka [14]

Answer:

a. A = 0.1656 m

b. % E = 1.219

Explanation:

Given

mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150 m/s , k = 500 N / m

To find the amplitude of the resulting SHM using conserver energy

ΔKe + ΔUg + ΔUs = 0

¹/₂ * m * v² - ¹/₂ * k * A² = 0

A = √ mB * vₓ² / k

vₓ = mb * u₁ / mb + mB

vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518

A = √ 4.0 kg * (1.852 m/s)² / (500 N / m)

A = 0.1656 m

The percentage of kinetic energy

%E = Es / Ek

Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5

Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N

% E = 13.72 / 1125 = 0.01219 *100

% E = 1.219

6 0

3 years ago