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3 years ago

Find the energy released when there is a decrease of 0.3kg of material in a nuclear reaction

Physics

1 answer:

MrRissso [65]3 years ago

4 0

Answer:

E = 2.7 x 10¹⁶ J

Explanation:

The release of energy associated with the mass can be calculated by Einstein's mass-energy relation, as follows:

$E = mc^2$

where,

E = Energy Released = ?

m = mass of material reduced = 0.3 kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (0.3\ kg)(3\ x\ 10^8\ m/s)^2$

E = 2.7 x 10¹⁶ J

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the maximum intensity levels of a trumpet, trombone, and a bass drum, each at a distance of 3m are 94 dB, 107dB, and 113dB respe

Gwar [14]

Answer:

β = 114 db

Explanation:

The intensity of sound in decibles is

β = 10 log $\frac{I}{I_{o}}$

in most cases Io is the hearing threshold 1 10-12 W / cm²

let's calculate the intensity of each instrument

I / I₀ = 10 (β / 10)

I = I₀ 10 (β / 10)

trumpet

I1 = 1 10⁻¹² 10 (94/10)

I1 = 2.51 10⁻³ / cm²

Thrombus

I2 = 1 10⁻¹² 10 (107/10)

I2 = 5.01 10-2 W / cm²

low

I3 =1 1-12 (113/10) W/cm²

I3 = 1,995 10-1 W / cm²

when we place the three instruments together their sounds reinforce

I_total = I₁ + I₂ + I₃

I_ttoal = 2.51 10-3 + 5.01 10-2 + 1.995 10-1

I_total = 0.00251 + 0.0501 + 0.1995

I_total = 0.25211 W / cm²

let's bring this amount to the SI system

β = 10 log (0.25211 / 1 10⁻¹²)

β = 114 db

7 0

3 years ago

A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles

Dimas [21]

A) The total energy of the system is sum of kinetic energy and elastic potential energy:
$E=K+U= \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
$x=A=4.00 cm = 0.04 m$
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
$E=U= \frac{1}{2}kA^2 = \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J$

b) When the position of the object is
$x=1.00 cm = 0.01 m$
the potential energy of the system is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J$
and so the kinetic energy is
$K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J$
since the mass is $m=50.0 g=0.05 kg$ , and the kinetic energy is given by
$K= \frac{1}{2}mv^2$
we can re-arrange the formula to find the speed of the object:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s$

c) The potential energy when the object is at
$x=3.00 cm=0.03 m$
is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$
Therefore the kinetic energy is
$K=E-U=0.028 J-0.016 J = 0.012 J$

d) We already found the potential energy at point c, and it is given by
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$

5 0

4 years ago

Does the KE of a car change more when it accelerates from 22 km/h to 32 km/h or when it accelerates from 32 km/h to 42 km/h

mote1985 [20]

Answer:

The change in kinetic energy (KE) of the car is more in the second case.

Explanation:

Let the mass of the car = m

initial velocity of the first case, u = 22 km/h = 6.11 m/s

final velocity of the first case, v = 32 km/h = 8.89 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

ΔK.E = ¹/₂m(8.89² - 6.11²)

= 20.85m J

initial velocity of the second case, u = 32 km/h = 8.89 m/s

final velocity of the second case, v = 42 km/h = 11.67 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

ΔK.E = ¹/₂m(11.67² - 8.89²)

= 28.58m J

The change in kinetic energy (KE) of the car is more in the second case.

6 0

3 years ago

Sophie says that geologic maps do not matter because she gets no benefits from them. Why is Sophie wrong? a. She can see the ran

astra-53 [7]

The Answer is Option C
I think...
Sorry If i am wrong...

4 0

3 years ago

Read 2 more answers

A carbon resistor is 4 mm long and has a constant cross section of 0.2 mm2. The conductivity of carbon at room temperature is 3

tangare [24]

Answer:

The resistance is found to be 6Ω

The current is found to be 0.66 A

Explanation:

The resistance of a conductor in terms of its dimensions is given as:

R = ρL/A

where,

R = resistance = ?

ρ = resistivity = 3 x 10⁴ Ω.m

L = Length = 4 mm = 0.004 m

A = Cross-sectional area = 0.2 mm² = 0.2 x 10⁻⁶ m²

Therefore,

R = (3 x 10⁴ Ω.m)(0.004 m)/(0.2 x 10⁻⁶ m²)

R = 6 Ω

Now, the potential difference between both ends of the resistor is:

ΔV = 16 V - 12 V = 4 V

Now, from Ohm's Law:

V = IR

I = V/R

I = 4 V/ 6 Ω

I = 0.66 A

7 0

3 years ago

Read 2 more answers