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2 years ago

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The half-life of Cs-137 is 30.2 years. If the initial mass of the sample is 1.00 kg, how much will remain after 151 years?

1 answer:

Andru [333]2 years ago

8 0

Answer:can someone help

Explanation:

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An internal combustion engine uses fuel, of energy content 44.4 MJ/kg, at a rate of 5 kg/h. If the efficiency is 28%, determine

Fudgin [204]

Efficiency = Power Output / Power Input

Power Input = Rate of Energy input = 44.4 MJ/kg * 5 kg/h

= 222 MJ/h

But 1 hour = 3600seconds

222 MJ/h = 222 MJ/3600s = 0.061667 MW J/s = Watts

Power input = 0.061667 MW = 61 667 W

From Efficiency = Power Output / Power Input

   28% = Power Output / 61667

   Power Output = 0.28 * 61667

   Power Output = 17266.76 W

Power Output = 17 267 W

Rate of heat Rejection = Power input - Power output

      = 61667 - 17267 = 44400 W

Rate of heat Rejection = 44 400 W.

C- Copyright.

5 0

3 years ago

A test of the prototype of a new automobile shows that the minimum distance for a controlled stop from 95 km/h to zero is 55 m.

iris [78.8K]

Answer:

-0.64525g

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 95 km/h

v = Final velocity = 0 km/h

s = Displacement

a = Acceleration

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-95^2}{2\times 0.055}\\\Rightarrow a=-82045.45\ km/h^2$

Converting to m/s²

$a=82045.45=\frac{82045.45\times 1000}{3600\times 3600}=-6.33\ m/s^2$

g = Acceleration due to gravity = 9.81 m/s²

Dividing both the accelerations, we get

$\frac{a}{g}=\frac{-6.33}{9.81}=-0.64525\\\Rightarrow a=-0.64525g$

Hence, acceleration of the car is -0.64525g

8 0

3 years ago

Whats the definition of deposition

Luda [366]

Deposition:

- when a gas changes directly to a solid

- latent heat is released

- physical change, NOT a chemical change

7 0

3 years ago

Read 2 more answers

Two long, straight parallel wires are placed 38 cm apart, one above the other. The top and bottom wires are carrying currents 4.

ElenaW [278]

Answer:

The force per unit length (N/m) on the top wire is 16.842 N/m

Explanation:

Given;

distance between the two parallel wire, d = 38 cm = 0.38 m

current in the first wire, I₁ = 4.0 kA

current in the second wire, I₂ = 8.0 kA

Force per unit length, between two parallel wires is given as;

$\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d }$

where;

μ₀ is constant = 4π x 10⁻⁷ T.m/A

Substitute the given values in the above equation and calculate the force per unit length

$\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d } = \frac{4\pi *10^{-7}*4000*8000 }{2\pi *0.38} = 16.842 \ N/m$

Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m

4 0

3 years ago

An isolated parallel-plate capacitor has a surface charge density. If the space between the plates is filled with a material of

anygoal [31]

Answer:

Explanation:

$2\sqrt{34}$

3 0

3 years ago