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2 years ago

8

Which object has the least thermal energy?

2 answers:

hichkok12 [17]2 years ago

5 0

Answer:

2kg brick

Explanation:

2kg brick at 25 degrees.

lana66690 [7]2 years ago

5 0

Answer: A 2kg brick at 20*C

Explanation:

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A light horizontal spring has a spring constant of 138 N/m. A 3.35 kg block is pressed against one end of the spring, compressin

BARSIC [14]

Answer:

$U_k = 0.113$

Explanation:

using the law of the conservation of energy:

$E_i -E_f=W_f$

$\frac{1}{2}Kx^2=NU_kd$

where K is the spring constant, x is the spring compression, N is the normal force of the block, $U_k$ is the coefficiet of kinetic friction and d is the distance.

Also, by laws of newton, N is calculated by:

N = mg

N = 3.35 kg * 9.81 m/s

N = 32.8635

So, Replacing values on the first equation, we get:

$\frac{1}{2}(138)(0.123)^2= (32.8635)U_k(0.281m)$

solving for $U_k$ :

$U_k = 0.113$

8 0

3 years ago

Steam types of<br> forces in<br> nature

max2010maxim [7]

Answer:

Gravity, Weak, Electromagnetic and Strong.

7 0

3 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago

Determine the elastic energy U stored in<br> the compressed spring.

hoa [83]

Answer:

Answer: The spring constant of the spring is k = 800 N/m, and the potential energy is U = 196 J. To find the distance, rearrange the equation: The equation to find the distance the spring has been compressed is therefore: The spring has been compressed 0.70 m, which resulted in an elastic potential energy of U = 196 J being stored.

Explanation:

7 0

3 years ago

How many significant figures does the value 0.080 have? 1 2 3 4

lara [203]

2. The zeros in front do not matter

3 0

4 years ago