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2 years ago

Uranium-235 undergoes fission, forming krypton-92, barium-141, and 3

Physics

2 answers:

Ahat [919]2 years ago

8 0

Answer:

Explanation:

Some of the mass

GarryVolchara [31]2 years ago

6 0

Answer:

D. Mass was transformed into energy during the process.

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A 24 kg crate is moving at a constant speed because it is being pushed with a force of 53n. What would be the coefficient of kin

____ [38]

Answer:

The coefficient of kinetic friction between the crate and the floor can be calculated using the formula μ = Ff / N, where Ff is the frictional force, N is the normal force, and μ is the coefficient of kinetic friction.

In this case, the normal force is equal to the weight of the crate, which is 24 kg * 9.8 m/s2 = 235.2 N. The frictional force can be calculated using the formula Ff = μ * N, where μ is the coefficient of kinetic friction and N is the normal force.

If we substitute the values for N and Ff into the formula for the coefficient of kinetic friction, we get:μ = 53 N / 235.2 N = 0.225

Therefore, the coefficient of kinetic friction between the crate and the floor is 0.225.

6 0

9 months ago

The burner on a stove is 325° f. given that the burner emits electromagnetic radiation as a blackbody, what is the maximum wavel

lesantik [10]

To solve this, we use the Wien's Displacement Law as shown in the attached picture. First, convert the temperature to Kelvin.

C to F:
C = (F - 32)*5/9
C = (325 - 32)*5/9 = 162.78 °C

C to K:
K = C + 273
K = 162.78 + 273 = 435.78 K

λmax = 2898/435.78 = 6.65 μm

5 0

2 years ago

A railroad car of mass 2.50 3 104 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railro

ipn [44]

Answer:2.5 m/s

37.5KJ

Explanation:

Let $u_1, u_2 , v_f$ be the initial velocity of rail road car ,coupled cars & Final velocity of system respectively.

$m=2.50\times 10^{4}$

Conserving momentum

$mu_1+3mu_2=4mv_f$

$4m+6m=4mv_f$

$v_f=2.5 m/s$

Therefore Final velocity of system is 2.5m/s

(b)Mechanical Energy lost =Initial Kinetic Energy -Final Kinetic Energy

$Initial Kinetic Energy=\frac{1}{2}m\left ( 4^2\right )+\frac{1}{2}m\left ( 2^2\right )=14m J$

$Final Kinetic Energy=\frac{1}{2}4m\left ( 2.5^2\right )=12.5m J$

$Mechanical Energy lost=14m-12.5m=3.75\times 10^4=37.5 KJ$

4 0

3 years ago

The mass of a string is 5.9 × 10-3 kg, and it is stretched so that the tension in it is 200 n. a transverse wave traveling on th

bagirrra123 [75]

The velocity of the wave on the string is given by

$v=\sqrt{\frac{T}{\frac{m}{L}}} \\ v=\sqrt{\frac{TL}{m}}$

Solving the above equation,

$v^2=\frac{TL}{m} \\ L=\frac{v^2m}{T}$

The frequency of the wave $f=300$ and wave length is $0.76$

The velocity is $v=(300)(0.76)=228$

Substituting numerical values,

$L=\frac{228^2(0.0059)}{200}\\ T=1.534$

The length of the string is $1.534 m$

4 0

3 years ago

What is the importance of time ?

lesantik [10]

I think that the importance of time is to allow there to be order. People need to be somewhere at a certain time and not just whenever they want. It allows people to get things done more efficiently.

8 0

2 years ago