Pure sodium metal placed in water will spark and ignite, as well as form bubbles and gas. What are the signs that this is a chem

Answer:

2.5 m

Explanation:

Billboard worker's weight = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension present in rope = 550 N

The total torques will be

The worker is positioned at 2.5 m

Answer:

H = 109.14 cm

Explanation:

Given,                                                            

Assume that the total energy equals 1 unit.                                

Energy remaining after the first collision = 0.78 x 1 unit

Balance after the first impact = 0.78 units

Remaining energy after the second impact = 0.78 ^2 units

Balance after the second impact = 0.6084 units

Remaining energy after the third impact = 0.78 ^3 units

Balance after the third impact = 0.475 units

The height reached after the third collision is equivalent to the remaining energy.

Let H denote the height achieved after three bounces.

0.475 (m g h) = m g H                  

H = 0.475 x h                                    

H = 0.475 x 2.3 m                          

H = 1.0914 m                      

H = 109.14 cm                      

Response:

Clarification:

We need an expression that shows how much water has been drained from the tub. This is represented by v, which indicates how many gallons have flowed out since the plug was taken out. Each gallon removed equates to 8.345 pounds of water, so the weight of the drained water Q in pounds as a function of v can be expressed as:

Where v signifies the number of gallons emptied from the tub.

Have a great day! Let me know if there's anything else I can assist with.

Answer:3.87*10^-4

Explanation:

To determine the mass reduction, delta mass Xe, of the xenon nucleus due to its decay, we first use the provided wavelength of the gamma radiation to calculate its frequency via c = freq*wavelength.

From C=f*lambda we set up: 3*10^8=f*3.44*10^-12.

Solving gives frequency F=0.87*10^20 Hz.

Next, we calculate the emitted energy using the equation E=hf, which translates to E=f*Planck's constant.

Thus, E=0.87*10^20*6.62*10^-34, resulting in E=575.94*10^(-16).

This energy is then converted from joules to MeV.

Utilizing the formula E=mc^2, with c^2 = 931.5 MeV/u, enables us to find the reduction in mass, yielding

3.87*10^-4 u.

Answer:

The area that remains is 4.201 m²

Explanation:

Provided that

AB=D=  4 m  (R=2 m)

The area of the half-circle AB is

A=2π

The area of the smaller half-circle is

a=5π/6 + 2√3/4  m²

a=5π/6 + √3/2  m²

                                     = 2π - (5π/6 + √3/2) m²

The remaining area is expressed as  2π - (5π/6 + √3/2) m²