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Darina [25.2K]
2 years ago
7

Protective equipment and protective measures help keep all types of workers safe on the job.

Physics
2 answers:
velikii [3]2 years ago
8 0
Answer: it is true
explanation: safety for everyone
tigry1 [53]2 years ago
6 0
Yes it is true this is a very logical question use your brain mate
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You drop a small ball, and then a second small ball. When you drop the second ball, the distance between them is 3 cm. What stat
Alja [10]

Answer:

c) The distance between the balls increases.

Explanation:

If you drop the balls at the same time, regardless of their masses they accelerate equally, since they will be in free fall.

However, if you drop one of the balls earlier, then that ball will gain velocity, whereas the second ball has zero initial velocity. At the time the second ball is dropped, both balls have the same acceleration but different initial velocities.

According to the below kinematics equation:

x = v_0t + \frac{1}{2}at^2

The initial velocity of the first ball will make the difference, and the first ball will travel a greater distance than the second ball. Hence, their distance increases.

3 0
3 years ago
Which property of a star can help us deduce its surface temperature?
horrorfan [7]

I think I has to be C

8 0
3 years ago
Read 2 more answers
If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wave
kotegsom [21]

Answer:

0.82 mm

Explanation:

The formula for calculation an n^{th} bright fringe from the central maxima is given as:

y_n=\frac{n \lambda D}{d}

so for the distance of the second-order fringe when wavelength \lambda_1 = 745-nm can be calculated as:

y_2 = \frac{n \lambda_1 D}{d}

where;

n = 2

\lambda_1 = 745-nm

D = 1.0 m

d = 0.54 mm

substituting the parameters in the above equation; we have:

y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y_2 = 0.00276 m

y_2 = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength \lambda_2 = 660-nm is as follows:

y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y^'}_2 = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

\delta y = y_2-y^{'}_2

\delta y = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

\delta y = 10 ⁻³ (2.76 - 1.94)

\delta y = 10 ⁻³ (0.82)

\delta y = 0.82 × 10 ⁻³ m

\delta y =  0.82 × 10 ⁻³ m (\frac{1.0mm}{10^{-3}m} )

\delta y = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

6 0
3 years ago
A horizontal line labeled B has an arrow labeled A strike it from right and above and then another arrow D emerges from the stri
patriot [66]

Answer:

c is the actual answer.

Explanation:

7 0
3 years ago
Read 2 more answers
A police car with its 300-Hz siren is moving toward a warehouse at 30 m/s, intending to crash through the door. The sound bounce
Lady bird [3.3K]

Answer: The frequency heard will be f = 275.675Hz

Explanation: When an object emitting sound is moving, it occurs a phenomenon called Doppler shift or Doppler effect. What happens is that the sound gets higher when the moving object comes closer the observer and becomes lower after it passes, This change is due to the quantity of waves that passes through an area in an unit of time.

The formula to calculate the Doppler effect is as follows

f = (\frac{c}{c+Vs}) · f₀

f is the observed frequency;

c is the speed of sound;

Vs is velocity of the source;

f₀ is the emitted frequency of source;

Substituting and calculating,

f = \frac{340}{340+30} · 300

f = 275.675 Hz

Thus, the frequency heard by the police officer is 275.675Hz.

8 0
3 years ago
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