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Gwar [14]
3 years ago
7

You must record your daily workouts on a

Physics
2 answers:
Doss [256]3 years ago
6 0

Answer:

Fitness Log (apex)

Explanation:

larisa [96]3 years ago
5 0

I do a yoga mat that could be different to your answer

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If you dont wear protection while shooting a rifle it will damge your hear
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A television camera lens has a 17-cm focal length and a lens diameter of 6.0 cm. what is its number?
IRINA_888 [86]

Answer:

= 2.83

Explanation:

F number (N) is given by the formula;

  F- number = f/D

where f = focal length of lens and D = diameter of the aperture  

Therefore;

F number = 17 cm/6 cm

                <u> = 2.83</u>

3 0
3 years ago
Please help I wasn’t here for this lesson
katovenus [111]

Answer:

27 cm squared

Explanation:

have a good rest of ur day

8 0
2 years ago
A 570 kg elevator accelerates downwards at 1.5 m/s2 for the first 13 m of its motion.
jeka94
  • Mass of the elevator (m) = 570 Kg
  • Acceleration = 1.5 m/s^2
  • Distance (s) = 13 m
  • Let the force be F.
  • We know, F = ma,
  • Therefore, F = (570 × 1.5) N = 855 N
  • Angle between distance and force (θ) = 0°
  • We know, work done = F s Cos θ
  • Therefore, work done by the cable during this part
  • = (855 × 13 × Cos 0°) J
  • = (855 × 13 × 1) J
  • = 11115 J

<u>Answer</u><u>:</u>

<u>1</u><u>1</u><u>1</u><u>1</u><u>5</u><u> </u><u>J</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0
3 years ago
A light bulb emits light that travels uniformly in all directions. Detailed measurements show that at a distance of 56 m from th
vladimir1956 [14]

Complete question:

A light bulb emits light that travels uniformly in all directions. Detailed measurements show that at a distance of 56 m from the bulb, the amplitude of the electric field is 3.78 V/m. What is the average intensity of the light?

Answer:

The average intensity of the light is 0.02 W/m²

Explanation:

Given;

Amplitude of the electric field, E₀ = 3.78 V/m

The average intensity of the light is calculated as follows;

I_{avg} = \frac{c\epsilon_0 E_0^2}{2}

where;

I_{avg} is the average intensity of the light

c is speed of light = 3 x 10⁸ m/s

I_{avg} = \frac{(3\times 10^8)(8.85 \times 10^{-12}) (3.78)^2}{2} \\\\I_{avg} = 0.01897 \ W/m^2\\\\I_{avg} = 0.02 \ W/m^2

Therefore, the average intensity of the light is 0.02 W/m²

4 0
3 years ago
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