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swat32
2 years ago
7

A 20 kg box rests on the ground. Round all answers to the hundredths, if necessary. What is the weight of the box?​

Physics
2 answers:
BabaBlast [244]2 years ago
6 0

Answer:

Weight=200

Explanation:

Given,

\sf\longmapsto \: mass = 20 \: kilograms

We know that–

\sf\longmapsto \: acceleration \:  \: because \: of \: gravity = 10m/s^2

\sf\longmapsto \: Weight= \:  is  \: F = m•g

Let's start solving!

Weight = 20×10

Weight=200

cricket20 [7]2 years ago
5 0
  • Mass=m=20kg
  • Acceleration due to gravity=10m/s^2

Weight=Applied force

\\ \sf\longmapsto F=mg

\\ \sf\longmapsto F=20(10)

\\ \sf\longmapsto F=200N

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Correct Answer:

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Answer: 3.1375

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Two waves meet and overlap. The first wave has an amplitude of 4.6 centimeters and the second wave has an amplitude of 2 centime
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Explanation:

The amplitude of resultant wave as the result of  overlap of two waves depends upon the phase difference between the two.  If the waves meet crest to trough , the phase difference is 180 degree or they are in opposite phase . Hence they will destroy each other . The amplitude of resultant wave can be obtained by subtracting the amplitudes of two waves. They will interfere destructively.

Amplitude of resultant gives waves = 4.6 - 2 = 2.6 cm.

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3 years ago
A rotating light is located 13 feet from a wall. The light completes one rotation every 3 seconds. Find the rate at which the li
saveliy_v [14]

Answer:

29.2 ft/s

Explanation:

The distance of the light's projection on the wall

y = 13 tan θ

where θ is the light's angle from perpendicular to the wall.

The light completes one rotation every 3 seconds, that is, 2π in 3 seconds,

Angular speed = w = (2π/3)

w = (θ/t)

θ = wt = (2πt/3)

(dθ/dt) = (2π/3)

y = 13 tan θ

(dy/dt) = 13 sec² θ (dθ/dt)

(dy/dt) = 13 sec² θ (2π/3)

(dy/dt) = (26π/3) sec² θ

when θ = 15°

(dy/dt) = (26π/3) sec² (15°)

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5 0
3 years ago
A solid concrete block weighs 150. N and is resting on the ground. Its dimensions are 0.400 m ✕ 0.200 m ✕ 0.100 m. A number of i
N76 [4]

Answer:

27 blocks

Explanation:

First, the expression to use here is the following:

P = F/A

Where:

P: pressure

F: Force exerted

A: Area of the block.

Now , we need to know the number of blocks needed to exert a pressure that equals at least 2 atm. To know this, we should rewrite the equation. We know that certain number of blocks, with the same weight and dimensions are putting one after one over the first block, so we can say that:

P = W/A

P = n * W1 / A

n would be the number of blocks, and W1 the weight of the block.We have all the data, and we need to calculate the area of the block which is:

A = 0.2 * 0.1 = 0.02 m²

Solving now for n:

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The pressure has to be expressed in N/m²

P = 2 atm * 1.01x10^5 N/m² atm = 2.02x10^5 N/m²

Finally, replacing all data we have:

n = 2.02x10^5 * 0.02 / 150

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We can round this result to 27. So the minimum number of blocks is 27.

5 0
3 years ago
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