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irakobra [83]
4 years ago
14

How can you produce more power than an excavator?

Physics
2 answers:
Tanya [424]4 years ago
8 0
Just do energy spent divided by time to get your answer. With this we can say a human might be able to!
sukhopar [10]4 years ago
8 0

Answer:

If the work done by us is greater than the excavator in the given time then we can produce more power than excavator.

Explanation:

Power is defined as work done per unit time. Let’s take an example:

Case I:

If we do 300J work in 2 seconds and excavator 500J work in the same time then:

Power produced by us = 300/2 = 150 watt

Power produced by excavator = 500/2 = 250 watt

In this case power produced by excavator is more than us.

Case II:

If we do 500J work in 2 seconds and excavator 300J work in the same time then:

Power produced by us = 500/2 = 250 watt

Power produced by excavator = 300/2 = 150 watt

In this case power produced by us is more than excavator.

Hence, If the work done by us is greater than the excavator in the given time then we can produce more power than excavator. But it is very difficult to believe in reality.

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Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m
Natali [406]
<h2>Answer:</h2>

D. (1m, 0.5m)

<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M       ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

7 0
3 years ago
An Olympic high jumper, with a mass of 82 kg, has a
Digiron [165]

Answer:

I don't really know

Explanation:

I really wanted to help you, but then I realized i didnt know how to

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3 years ago
A hammer exerts 49.8 N of force on the head (r=0.00510 m) of a nail. How much pressure does it exert on the nail?
Kisachek [45]

Answer:

609547.12 Pa ≈ 6.10×10^5 Pa

Explanation:

Step 1:

Data obtained from the question. This include the following:

Force (F) = 49.8 N

Radius (r) = 0.00510 m

Pressure (P) =..?

Step 2:

Determination of the area of the head of the nail.

The head of a nail is circular in nature. Therefore, the area is given by:

Area (A) = πr²

With the above formula we can obtain the area as follow:

Radius (r) = 0.00510 m

Area (A) =?

A = πr²

A = π x (0.00510)²

A = 8.17×10^-5 m²

Therefore the area of the head of the nail is 8.17×10^-5 m²

Step 3:

Determination of the pressure exerted by the hammer.

This is illustrated below:

Force (F) = 49.8 N

Area (A) = 8.17×10^-5 m²

Pressure (P) =..?

Pressure (P) = Force (F) /Area (A)

P = F/A

P = 49.8/8.17×10^-5

P = 609547.12 N/m²

Now, we shall convert 609547.12 N/m² to Pa.

1 N/m² = 1 Pa

Therefore, 609547.12 N/m² = 609547.12 Pa.

Therefore, the pressure exerted by the hammer on the nail is 609547.12 Pa or 6.10×10^5 Pa

8 0
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Answer:

1. It undergoes reflection. 2. It undergoes refraction. 3. It undergoes diffraction.

Explanation:

1. It undergoes reflection. This is because it bounces off surfaces when incident on them.

2. It undergoes refraction. This is because it changes direction when it passes from one medium to another

3. It undergoes diffraction. This is because it spreads out when it passes through doors and windows similar in dimension to the dimensions of its wavelength

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Eric is writing about the cell wall.
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