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3 years ago

How can you produce more power than an excavator?

Physics

2 answers:

alina1380 [7]3 years ago

5 0

Just do energy spent divided by time to get your answer. With this we can say a human might be able to!

Andre45 [30]3 years ago

4 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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Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago

a racing car undergoes a uniform acceleration of 4.00m/s2. if the net force causing the acceleration is 3.00 times 10^3 N, what

yKpoI14uk [10]

Answer: 750Kg

Explanation:

Recall that force is the product of the mass M, of an object moving at a uniform acceleration.

i.e Force = Mass x Acceleration

In this case, Mass = ?

Force = 3.00 x 10^3 N = (3.00 x 1000N)

= 3000N

Uniform acceleration = 4.00m/s^2

Force = Mass x Acceleration

3000N = Mass x 4.00m/s^2

Mass = (3000N/4.00m/s^2)

Mass = 750Kg (The SI unit of mass is kilograms)

Thus, the mass of the car is 750Kg

4 0

3 years ago

The graph shows a heating curve for water. Between which points on the graph would condensation occur?

Radda [10]

I would have to see the graph.. but by looking at one one online, they are between points D and E.

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3 years ago

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What’s the difference between gravitational force & gravitational field strength?

Zepler [3.9K]

Answer:

Gravitational field strength is the force experienced by a unit mass. Gravitational force is the amount of force acting on a body. It is the product of field strength times the mass under consideration. Gravitational pull is just a more colloquial name for gravitational force.

Explanation:

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3 years ago

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a 3.2 kg durian fruit is pushed across the table.If the acceleration of the durian is 3.1 m/s/s to the right,what is the net for

Gala2k [10]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 3.2 × 3.1

We have the final answer as

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3 years ago

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