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mario62 [17]
3 years ago
11

What must you know to find the amount of work done on an object

Physics
2 answers:
Anton [14]3 years ago
8 0
Work(J)=force(N) x distance(m)
Cerrena [4.2K]3 years ago
5 0

Answer:

The work is calculated by multiplying the force by the amount of movement of an object (W = F * d). A force of 10 newtons, that moves an object 3 meters, does 30 n-m of work. A newton-meter is the same thing as a joule, so the units for work are the same as those for energy – joules.

Explanation:

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Explanation:

Most waves appear complex because they result from two or more simple waves that combine as they come together at the same place at the same time—a phenomenon called superposition. Waves superimpose by adding their disturbances; each disturbance corresponds to a force, and all the forces add

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A jet transport has a weight of 2.25 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel
Rudik [331]

Answer:

Explanation:

Given that,

Weight of jet

W = 2.25 × 10^6 N

It is at rest on the run way.

Two rear wheels are 16m behind the front wheel

Center of gravity of plane 10.6m behind the front wheel

A. Normal force entered on the ground by front wheel.

Taking moment about the the about the real wheel.

Check attachment for better understanding

So,

Clock wise moment = anti-clockwise moment

W × 5.4 = N × 16

2.25 × 10^6 × 5.4 = 16•N

N = 2.25 × 10^6 × 5.4 / 16

N = 7.594 × 10^5 N

B. Normal force on each of the rear two wheels.

Using the second principle of equilibrium body.

Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces

ΣFy = 0

Nr + Nr + N — W = 0

2•Nr = W—N

2•Nr = 2.25 × 10^6 — 7.594 × 10^5

2•Nr = 1.491 × 10^6

Nr = 1.491 × 10^6 / 2

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3 years ago
Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted out
blondinia [14]

Answer:

\theta = 20.98 degree

Explanation:

As we know that the speed of the sound is given as

v = 332 + 0.6 t

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a sin\theta = N\lambda

a sin\theta = N(\frac{v_1}{f})

now when temperature is changed to 313 K we have

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sin\theta = 0.358

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