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2 years ago

A box is pushed 40 m by amover. The amount of work done was 2,240 j. How much force was exerted on the box?

Physics

2 answers:

Serga [27]2 years ago

6 0

56 Newtons bc w=F×D so if you divide by D on both side you get w/D=F

Furkat [3]2 years ago

5 0

Work done = force × distance
force =work done/distance
f =2240/40
=56N

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enot [183]

You can't change time to a measurement of distance

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3 years ago

One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor

Bess [88]

Answer:

a) The final velocity is 20 m/s when the large-mass object is the one moving initially.

b) The final velocity is 9.0 m/s when the small-mass object is the one moving initially.

Explanation:

The momentum of the system is calculated as the sum of the momenta of each object. Each momentum is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass.

v = velocity.

Then, the momentum of the system is the following:

m1 · v1 + m2 · v2 = (m1 + m2) · v

Where:

m1 = mass of the bigger object.

v1 = velocity of the bigger object.

m2 = mass of the smaller object.

v2 = velocity of the smaller object.

v = final velocity of the two objects after the collision.

Solving the equation for the final velocity:

(m1 · v1 + m2 · v2)/ (m1 + m2) = v

a) Let´s calculate the final velocity when the bigger object is moving:

(7.1 kg · 29 m/s + 3.2 kg · 0)/(7.1 kg + 3.2 kg) = v

b) When the smaller object is moving:

(7.1 kg · 0 m/s + 3.2 kg · 29 m/s) / (7.1 kg + 3.2 kg) = v

7 0

3 years ago

To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you

Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

β = 10 log $\frac{I}{I_o}$

let's write this expression for our case

β₁ = 10 log \frac{I_1}{I_o}

β₂ = 10 log \frac{I_2}{I_o}

β₂ -β₁ = 10 ( $log \frac{I_2}{I_o} - log \frac{I_1}{I_o}$ )

β₂ - β₁ = 10 $log \frac{I_2}{I_1}$

log \frac{I_2}{I_1} = $\frac{60 - 20}{10}$ = 3

$\frac{I_2}{I_1}$ = 10³

I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

I = P / A

P = I A

the area is of a sphere

A = 4π r²

the power of the sound does not change, so we can write it for the two points

P = I₁ A₁ = I₂ A₂

I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

I₁ r₁² = (10³ I₁ ) r₂²

r₁² = 10³ r₂²

r₂ = r₁ / √10³

we calculate

r₂ = $\frac{10.0}{\sqrt{10^3} }$

r₂ = 0.316 m

8 0

3 years ago

The spreading of waves behind an aperture is more for long wavelengths and less for short wavelengths.Less for long wavelengths

Allisa [31]

Answer:

Increase in wavelength of incident wave also increases the spread angle or spread of the interference pattern.

Explanation:

Solution:-

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- Moreover, waves diffract best when the size of the diffraction opening (or grting or groove) corresponds to the size of the wavelength. Hence, light diffracts more through small openings than through larger openings.

- The formula for diffraction shows a direct relationship between the angle of diffraction (theta) and wavelength:

d sin (θ) = m λ

Where,

λ : Wavelength , θ : The spread angle , d : Slit opening or grating

- We can see that the wavelength λ and spread angle θ are related proportionally. So if we increase the wavelength of incident wave we also increase the spread angle or spread of the interference pattern.

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3 years ago

In noisy factory environments, it's possible to use a loudspeaker to cancel persistent low-frequency machine noise at the positi

Vaselesa [24]

Answer:

7.39 m or 3.61 m

Explanation:

$\lambda$ = Wavelength