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nydimaria [60]
2 years ago
8

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
5 0
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

<u>v = 20 m/s</u>

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

<u>v = 9.0 m/s</u>

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:-

- The diffraction occurs when light bends in the same medium. The bending is the result of light waves "squeezing" through small openings or "curving" around sharp edges.

- 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.

5 0
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

f = Frequency = 90 Hz

v = Speed of sound = 340 m/s

Path difference of the two waves is given by

s_1-s_2=\frac{\lambda}{2}

Velocity of wave

v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{340}{90}\\\Rightarrow \lambda=3.78\ m

s_1=s_2\pm\frac{\lambda}{2}\\\Rightarrow s_1=5.5\pm \frac{3.78}{2}\\\Rightarrow s_1=7.39\ m, 3.61\ m

So, the location from the worker is 7.39 m or 3.61 m

7 0
3 years ago
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