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Molodets [167]
2 years ago
8

What is the final speed of an object that starts from rest and accelerates uniformly at 3.0 meters per second2 over a distance o

f 6.0 meters? (1) 8.0 m/s (2) 6 m/s (3) 16 m/s (4) 64 m/s
Physics
1 answer:
jasenka [17]2 years ago
7 0

Answer:

16 m/s sir

Explanation:

because i am in collage

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If cells are placed in a hypertonic solution containing a solute to which the membrane is impermeable what could happena)cells w
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cell swell and burst(b)

Explanation:

This process is called hemolysis. it also occurs in a red blood cell.

6 0
3 years ago
Which of the following accurately describes the behavior of these two mechanical waves when they intersect?
Lerok [7]
D because I learned this 2 years ago
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3 years ago
Why must mine tailings be stored and disposed of carefully?
gavmur [86]
After thorough researching, the mine tailings must be stored and disposed of carefully because they have lots of chemical and various toxic materials. They can also leach to the aquifers. The correct answer to the following given statement above is they have chemicals which are dangerous.
7 0
3 years ago
A 0.500-nm x-ray photon is deected through 134 in a Compton scattering event. At what angle (with respect to the incident beam)
natita [175]

Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

Compton Scattering is the process of scattering of X-rays by a charge particle like electron.

The angle of the recoiling electron with respect to the incident beam is determine by the relation :

\cot\phi = (1+\frac{hf}{m_{e}c^{2}  })\tan\frac{\theta }{2}      ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, m_{e} is mass of electron, c is speed of light and f is the frequency of the x-ray photon.

We know that, f = c/λ      ......(2)

Here λ is wavelength of x-ray photon.

Rearrange equation (1) with the help of equation (1) in terms of  λ .

\cot\phi = (1+\frac{h}{m_{e}c\lambda  })\tan\frac{\theta }{2}

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for m_{e}, 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ  and 134° for θ in the above equation.

\cot\phi = (1+\frac{6.6\times10^{-34} }{9.1\times10^{-31}\times3\times10^{8}\times0.5\times10^{-9}  })\tan\frac{134 }{2}

\cot\phi=2.37

\phi = 22.90°

8 0
3 years ago
An ice chest at a beach party contains 12 cans of soda at 4.05 °C. Each can of soda has a mass of 0.35 kg and a specific heat ca
BARSIC [14]
Below is the solution:

Heat soda=heat melon 
<span>m1*cp1*(t-t1)=m2*cp2*(t2-t); cp2=cpwater </span>
<span>12*0.35*3800*(t-5)=6.5*4200*(27-t) </span>
<span>15960(t-5)=27300(27-t) </span>
<span>15960t-136500=737100-27300t </span>
<span>43260t=873600 </span>
<span>t=873600/43260 </span>
<span>t=20.19 deg celcius</span>
4 0
3 years ago
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