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frutty [35]
3 years ago
8

The vertical component of a velocity vector points up and has a magnitude of 20, ms'. The velocity vector Itself makes an angle

of 25' with the horizontal. Find the magnitude of the velocity vector.
Physics
1 answer:
kirill115 [55]3 years ago
3 0

Answer:

Kindly check explanation

Explanation:

Given the following :

Vertical component has Magnitude = 20m/s

Angle made with the horizontal (θ) = 25°

Vy = Vsinθ = 20sin25

Vy = 20 × 0.4226182 = 8.4523652 m/s

Horizontal component :

Vx = Vcosθ = 20 × cos25

Vx = 20 × 0.9063077 = 18.126155 m/s

Hence, magnitude :

V = sqrt(Vx² + Vy²)

V = sqrt(8.4523652² + 18.126155²)

V = sqrt (399.99999)

V = 20m/s

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SSSSS [86.1K]
You said that she's losing 1.9 m/s of her speed every second.

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A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.
Tcecarenko [31]

Answer:

a) 1.89 m/s  b) 172.1 N

Explanation:

a)

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  • This work, is just 4,475 J.
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        \Delta K = \frac{1}{2} * m*v^{2} = 4,475 J

  • where m= mass of the truck = 2.5*10³ kg.
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        v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} }  = 1.89 m/s

b)

  • The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

        W = F*d

  • We can solve for F, as follows:

        F = \frac{W}{d} = \frac{4,475 J}{26.0m} =  172.1 N

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3 years ago
HELP...<br> 3.00 amu = _____ Mev.<br><br> 3.22 x 10-3<br> 2.79 x 103<br> 3.10 x 102
DanielleElmas [232]
The correct answer is:
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1 amu corresponds to the mass of the proton, which is:
m_p = 1.66 \cdot 10^{-27} kg
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E=mc^2 = (1.66 \cdot 10^{-27} kg)(3\cdot 10^8 m/s)^2 = 1.49 \cdot 10^{-10} J
Now we can convert it into electronvolts:
E= \frac{1.49 \cdot 10^{-10}kg}{1.6 \cdot 10^{-19} J/eV} =9.34 \cdot 10^9 eV = 934 MeV

So, 1 amu = 934 MeV. Therefore, 3 amu corresponds to 3 times this value:
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Answer:

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The formula for the uniform speed of an object is given as follows:

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t = time required

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