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3 years ago

8

An electron, starting from rest and moving with a constant acceleration, travels 2.0 cm in 5.0 ms. What is the magnitude of this

acceleration?

1 answer:

IrinaK [193]3 years ago

7 0

<span>s=2.7 centimeters = 0.027 meters
t=3.9 milliseconds = 0.0039 seconds

s=(1/2)a*t^2
so
a=(2.7*2)/(0.0039)^2
= 355,029.58 m/s^2

a=355,029.58 m/s^2 = 355.02958 km / s^2</span>

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Answer:

7 deg

Explanation:

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$k_{R}$ = spring constant for right spring = $33 Nm^{-1}$

$x_{L}$ = stretch in the left spring

$x_{R}$ = stretch in the right spring

$L$ = length of the rod = 0.75 m

$\theta$ = Angle the rod makes with the horizontal

Using equilibrium of force in vertical direction for left spring

$k_{L} x_{L} = (0.5) W\\(59) x_{L} = (0.5) (13.72)\\x_{L} = 0.116 m$

Using equilibrium of force in vertical direction for right spring

$k_{R} x_{R} = (0.5) W\\(33) x_{R} = (0.5) (13.72)\\x_{R} = 0.208 m$

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4 years ago

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3 years ago

Add a vector whose magnitude is 13 with angle 27 degrees to one whose magnitude is 11 with angle 45 degrees? Put the length firs

mafiozo [28]

Answer:

Magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

Explanation:

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$A_{x}=|A|\cos\theta_1\\\Rightarrow A_x=13\cos27^{\circ}\\\Rightarrow A_x=11.6\ \text{units}$

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$A_{y}=|A|\sin\theta_1\\\Rightarrow A_y=13\sin27^{\circ}\\\Rightarrow A_y=5.9\ \text{units}$

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$B_{x}=|B|\cos\theta_2\\\Rightarrow B_x=11\cos45^{\circ}\\\Rightarrow B_x=7.8\ \text{units}$

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Adding the magnitudes

$C_x=A_x+B_x=11.6+7.8\\\Rightarrow C_x=19.4\ \text{units}$

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$|C|=\sqrt{C_x^2+C_y^2}\\\Rightarrow |C|=\sqrt{19.4^2+13.7^2}=23.75\ \text{units}$

The direction would be

$\theta=\tan^{-1}\dfrac{C_y}{C_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{13.7}{19.4}\\\Rightarrow \theta=35.23^{\circ}$

The magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

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