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Lady_Fox [76]
3 years ago
10

Please help me with this question, especially the angle part. I have solved the initial velocity already. I don't know what to d

o. Thank you.

Physics
1 answer:
Helga [31]3 years ago
5 0
I think you almost got it.

At the top, the velocity only has horizontal component, so v=12 m/s is v_x, which is v*cos(theta), because v_x is constant, so the same when it was launched or now.

With the value of the initial speed (28 m/s, which is the total speed), you can set

v_x = v * cos( theta ) ---> 12 = 28*cos(theta) --> cos(theta)=12/28=3/7

or theta = 64.62 deg, it is D. Think about it. I hope you see it.
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We want a ratio of 3000000 to 1, or 3 million to 1. 

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a one bucket with full of sand because of it's mass which plays role in momentum

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As in the video, we apply a charge + to the half-shell that carries the electroscope. This time, we also apply a charge – to the
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Four uniform spheres, with masses mA = 200 kg, mB = 250 kg, mC = 1700 kg, and mD = 100 kg, have (x, y) coordinates of (0, 50 cm)
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Answer:

F_2=2.43\times10^{-19}\ N

\theta=21.496^{\circ} in the anticlockwise direction from the positive x- axis.

Explanation:

Given are the masses of various spheres with their names in subscript:

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  • m_B=250\ kg
  • m_C=1700\ kg
  • m_D=100\ kg

Now their respective coordinate positions (in cm) are as given below:

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  • P_B=(0,0)
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<u>Now force on B due to A:</u>

F_{BA}=G\times \frac{200\times 250}{50^2}

F_{BA}=20G\ N

<u>Now force on B due to C:</u>

F_{BC}=G\times \frac{1700\times 250}{80^2}

F_{BC}=66.406G\ N

<u>Now force on B due to D:</u>

F_{BD}=G\times \frac{100\times 250}{40^2}

F_{BD}=15.625G\ N

<em>We observe that the forces due to masses  C&D act opposite in direction.</em>

<u>So, the net force in the x-direction:</u>

F_x=F_{BC}-F_{BD}

F_x=66.406G-15.625G

F_x=50.781G\ N in the positive x-direction

<em>We have only one force in y-direction due to mass A.</em>

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F_y=20G\ N in the positive y-direction.

<u>Now the net force:</u>

F_2=\sqrt{F_y^2+F_x^2}

F_2=\sqrt{(20G)^2+(50.781G)^2}

F_2=54.5776G^2\ N

F_2=2.43\times10^{-19}\ N

<u>Now the direction of this force with respect to x-axis:</u>

tan\ \theta=\frac{F_y}{F_x}

tan\ \theta=\frac{20G}{50.781G}

\theta=21.496^{\circ} in the anticlockwise direction from the positive x- axis.

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