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GuDViN [60]
3 years ago
13

PLEASE HELP MEE PLEASE I BEG

Physics
1 answer:
TEA [102]3 years ago
8 0

Explanation: (I think)

Plug your values into the momentum equation.

So m1= 63kg

m2 = 10 kg

V1 = 12 m/s

And then plug in your values and solve for your unknown (v2)

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A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 6.0 m, y = 10.0 m, and has vel
mihalych1998 [28]

Explanation:

Given that,

Position of the particle at t = 0,

y=(6i+10j)\ m

Velocity of the particle at t = 0

u=(1i+6j)\ m

Acceleration of the particle,

a=(5i+7j)\ m/s^2

Solution,

(a) Let v is the velocity at t = 10 s. Using the equation of kinematics as :

v=u+at

v=(1i+6j)+(5i+7j)10

v=(51i+76j)\ m/s

(b) Let y' is the position at t = 1 s. Again using second equation of kinematics as :

y'=y+ut+\dfrac{1}{2}at^2

y'=(6i+10j)+(1i+6j)1+\dfrac{1}{2}\times (5i+7j)1^2

y'=\dfrac{19}{2}i+\dfrac{39}{2}j

(c) Magnitude of y',

|y'|=\sqrt{(\dfrac{19}{2})^2+(\dfrac{39}{2})^2}

|y'| = 21.69 meters

Direction of the y',

tan\theta=\dfrac{y}{x}

tan\theta=\dfrac{39/2}{19/2}

\theta=64.02^{\circ}

Hence, this is the required solution.

4 0
3 years ago
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What Could be the control in an experiment to measure the effects of gas additives on fuel
alina1380 [7]
Well,

A control in an experiment would basically be the "normal" version of your test subjects.

In a drug testing experiment with people, the control group would be the people who don't take the drug.

In an experiment on the effects of salt on potatoes, the control group would be a potato without salt on it.

So in an experiment to measure the effects of gas additives on fuel, the control would be fuel without additives.
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Answer:

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Explanation:

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3 0
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Read 2 more answers
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