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pantera1 [17]
3 years ago
5

The components of a 15 meter per second velocity at an angle of 60 degres above the horizontal

Physics
1 answer:
Neko [114]3 years ago
6 0
X= 7.5 m/s
Y= 12.99 m/s 
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Solid barium sulfate is placed into a beaker to form a saturated solution of barium sulfate. the solution has a barium concentra
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Ksp = [Ba⁺²][SO₄⁻²]

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5 0
3 years ago
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A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the
Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge 15\times10^-6 C is placed which is the origin.

Let B be the point where the charge 9\times 10^-6 C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, \frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2},

Q_A = 15\times 10^-6 C

Q_B=9\times10^-6C

d_A = 1+x cm

d_B=x cm

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

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3 0
1 year ago
Chloe is playing the flute some distance away from a crowd. If the atmospheric temperature is 15°C, what is the speed of the sou
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I believe the answer to your question is A. 340 meters/second hope i helped
7 0
3 years ago
NEED HELP FAST ILL MARK BRAINLIEST AND RATE 5 STARS AND SAY THANK YOU
sertanlavr [38]
3 is the answer to your question 
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3 years ago
Read 2 more answers
PLEASE HELP!!! <br><br> i’ll mark brainliest if you’re correct
dedylja [7]

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

The net force acting on the block is ~

\qquad \sf  \dashrightarrow \: 10 + 5

\qquad \sf  \dashrightarrow \: 15 \:N

So, the Answer in the boxes will be ~

\boxed{ \sf15} \:  \boxed{ \sf N}

7 0
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