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Dominik [7]
3 years ago
11

In the figure, the magnitude of vector a⃗ is 18.0 units, and the magnitude of vector b⃗ is 12.0 units. what vector c⃗ must be ad

ded to the vectors a⃗ and b⃗ so that the resultant of these three vectors points in the − x direction and has a magnitude of 7.50 units?
Physics
1 answer:
a_sh-v [17]3 years ago
3 0
0 degrees west of the coast in magnitude
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Useable energy, electricity

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1. A 46000N helicopter feels a net force of 9200N. What litt force is exerted by the air on the propellers?
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I need help with two more Physics problems​
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Answer:

c. They hit at the same time

b. BGS

Explanation:

A marble dropped (initial vertical velocity is 0) will land at the same time as a marble launched horizontally (initial vertical velocity is 0) from the same height.

Boat S has a net speed of 5 m/s (10 − 5).

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Boat G has a net speed of ≈11.2 m/s (√(10² + 5²)).

8 0
3 years ago
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A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.16
ycow [4]
We assign the variables: T as tension  and x the angle of the string
 The  <span>centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = </span><span>T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx. 
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<span>(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. </span>
<span>(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. </span>
<span>(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. </span>
<span>(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.

</span><span>This minimum v is obtained when T=0 </span><span>and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.</span>
4 0
3 years ago
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The equation r (t )=(2t + 4)⋅i + (√ 7 )t⋅ j + 3t ²⋅k the position of a particle in space at time t. Find the angle between the v
velikii [3]

Answer:

\theta = n\pi/2, {\rm where~n~is~an~integer.}

Explanation:

We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = (2)\^i + (\sqrt{7})\^j + (6t)\^k

Similarly,

\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = (6)\^k

Now, the angle between velocity and acceleration vectors can be found.

The angle between any two vectors can be found by scalar product of them:

\vec{A}.\vec{B} = |\vec{A}|.|\vec{B}|.\cos(\theta)

So,

\vec{v}(t).\vec{a}(t) = |\vec{v}(t)|.|\vec{a}(t)|.\cos(\theta)\\36t = \sqrt{4 + 7 + 36t^2}.6.\cos(\theta)

At time t = 0, this equation becomes

0 = 6\sqrt{11}\cos(\theta)\\\cos(\theta) = 0\\\theta = n\pi/2, {\rm where~n~is~an~integer.}

7 0
3 years ago
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