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andrew11 [14]
2 years ago
15

Help me solve this please

Physics
1 answer:
Leto [7]2 years ago
6 0
Since it's a projectile being launched the only force acting upon it is gravity, since the object is in free fall once it's launched

so to calculate time you'd utilize the general formula of
xf = xi + vxi(t) + \frac{1}{2} (a) {t}^{2}
and then solve using time and make it into the y axis, so change the x's to y's, which will change a to g.
since Vyi is always usually 0, you can drop that out of this equation so the formula to find time would be
t =  \sqrt{ \frac{2(y)}{g} }
So you'll plug in and it'll be
t =  \sqrt{ \frac{2( - 49m)}{ - 9.81 \frac{m}{s {}^{2} } } }
to find the maximum height you'll have to do some trigonometry to solve it.
To make it easier draw a triangle
put the 60° mark as shown in the picture.
Then you'll need to find the hypotenuse or horizontal to find the vertical
So the hypotenuse would be the 113m/s
so then you'll use
\ \sin( 60) = ( \frac{o}{h})

plug in the numbers
113( \ \sin (60) ) = o
now that you have the vertical

use the formula
{vf}^{2}  =  {vi}^{2}  + 2gd
solve for d which will give you the hypotenuse
d = { (-vertical \frac{m}{s} )}^{2}  \div 2( - 9.81 \frac{m}{ {s}^{2} } )
The "vertical" is what you found in the previous step.
Vf^2 is equal to 0 so you can just drop that number out since it's 0
then once you have that then youre not done yet
since you're on a cliff of 49 m you'll have to add 49m to the previous answer that you found d to find the maximum height.

I hope this helps!

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What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0∘ downward from the horizo
Andreyy89

Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here

A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.

What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees  downward from the horizontal?

Answer:

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Explanation:

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Charge q=28 nC

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Angle α=45°

To find

Work done by electric force

Solution

W_{work}=F_{force}*D_{distance}Cos\alpha  \\where\\F_{force}=q_{charge}*E_{Electric-Field}\\So\\W_{work}=qE*D*Cos\alpha \\W_{work}=(28*10^{-9}C )(5.00*10^{4}V/m )(2.70m)Cos(45)\\W_{work}=2.67*10^{-3}J

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ivann1987 [24]
This question is incomplete, but I can do it for you, considering the equation to be *In its most famous form*:
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