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Ket [755]
3 years ago
7

You are camping in the breathtaking mountains if Colorado. You spy an unopened diet soda can floating motionless below the surfa

ce of a lake. What is the direction and amount of force the water exerts on it?A. ZeroB. Down, equal to the can's weightC. Up, equal to the can's weightD. Not enough information is given
Physics
1 answer:
liubo4ka [24]3 years ago
5 0

Answer:

C

Explanation:

When the can is floating below the surface of the lake the principle of buoyancy applies. The water will be exerting force in the upward direction equals to the weight if the can.

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Help me solve this please
Leto [7]
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!

6 0
3 years ago
In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst
Archy [21]

Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

E = \frac{1}{2}mv^{2}

So, v = \sqrt{\frac{2E}{m}}   .... (1)

Now,

r = \frac{mv}{Bq}

Substitute the value of v from equation (1), we get

r = \frac{\sqrt{2mE}}{Bq}

Let the radius of the alpha particle is r2.

For proton

So, r_{1} = \frac{\sqrt{2m_{1}E}}{Bq_{1}}    ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, r_{2} = \frac{\sqrt{2m_{2}E}}{Bq_{2}}    ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

\frac{r_{1}}{r_{2}}=\frac{q_{2}}{q_{1}}\times \sqrt{\frac{m_{1}}{m_{2}}}

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

\frac{r_{1}}{r_{2}}=\frac{2q}}{q}}\times \sqrt{\frac{m}}{4m}}=1

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

8 0
3 years ago
Read 2 more answers
Given the equation E = P/N , solve for P
Mars2501 [29]

Answer:

E=P/N

multiply both sides by N

P=EN

6 0
3 years ago
What is the wave speed of a 6 m wave with a frequency of 20 Hz?
andrew-mc [135]

Answer: Wave speed= frequency x wavelength

=20 x 3

=60 m/s

Explanation:

8 0
3 years ago
Bob runs 1800 seconds at an average speed of 1.5 m/sec. How far did he go? 25 Points!!!!
german

Distance= Time×Speed

= 1800×1.5

= 2700 m

I am not sure it's right. the question itself is confusing.

4 0
2 years ago
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