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jonny [76]
3 years ago
15

What is the minimum speed the rock must have at the top of the circle if it is to always stay in contact with the bottom of the

bucket?
Physics
1 answer:
Naddika [18.5K]3 years ago
7 0
 <span>First lets determine the equation. Well at the top of the circle both the normal force and the weight are in the same direction. So we have Fnet=N+mg. Since this is a circular path the Fnet is also = to (mv^2)/r. 

We convert the situation where the rock is no longer in contact with the bottom to terms relevant to the equation. So, what is a requirement for normal force? The object must be in contact with the surface, meaning it can't be in free fall. Realizing this means that the instant when the object does not touch the bucket is where the normal force = 0. 

Now we have N+mg=(mv^2)/r where N=0 is the case we are interested in. This leaves 0+mg=(mv^2)/r 
Solve for v: 
v=(gr)^(1/2) or v=3.28m/s 
</span>
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a swimmer can swim in still water at a speed of 9.50 m/s. he intends to swim directly across the river that has a downstream cur
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Refer to the diagram shown below.

Still-water speed  = 9.5 m/s
River speed = 3.75 m/s down stream.

The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.

The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s

The downstream angle is
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Answer:  10.2 m/s at 21.5° downstream.

7 0
3 years ago
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X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m

b) The change in photon energy is given by:

\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\

E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV

5 0
3 years ago
Arrange these metric prefixes from smallest to largest.<br> Deca <br> Deci<br> Micro<br> milli
Ahat [919]

Answer:

micro

milli

deci  

deca

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Volume is a measurement of the height of the wave known as what? <br><br><br> this is science
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I’m gonna have to say “Ocean waves” as the answer
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klasskru [66]

Answer:

67.44 V

Explanation:

Number of turns N =123

Radius = 2.41 cm =0.0241 m

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Time interval dt = 0.151 sec

We know that the induced emf e=-NA\frac{dB}{dt}

Area A=\pi r^2=3.14\times 0.0241^2=1.8237\times 10^{-3}m^2

Putting all these values in emf equation e=-123\times 1.8237\times 10^{-3}\times \frac{45.4}{0.151}=-67.44\ V here negative sign indicates that it opposes the cause due to which it is produced

8 0
3 years ago
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