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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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4 This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive
zubka84 [21]

Answer:at 21.6 min they were separated by 12 km

Explanation:

We can consider the next diagram

B2------15km/h------->Dock

|

|

B1 at 20km/h

|

|

V

So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.

Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.

3 0
3 years ago
Which of the following means that an image is real?
DochEvi [55]
To declare an image is real you would call it “authentic”
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A bicyclist is in a 50-km race. She says she had an average velocity of 35.
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3 years ago
List at least three pieces of evidence to support the idea that sound is a wave.
Schach [20]

Answer:

1. It undergoes reflection. 2. It undergoes refraction. 3. It undergoes diffraction.

Explanation:

1. It undergoes reflection. This is because it bounces off surfaces when incident on them.

2. It undergoes refraction. This is because it changes direction when it passes from one medium to another

3. It undergoes diffraction. This is because it spreads out when it passes through doors and windows similar in dimension to the dimensions of its wavelength

8 0
3 years ago
Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of
wariber [46]

Answer:

For left = 0  N/C

For right = 0  N/C

At middle = -7.6836 * 10^{-11} \vec{i}  N/C

Explanation:

Given data :-

б =6.8 * 10^{-22} C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

E = \frac{\sigma }{2\varepsilon }

1) To the left of the plate

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

2) To the right of them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

3) Between them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(-\vec{i}) = (\frac{\sigma }{\varepsilon })(-\vec{i}) = -\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} }  \vec{i} =   -7.6836 * 10^{-11} \vec{i} N/C

5 0
3 years ago
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