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vodomira [7]
3 years ago
6

A student swings a container of water in a vertical circle of radius 1.0 m. Calculate the minimum speed of the container so that

the water remains in the container at the top of the circle. ANS. 3.1 m/s E-4, P-12?​
Physics
1 answer:
12345 [234]3 years ago
6 0

Answer:

Explanation:

The centripetal acceleration requirement must equal gravity at the top of the circle

mg = mv²/R

  v = √Rg

  v = √(1.0(9.8))

  v = 3.1304951...

  v = 3.1 m/s

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I believe the answer is stored energy.
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Explain the phenomenon of polarization when a dielectric slab is subjected to an electric<br> field?
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A dielectric, insulating material, or an extremely bad conductor of electrical current. Due to the absence of loosely bound, or free, electrons that could wander through the material, unlike metals, dielectrics practically do not conduct current when exposed to an electric field. Electric polarization takes place instead.

<h3>What is an Electric field?</h3>
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The phenomenon of polarization when a dielectric slab is subjected to an electric field:

A dielectric, insulating material, or an extremely bad conductor of electrical current. Due to the absence of loosely bound, or free, electrons that could wander through the material, unlike metals, dielectrics practically do not conduct current when exposed to an electric field. Electric polarization takes place instead.

To learn more about the electric field, refer to:

brainly.com/question/14372859

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8 0
2 years ago
Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
netineya [11]

Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{2}} and C is a constant

6 0
3 years ago
A bucket filled with water has a mass of 54 kg and is hanging from a rope that is wound around a 0.050 m radius stationary cylin
Lubov Fominskaja [6]

Answer:

The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.

Explanation:

Given that,

Mass of bucket = 54 kg

Radius = 0.050 m

We need to calculate the magnitude of the torque the bucket produces around the center of the cylinder

Using formula of torque

\tau=F\times r

\tau=mg\times r

Where, m = mass

g = acceleration due to gravity

r = radius

Put the value into the formula

\tau=54\times9.8\times0.050

\tau=26.46\ N-m

Hence, The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.

3 0
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From New Moon to Full Moon is half of that . . . 14.765 days,
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