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Pavlova-9 [17]
3 years ago
14

Include units with all of your answers. A 1.84 kg bucket full of water is attached to a 0.76 m long string.

Physics
1 answer:
aleksandrvk [35]3 years ago
7 0
<span>c. What is the magnitude of the tension in the string at the bottom of the circle if you are swinging it at 3.37 m/s? </span>
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In most circumstances, the normal force acting on an objectand
Lynna [10]

Answer:

Normal Force is usually perpendicular to the movement and static friction usually means that there is no movement.

Explanation:

The work donde by any force on an object is equal to the displacement of the object multiplied by the component of the force that is in the direction of the displacement.

Normal force is usually perpendicular to the movement, so there is no component in the direction of the displacement. This is why it is zero in most circumstances.

<em>Static</em> friction on the other hand, usually means that there is no movement at all (it's static). It means that there is no displacement between the object and ground (in most cases). If there is no displacement, there is no work.

4 0
3 years ago
Can someone please help me match these up
yaroslaw [1]

Answer:

you need to give the answers choices

Explanation:

4 0
3 years ago
Which of the following are basic solutions? Check all that apply. A. Laundry detergent B. Vinegar C. Drain cleaner D. Antacids
ratelena [41]
A laundry detergent
B vinegar
4 0
3 years ago
When electrical energy is being used by an electric light, what really happens to the energy?
vitfil [10]

Energy is not created and not  destroyed it will only change form


So heres your answer ; It is given off as other forms of energy/light and heat !!


=answer 2nd one

7 0
3 years ago
Read 2 more answers
A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the
Shalnov [3]

Answer:

x ’= 1,735 m,  measured from the far left

Explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

             

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

           x_{cm} = 1.2 -1

          x_ {cm} = 0.2 m

          Σ τ = 0

          w₁ 1.2 + mg 0.2 - W₂ x = 0

          x = \frac{m_1 g\ 1.2 \ + m g \ 0.2}{M_2 g}

          x = \frac{m_1 \ 1.2 \ + m \ 0.2 }{M_2}

let's calculate

          x = \frac{2.9 \ 1.2 \ + 4 \ 0.2 }{8.00}2.9 1.2 + 4 0.2 / 8

           

          x = 0.535 m

measured from the pivot point

measured from the far left is

           x’= 1,2 + x

           x'=  1.2 + 0.535

           x ’= 1,735 m

8 0
3 years ago
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