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natulia [17]
3 years ago
11

Cylinder A has a mass of 2kg and cylinder B has a mass of 10kg. Determinethe velocity of A after it has displaced 2m from its or

iginal starting position. Neglect the mass of the cable and pulleys and assume that both cylinders start at res
Physics
1 answer:
Marta_Voda [28]3 years ago
5 0

The velocity of A is 5.16m/s²

<u>Explanation:</u>

Given-

mass of cylinder A, mₐ = 2kg

mass of cylinder B, mb = 10kg

Distance, s = 2m

Velocity of A, v = ?

Let acceleration due to gravity, g = 10m/s²

We know,

a = \frac{mb * g - ma * g}{ma + mb} \\\\a = \frac{10 * 10 - 2 * 10}{ 2 + 10} \\\\a = \frac{80}{12} \\\\a = 6.67m/s^2

We know,

v = \sqrt{2as}

v = \sqrt{2 X 6.67 X 2} \\\\v = \sqrt{26.68} \\\\v = 5.16m/s^2

Therefore, the velocity of A is 5.16m/s²

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what is the name of the tool that allows you to copy two or more shapes into a single part? when working with 3D
Anna [14]

Answer:

Modeling tool or Align tool. it depends what type of sandbox platform you use

Explanation:

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8 0
3 years ago
For years, the tallest tower in the United States was the Phoenix Shot Tower in Baltimore, Maryland. The shot tower was used fro
Mariulka [41]

Answer:

The velocity of the droplet right before it hits the ground is 40.08 m/s.

Explanation:

To determine the velocity of the droplet right before it hits the ground,

From one of the equations of kinematic for free fall motions,

v = u + gt

Where v is the final velocity

u is the initial velocity

g is acceleration due to gravity (take g = 9.8 m/s²)

and t is time

For the question, v is the velocity of the droplet right before it hits the ground.

u = 0 m/s (Since the molten lead was dropped from rest)

Therefore,

v = gt

First, we will determine the time t

Also, from one of the equations of kinematic for free fall motions,

h = ut + 1/2(gt²)

u = 0 m/s

From the question, the molten lead was dropped from the top of the 82.15 m tall tower, therefore

h = 82.15

Hence,

82.15 = 0×t + 1/2 (9.8 × t²)

82.15 = 1/2 (9.8 × t²)

82.15 = 4.9 t²

t² = 82.15/4.9

∴ t = 4.09 secs

Now, for the velocity v, of the droplet right before it hits the ground,

Recall

v = gt

Then,

v = 9.8 × 4.09

v = 40.08 m/s

Hence, the velocity of the droplet right before it hits the ground is 40.08 m/s.

5 0
3 years ago
What are advantages and disadvantages of apparent solar time? How is the situation improved by introducing mean solar time and s
Nutka1998 [239]

Answer:

Advantage:  

Apparent solar time gives the exact location of sun in the sky according to your precise location.

Disadvantage:  

As the apparent solar time changes with the change in longitude. It is very difficult to track these changes in longitude. Hence, it is almost impossible to make plan for events.

Improvement in situation:

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3 0
3 years ago
A man pulled a 13.0 kg object 11.8 cm vertically with his teeth. (a) How much work (in J) was done on the object by the man in t
jonny [76]

Answer:

(a)The work done by the man is -15.03J.

(b)The force exerted on the object is 127.4N.

Explanation:

Mass of the object pulled by the man is -13kg

Object is lifted 11.8 cm vertical with his teeth it means (displacement = +11.8cm = +0.118m)

Acceleration due to gravity is 9.8 \mathrm{m} / \mathrm{s}^{2}

(a) <u>Calculating the work done</u>:

Work done = mgh

Where "m" is mass of an object, "g" is acceleration due to gravity and "h" is the displacement.

\text { Work }=-13 \times 9.8 \times(+0.118 \mathrm{m})

\text { Work }=-15.03 \mathrm{J}

The work done by the man is -15.03J.

(b) <u>Calculating the force</u>:

Probably the man and the object are close to the exterior of the earth. If the rigidity required to maintained the object of consistent velocity interior the gravitational field of the earth is \mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}

Thus the weight of the object is balanced by the force of the man's teeth on the object. That is

F = mg

\mathrm{F}=13 \times 9.8

F = 127.4N

The force exerted on the object is 127.4N.

4 0
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The shock absorbers in a car act as a
Varvara68 [4.7K]

Answer: it's 69200

Explanation:

I gotchu guys

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