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valkas [14]
2 years ago
9

A cable with 19.0 N of tension pulls straight up on a 1.50 kg block that is initially at rest. What is the block's speed after b

eing lifted 2.00 m ? Solve this problem using work and energy
Physics
1 answer:
frez [133]2 years ago
7 0

The final speed of the block, after being lifted 2.00 m is 3.39 m/s

<h3>What is speed?</h3>

Speed can be defined as the rate of change in the distance of a body.

To calculate the speed of the block after being lifted 2.00 m,  first, we need to calculate the acceleration of the block using the formula below

Formula:

  • T-mg = ma......... Equation 1

Where:

  • T = Tension in the cable
  • m = mass of the cable
  • a = acceleration
  • g = acceleration due to gravity

Restructuring the formula above,

  • a = (T-mg)/m............... Equation 2

From the question,

Given:

  • T = 19 N
  • m = 1.5 kg
  • g = 9.8 m/s²

Substitute these values into equation 2

  • a = [(19)-(1.5×9.8)]/1.5
  • a = 4.3/1.5
  • a = 2.87 m/s²

Finally, to calculate the speed of the block, we use the formula below.

  • v² = u²+2as.......... Equation 3

Where:

  • v = Final speed
  • u = initial speed
  • a = acceleration
  • s = distance

From the question,

Given:

  • u = 0 m/s
  • a = 2.87 m/s²
  • s = 2.00 m

Substitute these values into equation 3

  • v² = 0²+(2×2×2.87)
  • v² = 11.48
  • v = √11.48
  • v = 3.39 m/s

Hence, The final speed of the block, after being lifted 2.00 m is 3.39 m/s.

Learn more about speed here: brainly.com/question/6504879

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Answer: C. 12.6

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astra-53 [7]

Answer:

a) There are 100 centimeters in 1 meter.

b) \texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}

Explanation:

a) We have the conversion

         1 m = 100 cm

   So there are 100 centimeters in 1 meter.

b) 1 inch = 2.54 cm

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   \texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}

8 0
3 years ago
Examine the cross-sectional slice of a stem of a mint plant. Which statement most specifically describes mint?
Maslowich

The available options are:

Mint is a dicot.

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Mint is a bulb plant.

Answer:

Mint is a dicot.

Explanation:

Given the fact that Mint is considered to be a member of Lamiaceae, an angiosperm plant which is characterized by typically having leaves that consist of reticulate vacation and appears like veins in structure. It also has a seed that contains two cotyledons.

Hence, it is considered a DICOT PLANT due to these characteristics. The botanical name of Mint is referred to as Mentha arvensis.

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2 years ago
A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cable
kompoz [17]

Answer:

4.44s

Explanation:

A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing? Assume that the child and swing set are very small compared to the length of the cables

since the mass of the child and that of the swing is negligible, the masses wont be involved in the calculation

T=2π√L/g

g=acceleration due to gravity which is 9.81m/s2

the length of the supporting cable is 4.9m

T the period

period is the time required to make a complete oscillation

T=2*π√4.9/9.81

T=2*π*0.706

T=4.44s

4.44s

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The object represented by this graph is moving toward the origin at constant velocity.

Option 3.

<u>Explanation:</u>

In the figure, x-axis is representing increase in the time and y-axis is presenting increase in the distance from bottom to up. But the line in the graph which is plotted is decreasing from high distance to small distance with increase in time. So this indicates that as the time is increasing, the distance is decreasing.

And the object is moving toward the origin as the distance of the object motion is found to decrease with increase of time as per the graph. But the slope of the graph is found to be almost constant, this indicates that the velocity of the object is constant. Thus, the object represented by this graph is moving toward the origin at constant velocity.

4 0
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