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3 years ago

What is the speed of a bobsled whose distance-time graph indicates that it traveled 116m in 29s?

Physics

2 answers:

ryzh [129]3 years ago

5 0

Speed = Distance ÷ Time
Speed = 116 ÷ 29
Speed = 4

mars1129 [50]3 years ago

5 0

Speed = (distance covered) / (time to cover the distance)

= (116 meters) / (29 seconds)

= (116/29) (meters per second)

= 4 meters per second (4 m/s)

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2 years ago

A horizontal 745 N merry-go-round of radius

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Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

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Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

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Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

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3 years ago

A horizontal force of magnitude 46.3 n pushes a block of mass 4.14 kg across a floor where the coefficient of kinetic friction i

IrinaVladis [17]

A) Calling F the intensity of the horizontal force and d the displacement of the block across the floor, the work done by the horizontal force is equal to
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b) The work done by the frictional force against the motion of the block is equal to:
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Part of these 105.1 Joules of work becomes increase of thermal energy of the block ( $\Delta E_B$ ), and part of it becomes increase of thermal energy of the floor ( $\Delta E_F$ ). We already know the increase in thermal energy of the block (38.2 J), so we can find the increase in thermal energy of the floor:
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c) The net work done on the block is the work done by the horizontal force F minus the work done by the frictional force (the frictional force acts against the motion, so we must take it with a negative sign):
$W_{net}=W-W_f=196.8 J-105.1 J=91.7 J$
For the work-energy theorem, the work done on the block is equal to its increase of kinetic energy:
$W_{net} = \Delta K$
So, we have $\Delta K=+91.7 J$

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