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3 years ago

15

A kayaker moves 22 meters northward, then 18 meters southward , and finally 24 meters northward what is the distance and magnitu

de and direction?

1 answer:

Elina [12.6K]3 years ago

8 0

Answer:

Distance = 64 metres

Displacement = 28 metres

Direction = northward

Explanation:

Given that a kayaker moves 22 meters northward, then 18 meters southward , and finally 24 meters northward

Distance covered is only about the magnitude of the length covered.

Distance = 22 + 18 + 24 = 64 metres

The direction will be considered when calculating the displacement

Let northward be positive and southward be negative

Displacement = 22 - 18 + 24 = 28 metres

Since the displacement is positive, the direction of the motion is northward.

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An 88 kg worker stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale reads 900 N. Fi

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Answer:

The magnitude is "3.8 m/s²", in the upward direction.

Explanation:

The given values are:

Mass,

m = 88 kg

Scale reads,

T = 900 N

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On substituting the given values, we get

⇒ $=88\times 9.8$

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Now,

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On substituting the given values in the above equation, we get

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On subtracting "862.4" from both sides, we get

⇒ $900-862.4=862.4-9.8 a-862.4$

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3 years ago

Interactive Solution 9.37 presents a method for modeling this problem. Multiple-Concept Example 10 offers useful background for

viva [34]

Answer:

21.67 rad/s²

208.36538 N

Explanation:

$\omega_f$ = Final angular velocity = $\dfrac{1}{6}78=13\ rad/s$

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$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

t = Time taken

r = Radius = 0.13

I = Moment of inertia = 1.25 kgm²

From equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{13-78}{3}\\\Rightarrow \alpha=-21.67\ rad/s^2$

The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=1.25\times -21.67\\\Rightarrow \tau=-27.0875$

Frictional force is given by

$F=\dfrac{\tau}{r}\\\Rightarrow F=\dfrac{-27.0875}{0.13}\\\Rightarrow F=-208.36538\ N$

The magnitude of the force of friction applied by the brake shoe is 208.36538 N

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atroni [7]

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3 years ago

7. A spring balance reads force in Newton. The scale is 20 cm long and reads from 0 to

GarryVolchara [31]

Answer:

The potential energy when it reads 40 N is $PE = 5.33 \ J$

Explanation:

From the question we are told that

The lowest reading of the spring balance is 0 N and this is at 0 cm = 0 m

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Generally the length corresponding to the reading of 40 N is mathematically represented as

$d = \frac{40 * 0.20 }{60 }$

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Here

$F = mg = 40 N$

So

$PE = 40 * 0.133$

=> $PE = 5.33 \ J$

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3 years ago