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2 years ago

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Quddus and Saadman start running together from the same point. Quddus runs at constant velocity 0.5ms−1. Saadman’s initial veloc

ity is 0 but he runs at constant acceleration 0.1ms−2. How long will it take for them to meet each other?
a. 13.33
b. 10s
c. 50s
d. 26.67

1 answer:

svp [43]2 years ago

8 0

Answer:

the answer is C.

Explanation:

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Answer:

D

Explanation:

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How does this picture apply to Newtons 3rd law.Please make it well developed!!!Thanks!

Orlov [11]

<u>Explanation for the given picture:</u>

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Newtons 3rd law applies in above picture.

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3 years ago

A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m

Dvinal [7]

(a) 638.4 J

The work done by a force is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

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The displacement of the suitcase is

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And the force is parallel to the displacement, so $\theta=0^{\circ}$ . Therefore, the work done by this force is

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b) -328.2 J

The magnitude of the gravitational force is

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where

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$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$W=(19.6)(9.8)=192.1 N$

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

$\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}$

Where $\alpha = 24^{\circ}$ is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

$W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J$

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

$R=mg cos \alpha$

And substituting

m = 19.6 kg

g = 9.8 m/s^2

$\alpha=24^{\circ}$

We find

$R=(19.6)(9.8)(cos 24)=175.5 N$

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

$\theta=90^{\circ}$

Therefore, the work done by the normal force is

$W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0$

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The magnitude of the force of friction is

$F_f = \mu R$

where

$\mu = 0.264$ is the coefficient of kinetic friction

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Substituting,

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d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

$\theta=180^{\circ}$

Therefore, the work done by friction is

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$W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J$

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First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

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$W_f =(46.3)(3.80)(cos 180)=-175.9 J$

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3 years ago

Which of the following types of energy are present at some point in the energy transfer process in a nuclear power plan? Select

Alexeev081 [22]

Solar Energy is the answer to the question tell me if i`m wrong

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3 years ago

The time taken for an object to fall from a height of h m from the earth'fs surface is t s.

telo118 [61]

Answer:

B. Longer than t s,

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now the time taken on earth will be lesser cuz from the same height if you drop the object from rest!

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✌️:)

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3 years ago