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Aleonysh [2.5K]
3 years ago
7

A 35.8 kg box initially at rest is pushed 2.38 m along a rough, horizontal floor with a constant applied horizontal force of 108

.915 N. If the coefficient of friction between box and floor is 0.256, find the work done by the applied force. The acceleration of gravity is 9.8 m/s 2 . Answer in units of J.
Physics
1 answer:
tiny-mole [99]3 years ago
6 0

Answer:

The work done by the applied force is 259.22 J.

Explanation:

The work done by the applied force is given by:

W = F*d

Where:

F: is the applied horizontal force = 108.915 N

d: is the distance = 2.38 m  

Hence, the work is:

W = F*d = 108.915 N*2.38 m = 259.22 J

Therefore, the work done by the applied force is 259.22 J.

I hope it helps you!                                                

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The velocity at impact is 222.2 m/s

Explanation:

The given information are;

The time at which the woman heard the sound of the coin after dropping it into the wishing well = 30 seconds

The average speed of sound in air = 344 m/s

The time in which the dime travel = t₁

The time in which the sound travel = t₂

1/2 × 9.8 × t₁² = 344 × t₂

t₁ + t₂ = 30

4.9·t₁² = 344 × (30 - t₁)

4.9·t₁² + 344·t₁ - 10320 = 0

Which gives -344±

t_1 = \dfrac{-344\pm \sqrt{344^{2}-4\times 4.9 \times 10320}}{2\times 4.9}

t₁  = 22.676 seconds or -92.9 seconds

Therefore the correct natural time is t₁  = 22.676 seconds

The velocity at impact, v = g×t₁ = 9.8 × 22.676  = 222.2 m/s

The velocity at impact= 222.2 m/s.

8 0
4 years ago
ifif it takes 1-minute for 45 c of charge to pass a point in an electric circuit what is the current through the circuit?​
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Two masses 1.2kg and 1.8kg are connected to the ends of a rod of length 2m. Find the moment of inertia about the axes, 1)going t
frutty [35]

Answers: 1) 3 kg m²

                2) 2.88 kg m²

Explanation: <u> </u><u>Question 1</u>

                      I = m(r)²+ M(r)²

                      I = 1.2 kg × (1 m )² +1.8 kg ×(1 m )²

                    ∴  I =   3 kg m²

                       

                     <u> </u><u>Question 2 </u>

ACCORDING TO THE DIAGRAM DRAWN FOR QUESTION 2

we have to decide where the center of gravity (G) lies and obviously it should lie somewhere near to the greater mass.<em> (which is 1.8 kg). S</em>ince we don't know the distance from center of gravity(G) to the mass (1.8 kg) we'll take it as 'x' and solve!!

<u>moments around 'G' </u>

F₁ d ₁ = F₂ d ₂

12 (2-X) = 18 (X)

24 -12 X =18 X

∴  X = 0.8 m

∴ ( 2 - x ) = 1.2 m

∴ Moment of inertia (I) going through the center of mass of two masses,

⇒ I = m (r)² +M (r)²

⇒ I = 1.2 × (1.2)² + 1.8 × (0.8)²

⇒ I = 1.2 × 1.44 + 1.8 × 0.64

⇒ I = 1.728 + 1.152

⇒ ∴ I = 2.88 kg m²

∴ THE QUESTION IS SOLVED !!!

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8 0
3 years ago
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