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noname [10]
1 year ago
10

Richardson pulls a toy 3.0 m across the floor by a string, applying a force of0.50 N. During the first meter, the string is para

llel to the floor. In the next two meters, the string makes an angle 0f 300 with the horizontal direction. What’s the total amount of work done by Richardson on the toy?
Physics
1 answer:
Anastasy [175]1 year ago
7 0

Answer:

Total Work done =0.65 joule

Explanation:

Work done is given Mathematically as

W=F *d

Where w=work done in joules

F=applied force

d= distance moved

The work done to move the toy accros the first meter is

W1=0.5*1

W1=0.5joule

The work done to move the toy across the next 2m at an angle of 30° is

.W2=0.5*2cos30

W2=0.5*2*0.154

W2=0.154joule

Hence total work done is

W1+W2=0.5+0.154

Total Work done =0.65 joule

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A resistor R1 is wired to a battery, then resistor R2 is added in series. Are (a) the potential difference across R1 and (b) the
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Potential difference on R₁ will become less .

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c )

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1 year ago
A basketball rolls off a deck that is 3.2 m from the pavement. The basketball lands 0.75 m from the edge of the deck. How fast w
astra-53 [7]

Answer:

d). The value of y should be -32m

Vx=0.92 m/s

Explanation:

Using equation of motion uniform to y motion

x_{f}=x_{o}+v_{o}*t+\frac{1}{2}*a*t^{2}\\x_{o}=0m\\v_{o}=0\\x_{f}=\frac{1}{2}*a*t^{2}\\x_{f}=-3.2m \\a=g=-9.8\frac{m}{s^{2}} \\

So to find t that is the same time for all the motion

t^{2}=\frac{2*x_{f}}{a} \\t=\sqrt{\frac{2*x_{f}}{a}}=\sqrt{\frac{2*-3.2m}{-9.8\frac{m}{s^{2}}}}\\t=0.808s

The value of Xf=-3.2m because the g is negative from the axis

Now in the axis 'x' to find Vx

Vx=\frac{0.75m}{0.808S}\\ Vx=0.92 \frac{m}{s}

5 0
2 years ago
Two window washers, Bob and Joe, are on a 3.00 m long, 395 N scaffold supported by two cables attached to its ends. Bob weighs 8
WINSTONCH [101]

Answer:

- the forces on the left hand side is 1.038 kN

- the forces on the right hand side is 1.483 kN

Explanation:

Given the data in the question, as illustrated in the image below;

Length of the scaffold = 3 m

weight of the scaffold = 395 N

Weight of Bob = 805 N and stands 1 m from the left end

weight of washing equipment = 500N and on sits 2 m from the left end

Weight of Joe = 820 N and stand 0.500 m from the right end

so the force on the left cable will be;

T_{left = \frac{1}{3m}[ (805 N)( (3-1) m) + ( 395 N )( \frac{3}{2} m) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]

T_{left =  \frac{1}{3m}[ 1610 + 592.5 + 500 + 410 ]

T_{left =  \frac{1}{3m}[ 3112.5 ]

T_{left =  1037.5 N

T_{left =  1.038 kN

Therefore, the forces on the left hand side is 1.038 kN

On the right hand side;

T_{Right =  ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N

T_{Right =  2520 N - 1037.5 N

T_{Right =  1482.5 N

T_{Right =  1.483 kN

Therefore, the forces on the right hand side is 1.483 kN

5 0
1 year ago
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