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noname [10]
3 years 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]3 years 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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7 0
2 years ago
A spring stretches by 15cm when a mass of 300g hangs down from it,if the spring is then stretched an additional 10cm and release
bearhunter [10]

Answer:

0.1 m

Explanation:

It is given that,

Mass of the object, m = 350 g = 0.35 kg

Spring constant of the spring, k = 5.2 N/m

Amplitude of the oscillation, A = 10 cm = 0.1 m

Frequency of a spring mass system is given by :

Time period:

4 0
3 years ago
Two speakers emit the same sound wave, identical frequency, wavelength, and amplitude. What other quantity would be necessary to
Komok [63]

Answer:

Phase Difference

Explanation:

When the sound waves have same wavelength, frequency and amplitude we just need the phase difference between them at a particular location to determine whether the waves are in constructive interference or destructive interference.

Interference is a phenomenon in which there is superposition of two coherent waves at a particular location in the medium of propagation.

When the waves are in constructive interference then we get a resultant wave of maximum amplitude and vice-versa in case of destructive interference.

  • For constructive interference the waves must have either no phase difference or a phase difference of nλ, where n is any natural number.
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6 0
3 years ago
A push broom is being pushed down across a rough floor. The broom moves to the right. What is the correct free body diagram of t
zysi [14]
The far right.

Fg is gravity which always acts down and since we assume the floor is flat the normal, Fn, acts opposite gravity, so straight up.

But you’re probably wondering about the pushing force, Fp, and the friction force, Ff. For the Fp, consider where the applied force is coming from. The head of the broom is on the floor and the man’s arms, where he’s applying the force from, is above and to the left, so when the man pushes the broom the force is down and to the right. The broom my not be moving down, but the applied force is still in that direction. And Ff always acts against motion so since the broom moves to the right, the friction is to the left.

5 0
3 years ago
Read 2 more answers
On a horizontal frictionless floor, a worker of weight 0.900 kN pushes horizontally with a force of 0.200 kN on a box weighing 1
Ad libitum [116K]

Answer:

D) The worker will accelerate at 2.17  m/s²  and the box will accelerate at 1.08  m/s² , but in opposite directions.

Explanation:

Newton's third law

Newton's third law or principle of action and reaction states that when two interaction bodies appear equal forces and opposite directions. in each of them.

F₁₂= -F₂₁

F₁₂: Force of the box on the worker

F₂₁: Force of the worker on the box

Newton's second law

∑F = m*a

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Formula to calculate the mass (m)

m =  W/g

Where:

W : Weight (N)

g : acceleration due to gravity  (m/s²)

Data

W₁ =1.8 kN   : box weight

W₂ = 0.900 kN : worker weight

F₂₁ = 0.200 kN

F₁₂ = - 0.200 kN

g = 9.8 m/s²

Newton's second law for the box

∑F = m*a

F₂₁ = m₁*a₁    m₁=W₁/g

0.2 kN = (1.8kN)/(9.8 m/s² ) *a₁

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{1.8 kN}

a₁= 1.08 m/s² : acceleration of the box

Newton's second law for the worker

∑F = m*a

F₁₂ = m₂*a₂ , m₂=W₂/g

- 0.2 kN =( (0.9 kN) /(9.8 m/s² ) )*a₂

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{0.9 kN}

a₂=  -2.17 m/s² : acceleration of the worker

5 0
3 years ago
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