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ruslelena [56]
2 years ago
13

Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

loor. The tension is a constant 32.0N and the coefficient of friction is 0.190.
Use work and energy to find Paul's speed after being pulled 3.20m .

Solve for vf.
Physics
1 answer:
elena-s [515]2 years ago
8 0

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

W_T+W_F=K_f-K_i (1)

Where:

W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to \frac{1}{2}mv^{2}, so if the initial velocity v_i is equal to zero, the initial kinetic energy K_i is equal to zero.

Then, replacing the values on the equation and solving for v_f, we get:

W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}

\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f

So, the speed after being pulled 3.2m is 2.4123 m/s

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Explanation:

Given:

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Ask Your Teacher A basketball player shoots toward a basket 5.8 m away and 3.0 m above the floor. If the ball is released 1.7 m
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Answer:

The answer to your question is    vo = 5.43 m/s

Explanation:

Data

distance = d= 5.8 m

height = 3 m

height 2 = 1.7 m

angle = 60°

vo = ?

g = 9.81 m/s²

Formula

              hmax = vo²sinФ/ 2g

Solve for vo²

              vo² = 2ghmax / sinФ

Substitution

              vo² = 2(9.81)(3 - 1.7) / 0.866

Simplification

              vo² = 19.62(1.3) / 0.866

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Result

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A rubber balloon is filled with 1 L of air at 1 atm and 300 K and is then put into a cryogenic refrigerator at 100 K. The rubber
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Answer:

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Explanation:

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at 100K the content of the balloon cools reducing the excitement of the gas content which also reduces the pressure, however, the balloon being perfectly elastic, contracts to maintain the 1 atmospheric pressure, hence the answer to (ii) is (c) It is constant,

For (i) since we know that the pressure of the balloon is constant

by Charles Law V1/T1 =V2/T2

or V2 = (V1/T1)×T2 =\frac{1L}{300K}× 100K= \frac{1}{3} × L = L/3 hence the correct answer to (i) is 1/3L

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Answer:

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