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polet [3.4K]
3 years ago
12

A 20 kg box is being pulled across a floor by a horizontal rope. The tension in the rope is 99 Newtons. The coefficient of frict

ion is 0.25. What is the force of friction on the box? What is the acceleration of the box?
Physics
2 answers:
fgiga [73]3 years ago
5 0

Answer:

friction = 49 N, a = 2.5 m/s/s

Explanation:

big brain

Burka [1]3 years ago
4 0
Yeah what that guy said lol
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Two velocity vectors, one is twice of the other, and separated by 90 Degree angle. If their resultant is calculated 90 m/s, what
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Answer:

Vy = 80.5 [m/s]

Explanation:

In order to solve this problem we must use the Pythagorean theorem.

V = 90 [m/s]

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Therefore:

v=\sqrt{v_{x}^{2} + v_{y}^{2}   }

where:

Vy = 2*Vx ; because one is twice of the other.

90 = \sqrt{v_{x}^{2}  +(2*v_{x})^{2} }\\ 90 =\sqrt{v_{x}^{2}+4*v_{x}^{2}} \\90 =\sqrt{5v_{x}^{2}} \\90=2.23*v_{x} \\v_{x}=40.25[m/s]

and the bigger vector is:

Vy = 40.25*2

Vy = 80.5 [m/s]

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3 years ago
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n isolated charged soap bubble of radius R0=7.45 cmR0=7.45 cm is at a potential of V0=307.0 volts.V0=307.0 volts. If the bubble
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Complete Question

An isolated charged soap bubble of radius R0 = 7.45 cm  is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

Answer:

The difference is    U_f -U_i = 16 *10^{-7} J

Explanation:

From the question we are told that

     The radius of the soap bubble  is  R_o =  7.45 \ cm =  \frac{7.45}{100} =  0.0745 \ m

      The potential of the soap bubble is  V_1  =307.0 V

      The new radius of the soap bubble  is R_1 =  0.19 * 7.45=1.4155\ cm = 0.014155 \ m

The initial electric potential is mathematically represented as

     U_i  = \frac{V_1^2 R_o }{2k }

The final  electric potential is mathematically represented as

    U_f  = \frac{V_2^2 R_1 }{2k }

The initial potential is mathematically represented as

     V_1 =  \frac{kQ}{R_o}

The final  potential is mathematically represented as

        V_2 =  \frac{kQ}{R_1}

Now  

         \frac{V_2}{V_1}  =  \frac{R_o}{R_1}

substituting values

        \frac{V_2}{V_1}  =  \frac{7.45}{1.4155} =   \frac{1}{0.19}

=>      V_2 =  \frac{V_1}{0.19}

    So

         U_f  = \frac{V_1^2 R_2 }{2k * 0.19^2}

Therefore

        U_f -U_i = \frac{V_1^2 R_2 }{2k * 0.19^2} - \frac{V_1^2 R_o }{2k }

       U_f -U_i =     \frac{V_1^2}{2k} [\frac{ R_1 }{ * 0.19^2} - R_o]

where k is the coulomb's constant with value 9*10^{9} \  kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

       U_f -U_i =     \frac{307^2}{9 * 10^{9}} [\frac{ 0.014155 }{ 0.19^2} - 0.0745]

       U_f -U_i = 16 *10^{-7} J

           

     

8 0
3 years ago
We wrap a light, nonstretching cable around a 10.0 kg kg solid cylinder with diameter of 38.0 cm cm . The cylinder rotates with
amm1812

Answer: 14.16

Explanation:

Given

d = 38cm

r = d/2 = 38/2 = 19cm = 0.19m

K.E = 510J

m = 10kg

I = 1/2mr²

I = 1/2*10*0.19²

I = 0.18kgm²

When it has 510J of Kinetic Energy then,

510J = 1/2Iω²

ω² = 1020/I

ω² = 1020/0.18

ω² = 5666.67

ω = √5666.67 = 75.28 rad/s

Velocity is the block, v = ωr

V = 75.28 * 0.19

V = 14.30m/s

The "effective mass" M of the system is

M = (14.0 + ½*10.0) kg = 19.0 kg

The motive force would be

F = ma

F = 14 * 9.8

F = 137.2N

so that the acceleration would be

a = F/m

a = 137.2/19

a = 7.22m/s²

Finally, using equation of motion.

V² = u² + 2as

14.3² = 0 + 2*7.22*s

204.49 = 14.44s

s = 204.49/14.44

s = 14.16m

6 0
3 years ago
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