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sergey [27]
3 years ago
5

An object with a mass of 32 kg has an initial energy of 500). At the end of the experimentthe velocity of the object is recorded

as 5.1 m/s . the object travelled 50 m to get to this point, what was the average force of friction on object during the tripAssume no potential energy Show all work
Physics
1 answer:
Alik [6]3 years ago
7 0

Answer:

 F = 1.68 N

Explanation:

Let's solve this exercise in parts.

Let's use the concept of conservation of the mechanical nerve

initial

    Em₀ = 500 J

The energy is totally kinetic

     Em₀ = K = ½ m v₀²

     v₀ = \sqrt{\frac{2 Em_{o} }{m} }

     v₀ = √ (2 500/32)

     v₀ = 5.59 m / s

now with kinematics we can find a space

      v² = v₀² - 2 a x

the negative sign is because the body is stopping

       a =( \frac{v_{o}^{2} - v^{2}   }{2x} )  

let's calculate

       a = (5.59² - 5.1²) / 2 50  

       a = 0.0524 m / s²

Finally let's use Newton's second law

     F = ma

     F = 32 0.0524

     F = 1.68 N

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I think it is A, B, C. :)
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7 0
3 years ago
A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring  73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E  =  2000 N/C

Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}

E1 = 3714.672 N/C

electric field due to point charge q

E  =\frac[kq}{x^2}

E = \frac{9\times 10^9 \times q}{0.70^2}

E2 = 1.837\times 10^{10}\times q

now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

7 0
4 years ago
Inserting the formulas you found for Xman(t) and Xbus(t) into the conditionXman(tcatch)=Xbus(tcatch) , you obtain the following-
lara [203]

Answer:

c > √(2ab)

Explanation:

In this exercise we are asked to find the condition for c in such a way that the results have been real

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we can see that this is a quadratic equation whose solution is

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for the results to be real, the square root must be real, so the radicand must be greater than zero

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3 0
3 years ago
Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that
elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

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W = 7.85 * 10^-17 J

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Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

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Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

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Answer: Kinetic Energy of an electron gained E_k = 490 eV

7 0
4 years ago
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