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4 years ago

Plz.. help me this out I am from nepal actually it is math

Physics

1 answer:

Ierofanga [76]4 years ago

4 0

Answer:

tan (x eˣ) + C

Explanation:

∫ (eˣ + x eˣ) / (cos² (x eˣ)) dx

First, use u-substitution.

u = x eˣ

du = (eˣ + x eˣ) dx

∫ 1 / (cos² u) du

∫ sec² u du

tan u + C

Substitute back:

tan (x eˣ) + C

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The Electric Potential Difference Created by Point Charges 13. Two point charges, +3.40 pC and -6.10 uC, are separated by 1.20 m

Alex Ar [27]

Answer:

- 91499.95 V

Explanation:

Let the mid point is P.

qA = 3.4 pC = 3.4 x 10^-12 C

qB = - 6.10 micro Coulomb = - 6.10 x 10^-6 C

AB = 1.20 m

AP = BP = 0.6 m

Let the potential at P due to the charge at A is VA and the potential at P due to the charge at B is VB.

The potential at P is V

V = VA + VB

$V = \frac{Kq_{A}}{AP}+\frac{Kq_{B}}{BP}$

$V = \frac{9\times 10^{9}\times 3.4\times 10^{-12}}{0.6}-\frac{9\times 10^{9}\times 6.10\times 10^{-6}}{0.6}$

V = 0.051 - 91500

V = - 91499.95 V

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3 years ago

A 30.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 6.00 cm on a frictionles

Naddika [18.5K]

Answer:

a) 72mJ

b) 0.33m/s

c) 54mJ

d) 18mJ

Explanation:

(a) The total energy of the system is given by:

$E_T=\frac{1}{2}kA^2$

where k is the force constant and A is the amplitude. By replacing we get:

$E_T=\frac{1}{2}(40N/m)(0.06m)^2=0.072J$ = 72mJ

(b) we can get the speed by the conservation of energy (the kinetic energy and potential energy must equal the total energy in any place):

$E_T=E_k+E_p\\\\\frac{1}{2}kA^2=\frac{1}{2}mv^2+\frac{1}{2}kx^2\\\\v=\sqrt{\frac{kA^2-kx^2}{m}}=\sqrt{\frac{(40N/m)((0.06m)^2-(0.0115m)^2)}{0.03kg}}=0.33m/s$

where we have used that x=1.15cm=0.0115m; A=6.00cm=0.06m

$E_k=\frac{1}{2}k(A^2-x^2)=\frac{1}{2}(40N/m)((0.06m)^2-(0.03m)^2)=0.054J=54mJ$

(d) $E_p=\frac{1}{2}kx^2=\frac{1}{2}(40.0N/m)(0.03m)^2=0.018J=18mJ$

hope this helps!!

6 0

3 years ago

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vodomira [7]

Answer:

the Answer is b

Explanation:

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Answer:

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3 years ago

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