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alex41 [277]
3 years ago
6

Two charges are located in the xx–yy plane. If q1=−4.10 nCq1=−4.10 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.080 m)

, and the second charge has magnitude of q2=3.60 nCq2=3.60 nC and is located at (x=1.20 m,y=0.600 m)(x=1.20 m,y=0.600 m), calculate the xx and yy components, ExEx and EyEy, of the electric field, E⃗ E→, in component form at the origin, (0,0)(0,0). The Coulomb force constant is 1/(4π????0)=8.99×109 N⋅m2/C21/(4πϵ0)=8.99×109 N⋅m2/C2.
Physics
1 answer:
Gala2k [10]3 years ago
8 0

Answer:

Explanation:

Due to first charge , electric field at origin will be oriented towards - ve of y axis.

magnitude

Ey = -8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j

= - 31.6 j N/C

Due to second charge electric field at origin

= 8.99 x 10⁹ x 3.6  x 10⁻⁹ / 1.2²+ .6²

= 8.99 x 10⁹ x 3.6  x 10⁻⁹ / 1.8

= 18 N/C

It is making angle θ where

Tanθ = .6 / 1.2

= 26.55°

this field in vector form

= - 18 cos 26.55 i - 18 sin26.55 j

= - 16.10 i - 8.04 j

Total field

= - 16.10 i - 8.04 j + ( - 31.6 j )

= -16.1 i - 39.64 j .

Ex = - 16.1 i

Ey = - 39.64 j .

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irakobra [83]
Underline the words then eliminate the ones that arent part of the problem!
4 0
3 years ago
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him
nataly862011 [7]

Answer:

a) u_c=0\ m.s^{-1}       &        m_c.v_c=m_b.v_b\times \cos\theta

b) v_c=0.0566\ m.s^{-1}

c) p_e=2.9218\ kg.m.s^{-1}

Explanation:

Given:

mass of the book, m_b=1.35\ kg

combined mass of the student and the skateboard, m_c=97\ kg

initial velocity of the book, v_b=4.61\ m.s^{-1}

angle of projection of the book from the horizontal, \theta=28^{\circ}

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

u_c=0\ m.s^{-1}

where:

u_c= initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

m_c.v_c=m_b.v_b\times \cos\theta

where:

v_c= final velcotiy of the student after throwing the book

b)

m_c.v_c=m_b.v_b\times \cos\theta

97\times v_c=1.35\times 4.61\cos28

v_c=0.0566\ m.s^{-1}

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

p_e=m_b.v_b\sin\theta

p_e=1.35\times 4.61\times \sin28^{\circ}

p_e=2.9218\ kg.m.s^{-1}

6 0
3 years ago
Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that th
lapo4ka [179]

Answer:

T_f=5.0116^{\circ}C

Explanation:

Given:

  • mass of water, m_w=0.25\ kg
  • initial temperature of water, T_i_w=20^{\circ}C
  • initial temperature of pan, T_i_p=173^{\circ}C
  • mass of pan, m_p=0.6\ kg
  • mass of water evapourated, m_v=0.03\ kg
  • specific heat of water, c_w=4186\ J.kg^{-1}.K^{-1}
  • specific heat of aluminium pan, c_a=900\ J.kg^{-1}.K^{-1}
  • latent heat of vapourization, L=2256000\ J.kg^{-1}

<u>Using the equation of heat:</u>

<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>

m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})

0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)

T_f=5.0116^{\circ}C

5 0
3 years ago
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tino4ka555 [31]

Answer:

0m/s²

Explanation:

Given parameters:

Initial velocity of the boat = 8m/s

Final velocity  = 8m/s

Time taken  = 4s

Unknown:

Acceleration of the boat = ?

Solution:

Acceleration is the rate of change of velocity with time.

It is mathematically expressed as;

        A = \frac{v - u}{t}

A is the acceleration

v is the final velocity

u is the initial velocity

t is the time taken

    Insert the parameters and solve;

  A = \frac{8-8}{4}   = 0m/s²

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Answer:

A) Availability

Explanation:

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