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3 years ago

Find it ...............

Physics

1 answer:

Sholpan [36]3 years ago

7 0

Answer:

9 N

Explanation:

Treat X and y as a single book with weight 9 N.

Draw a free body diagram of this book combination. There are two forces: weight force 9 N pulling down, and normal force F pushing up.

Sum of forces in the y direction:

∑F = ma

F − 9 N = 0

F = 9 N

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Answer:

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4 years ago

A point charge with a charge q1 = 4.00 μC is held stationary at the origin. A second point charge with a charge q2 = -4.10 μC mo

GrogVix [38]

Answer:

W = -0.480 J

Explanation:

given,

q₁ = 4 μC

q₂ = -4.10 μC

$W = kq_1q_2(\dfrac{1}{a}+\dfrac{1}{b})$

$b = \sqrt{(0.27-0)^2+(0.27-0)^2}$

b = 0.381

k = 8.99 × 10⁹ Nm²/C²

$W = 8.99\times 10^9\times 4\times 10^{-6}\times (-4.1 \times 10^{-6})(\dfrac{1}{0.17}+\dfrac{1}{0.381})$

$W = [-147.436\times (5.88-2.62)\times 10^{-3}]J$

W = -0.480 J

Work done by the electric force W = -0.480 J

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3 years ago

Quentin made a table to summarize uses of electromagnetic waves.

snow_lady [41]

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3 years ago

Read 2 more answers

An electron that has a velocity with x component 1.6 × 106 m/s and y component 2.4 × 106 m/s moves through a uniform magnetic fi

Sergio039 [100]

Answer:

(a) 5.056 x 10^-14 N

(b) 5.056 x 10^-14 N

Explanation:

X component of velocity of electron is 1.6 × 10^6 m/s

Y component of velocity of electron is 2.4 × 10^6 m/s

X component of magnetic field is 0.025 T

Y component of magnetic field is -0.16 T

charge on electron, q = - 1.6 x 10^-19 C

Write the velocity and magnetic field in the vector forms.

$\overrightarrow{v}=1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j}$

$\overrightarrow{B}=0.025\widehat{i}-0.16\widehat{j}$

The force on the charge particle when it is moving in the magnetic field is given by

$\overrightarrow{F}=q\left ( \overrightarrow{v}\times \overrightarrow{B} \right )$

(a) Force on electron is given by

$\overrightarrow{F}=-1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )$

$\overrightarrow{F}=5.056\times 10^{-14}\widehat{k}$

Magnitude of force is 5.056 x 10^-14 N.

(b) Force on a proton is given by

$\overrightarrow{F}=1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )$

$\overrightarrow{F}=-5.056\times 10^{-14}\widehat{k}$

Magnitude of force is 5.056 x 10^-14 N.

Thus, the magnitude of force remains same but the direction of force is opposite to each other.

Explanation:

4 0

3 years ago

A car moving at some speed hits the brakes and skids to a stop after 13 m on a level road. If the coefficient of friction for th

vesna_86 [32]

Answer:

12.974 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of friction =0.66

a = Acceleration = $\mu g$

$v^2-u^2=2as\\\Rightarrow -u^2=2\mu gs-v^2\\\Rightarrow u=\sqrt{v^2-2\mu gs}\\\Rightarrow u=\sqrt{0^2-2\times -(9.81\times 0.66)\times 13}\\\Rightarrow u=12.974\ m/s$

Car's original speed before braking was 12.974 m/s

6 0

4 years ago

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