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3 years ago

5

Hai xe chuyển động trên cùng một đoạn đường , khi xe (1) ở A thì xe (2) ở B phía trước với AB=5km . Xe (1) đuổi theo xe (2) . Tạ

i C cách B một đoạn BC=10km thì xe (1) đuổi kịp xe (2) . Tìm tỉ số vận tốc của hai xe

1 answer:

12345 [234]3 years ago

4 0

Answer:

b

Explanation:

because it shows you it

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A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 . If the charge is 0.030 m from the source

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Answer:

0.027 J

Explanation:

The formula of the potential energy in electrostatic (U) is

$U = q \times E \times d$

The values given in the question are

Charge : q = $4.5 \times 10^{-5} C$

Electric Field strength : E = $2.0 \times 10^{4} N/C$

Distance : d is 0.030 m

Insert in the formula , will give us

$U = 4.5 \times 10^{-5} C \times 2.0 \times 10^{4} N/C \times 0.030 m$

Further solving it

$U = 0.027 J$

Which is the required answer.

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3 years ago

What is necessary to designate a position? A. a reference point B. a direction C. fundamental units D. motion E. all of these

max2010maxim [7]

Answer:

E. all of these

Explanation:

The designation of a point in space all the points that necessary

- reference point

- a direction

- fundamental units

- a direction

- motion

all are necessary to designate a point in space. Hence option E is correct.

For example in simple harmonic motion we need to specify all the above factors of the object in order to designate the position of the object.

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On a planet where g = 10.0 m/s2 and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30°. The

Cloud [144]

Answer:

Explanation:

a is the acceleration

μ is the coefficient of friction

Acceleration of the object is given by

$a = g (\sin \theta -\mu \cos \theta)\\\\=10( \sin 30 - 0.4 \cos 30)\\\\=10(0.5-0.3464)\\\\=1.54m/s^2$

Velocity at the bottom

$v^2=u^2+2as\\\\u=0\\\\v^2=2as\\\\=2*1.54*8\\\\=24.576\\\\v=4.96m/s$

after travelling 4m , its velocity becomes 0

$a=\frac{v^2-u^2}{2s}$

$a=\frac{0-u^2}{2s}$

$a=\frac{-(-4.96)^2}{2*4} \\\\=-3.075m/s^2$

Coefficient of kinetic friction

μ = F/N

$=\frac{ma}{mg} \\\\=\frac{3.075}{10} \\\\=0.31$

Therefore, the Coefficient of kinetic friction is 0.31

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