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2 years ago

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elixir [45]2 years ago

8 0

Answer:

<em>1</em><em>)</em><em> </em><em>P</em><em>late</em><em> </em><em>tectonic</em><em> </em><em>theory</em>

<em>2</em><em>)</em><em> </em><em>Lithospheric</em>

<em>3</em><em>)</em><em> </em><em>Plates</em>

<em>4</em><em>)</em><em> </em><em>Heat</em>

<em>5</em><em>)</em><em> </em><em>Less</em><em> </em><em>dense</em>

<em>6</em><em>)</em><em> </em><em>Conventional current</em>

<em>7</em><em>)</em><em> </em><em>Magma</em>

<em>8</em><em>)</em><em> </em><em>Ocean</em><em> </em><em>crust</em>

<em>9</em><em>)</em><em> </em><em>Slowly</em>

<em>10</em><em>)</em><em> </em><em>Drifting</em><em> </em><em>away</em><em> </em>

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harkovskaia [24]

Answer:

m = 3 kg

The mass m is 3 kg

Explanation:

From the equations of motion;

s = 0.5(u+v)t

Making t thr subject of formula;

t = 2s/(u+v)

t = time taken

s = distance travelled during deceleration = 62.5 m

u = initial speed = 25 m/s

v = final velocity = 0

Substituting the given values;

t = (2×62.5)/(25+0)

t = 5

Since, t = 5 the acceleration during this period is;

acceleration a = ∆v/t = (v-u)/t

a = (25)/5

a = 5 m/s^2

Force F = mass × acceleration

F = ma

Making m the subject of formula;

m = F/a

net force F = 15.0N

Substituting the values

m = 15/5

m = 3 kg

The mass m is 3 kg

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3 years ago

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

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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

8 0

2 years ago

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The answer is 253.8 m/s

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Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

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Explanation:

Given

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and it starts at x=0 and t=0

Car B has to travel a distance of $D_A and d_a$

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$d_a=v_A\times t$

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$d_B=D_A+d_a$

$v_B\times t=D_A+v_A\times t$

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tia_tia [17]

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Explanation:

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