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3 years ago

A mineral that has _______ can be split fairly easily along planes with weak atomic attraction.

Physics

2 answers:

scoundrel [369]3 years ago

6 0

The answer would be b. cleavage

Stells [14]3 years ago

6 0

<span> A mineral that has _______ can be split fairly easily along planes with weak atomic attraction.
<span> D. cleavage</span></span>

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As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh

marshall27 [118]

Answer:

The resultant velocity is $v_t=10 knots$

Explanation:

Apply the law of conservation of momentum

$M_L *v_L + M_f * V_f = (M_L + M_f) v_t$

Where $M_L$ is the mass of the Luxury Liner = 40,000 ton

$v_L$ is the velocity of Luxury Liner = 20 knots due west

$M_f$ mass of freighter = 60,000

$v_f$ is the velocity of freighter = 10 knots due north

Apply the law of conservation of momentum toward the the west direction

$v_f = 0 \ knots$

So the equation would be

$M_L *v_L = (M_L + M_f) v_t$

Substituting values

$40000*20 = (40000+ 60000)v_t_w$

Where $v_t_w$ the final velocity due west

Making $v_t_w$ the subject

$v_t_w = \frac{40,000* 20}{(40000 + 60000)}$

$= 8 \ knots$

Apply the law of conservation of momentum toward the the north direction

$v_L = 0 \ knots$

So the equation would be

$M_f *v_f = (M_L + M_f) v_t_n$

Where $v_t_n$ the final velocity due north

Making $v_t_n$ the subject

$v_t_n = \frac{60,000* 10}{(40000 + 60000)}$

$= 6 \ knots$

The resultant velocity is

$v_t = \sqrt{v_t_w^2 + v_t_n^2}$

$= \sqrt{8^2 +6^2}$

$v_t=10 knots$

8 0

3 years ago

A 3.00 kg object is fastened to a light spring, with the intervening cord passing over a pulley. The pulley is frictionless, and

finlep [7]

Answer:

Part a)

k = 588.6 N/m

Part b)

v = 0.7 m/s

Explanation:

As we know that initially block is at rest

now if block is released from rest then it will go down by 10 cm and again comes to rest

so here we have

Part a)

Work done by gravity + work done by spring force = change in kinetic energy

$W_g + W_{spring} = 0 - 0$

$mg(0.10) + \frac{1}{2}k(0^2 - 0.10^2) = 0$

$3(9.81)(0.10) - \frac{1}{2}k(0.10)^2 = 0$

$k = 588.6 N/m$

Part b)

Now when spring is stretch by x = 5 cm then the speed of the block is given as

$mgx' + \frac{1}{2}k(0^2 - x'^2) = \frac{1}{2}mv^2 - 0$

here we have

$x' = 0.05 m$

$3(9.81)(0.05) - \frac{1}{2}(588.6)(0 - 0.05^2) = \frac{1}{2}(3) v^2$

$1.4715 - 0.736 = 1.5 v^2$

$v = 0.7 m/s$

3 0

3 years ago

assume that the brakes in your car create a constant deceleration regardless of how fast you are going. if you double your drivi

Monica [59]

If that happens to your car and you double the speed then it will take the car longer to come to a complete stop

5 0

3 years ago

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Divers found two substances on the bottom of the ocean. At room temperature, both substances are liquid. Scientists then transfe

shtirl [24]

They have different melting points

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3 years ago

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A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Su

aniked [119]

Answer:

The planet´s orbital period will be one-half Earth´s orbital period.

Explanation:

The planet in orbit, is subject to the attractive force from the sun, which is given by the Newton´s Universal Law of Gravitation.

At the same time, this force, is the same centripetal force, that keeps the planet in orbit (assuming to be circular), so we can put the following equation:

Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r

As we know to find out the orbital period, as it is the time needed to give a complete revolution around the sun, we can say this:

ω = 2*π / T (rad/sec), so replacing this in the expression above, we get:

Fg = Fc ⇒ G*mp*ms / r² = mp*(2*π/T)²*r

Solving for T²:

T² = (2*π)²*r³ / G*ms (1)

For the planet orbiting the sun in Andromeda, we have:

Ta² = (2*π)*r³ / G*4*ms (2)

As the radius of the orbit (distance to the sun) is the same for both planets, we can simplify it in the expression, so, if we divide both sides in (1) and (2), simplifying common terms, we finally get:

(Te / Ta)² = 4 ⇒ Te / Ta = 2 ⇒ Ta = Te/2

So, The planet's orbital period will be one-half Earth's orbital period.

7 0

3 years ago