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AlladinOne [14]

2 years ago

7

Suppose you found a mineral crystal that looked like topaz. what two minerals could you use in a scratch test to help determine

if the mineral is indeed topaz?

1 answer:

Vesna [10]2 years ago

4 0

The two most important minerals you need to do the test are <span>Corundum and quartz. You use that to do scratch test</span>

You might be interested in

An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of

Vsevolod [243]

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

v = √ ( V₀² + 2 g H )

6 0

3 years ago

At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building

Inessa [10]

Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

$x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration$

So now we have an equation and unkown value.

for the thrown rock

$\frac{1}{2}(9.8)*t^2+29*t-300=0$

for the dropped rock

$\frac{1}{2}(9.8)*t^2+0*t-300=0$

solving both equation with the quadratic formula:

$\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}$

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

6 0

2 years ago

You're driving a vehicle of mass 1350 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 71 m

sergey [27]

Answer:

$v=12.65\ m.s^{-1}$

Explanation:

Given:

mass of vehicle, $m=1350\ kg$

radius of curvature, $r=71\ m$

coefficient of friction, $\mu=0.23$

<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>

$m.\frac{v^2}{r} =\mu.N$

where:

$\mu=$ coefficient of friction

$N=$ normal reaction force due to weight of the car

$v=$ velocity of the car

$1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)$

$v=12.65\ m.s^{-1}$ is the maximum velocity at which the vehicle can turn without skidding.

5 0

3 years ago

a 10 kilogram steel ball is dropped from the top of a tower 100 meters high the kinetic energy of the ball just before it sttike

EleoNora [17]

Explanation:

If we assume negligible air resistance and heat loss, we can assume that all of the Gravitational potential energy of the ball will turn into Kinetic energy as it falls toward the ground.

Therefore our Kinetic energy = mgh = (10kg)(9.81N/kg)(100m) = 9,810J.

4 0

2 years ago

A hockey puck is sliding across a frozen pond with an initial speed of 9.3 m/s. It comes to rest after sliding a distance of 42.

kondaur [170]

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.11

Explanation:

Given;

initial speed, u = 9.3 m/s

sliding distance, S = 42 m

From equation of motion we determine the acceleration;

v² = u² + 2as

0 = (9.3)² + (2x42)a

- 84a = 86.49

a = -86.49/84

|a| = 1.0296

$F_k = \mu_k N$ = ma

where;

Fk is the frictional force

μk is the coefficient of kinetic friction

N is the normal reaction = mg

μkmg = ma

μkg = a

μk = a/g

where;

g is the gravitational constant = 9.8 m/s²

μk = a/g

μk = 1.0296/9.8

μk = 0.11

Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11

3 0

2 years ago