A gemstone is a hard, colorful mineral with a glassy or shiny luster. True or False?

what if l was standing and all of continent are coming together and becomes pangea.What will happen???

azamat

Answer:

not really

Explanation:

pangea took million of years to separate. Years by Years the continents move. for example The two continents are moving away from each other at the rate of about 2.5 centimeters (1 inch) per year. And there movent is due to the activity beneath the earths crust. The plates are moved by currents in the magma at the very surface of the earths mantle , which then in turn cause things like earthquakes and volcanic eruptions.

5 0

3 years ago

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A car of mass 800kg travels a distance of 40m at constant speed in a duration of 2.0s. The car exerts a forward force of 15kN.

Alex17521 [72]

W = F × s

W = 15kN × 40 m

W = 15.000 N × 40 m

W = 600.000 J

P = W/t

P = 600.000 J/2 s

P = 300.000 Watt

P = 300kWatt

#LearnWithEXO

6 0

3 years ago

Question 1<br> 2.5 cm=<br> mm

Nookie1986 [14]

There’s 10mm in a cm: 22mm

8 0

3 years ago

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A boat travels at 30 mph for a huge, solid cliff that is about 3,000 meters away. When the horn on the boat makes a toot, you ca

AleksandrR [38]

Answer:The frequency of the echo is slightly decreased

Explanation:

Given

speed of boat $=30\ mph$

cliff is $3000\ m$ away

when boat is still , suppose t is the time taken by the echo to reach observer on the boat

But as soon as boat starts moving the distance between cliff and boat decreasing and time for echo to reach observer also decreases

and we know $time \propto \frac{1}{frequency}$

therefore frequency of the echo slightly decreased.

5 0

3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).