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3 years ago

What mineral is most abundant in granite

Physics

1 answer:

77julia77 [94]3 years ago

3 0

I believe it would be orthoclase feldspar, quartz, micas, and amphiboles are the most abundant in granite.<span />

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Comparing X-ray to radio wave, which has a shorter wavelength?

Daniel [21]

Answer:

x-ray

Explanation:

X-rays can penetrate through matter because of their high frequencies while radio waves are not able to do so as much. If you think about it, taking an x-ray requires a lot more energy than for radio waves to allow you to listen to songs on the radio. Hope this helps :)

6 0

3 years ago

Explain why you cannot use mass or volume alone to identify substances.

Alex17521 [72]

Objects can have the same mass (but different <span>compositions). Only mass or volume cannot tell you if the object is solid or vo</span>lumes) or same volume (but different masses)

4 0

3 years ago

Juan's mother drives 12.5 miles southwest to her favorite shopping mall. What is the velocity of her

SVETLANKA909090 [29]

Answer:

Calculate using the formula

Explanation:

velocity= displacement (m)/time(s)

1 mile =1.6km

1km=1000m

7 0

2 years ago

A ball droped from a building. How fast is it traveling after falling 3.55s

algol [13]

Answer:

d = 61.75 m

Explanation:

Given that,

A ball droped from a building.

We need to find how fast is it traveling after falling 3.55 s.

As it is dropped, its initial velocity is equal to 0.

Let d is the distance it covers after falling 3.55 s.

We can use second equation of motion to find d.

$d=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a =g

$d=\dfrac{1}{2}gt^2\\\\d=\dfrac{1}{2}\times 9.8\times (3.55)^2\\\\d=61.75\ m$

So, it will cover 61.75 m after falling 3.55 seconds.

4 0

3 years ago

A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. What is the m

Akimi4 [234]

Answer:

8.6 m

Explanation:

The motion of a soccer ball is a motion of a projectile, with a uniform motion along the horizontal (x-) direction and an accelerated motion along the vertical (y-) direction, with constant acceleration $a=g=-9.8 m/s^2$ towards the ground (we take upward as positive direction, so acceleration is negative).

The initial velocity along the vertical direction is

$v_{y0} = v_0 sin \theta = (26 m/s)(sin 30^{\circ})=13 m/s$

Now we can consider the motion along the vertical direction only. the vertical velocity at time t is given by:

$v_y(t)=v_{y0} +at$

At the point of maximum height, $v_y(t)=0$ , so we can find the time t at which the ball reaches the maximum height:

$0=v_{y0}+at\\t=-\frac{v_{y0}}{a}=-\frac{13 m/s}{-9.8 m/s^2}=1.33 s$

And now we can use the equation of motion along the y-axis to find the vertical position of the ball at t=1.33 s, which corresponds to the maximum height of the ball:

$y(t)=v_{y0}t + \frac{1}{2}at^2=(13 m/s)(1.33 s)+\frac{1}{2}(-9.8 m/s^2)(1.33 s)^2=8.6 m$

4 0

2 years ago