All categories

4 years ago

Why is it wrong that igneous always has to become sedimentary because of the rock cycle

Physics

1 answer:

Rama09 [41]4 years ago

3 0

Because the rock cycle shows that any rock can become any other type of rock by different processes

You might be interested in

light of wavelength 520 nm falls on a slit that is 3.20 um wide. Estimate how far the first bright-sh difrrection fringe is

Westkost [7]

Answer:

because of the gravity of the earth

6 0

2 years ago

Two objects are dropped from different heights at the same instant. If the first object takes 10.7 seconds to hit the ground and

Lubov Fominskaja [6]

Answer:

The answer to your question is : 521.8 m

Explanation:

Data:

Different heights

Time first object (tfo) = 10.7 s

Time second object (tso)= 14.8 s

Initial speed of both objects(vo) = 0 m/s

a = 9.81 m/s²

Formula:

h = vot + 1/2 (a)(t)² but vo = 0 so, h = 1/2 (a)(t)²

Then, height fo h = 1/2 (9.81)(10.7)² = 561.6 m

height so h = 1/2(9,81)(14.8)² = 1074.4 m

Difference in their heights = 1074.4 m - 561.6 m = 521.8 m

5 0

3 years ago

How do you know the speed of an electromagnetic wave in vacuum?

larisa [96]

Answer:

when huburxvkruxv hfgy

8 0

3 years ago

Read 2 more answers

Can carrying an object be said to be doing work on that object?

Schach [20]

Answer:

Only if that object is being moved while carried

Explanation:

Work = Force • distance travelled

While holding an object would need force equal to that objects mass multiplied by 9.8, if it hasn't moved, then no work is done.

if you were to carry the object and have it travel some distance, then work would be done on that object.

7 0

3 years ago

Read 2 more answers

Suppose your car accelerates from rest to 9 m/s in 2 s. Assume that the acceleration is constant in this time interval. A. What

Stolb23 [73]

Answer :

(a) The acceleration of the car is, $4.5m/s^2$

(b) The distance covered by the car is, 9 m

Explanation :

By the 1st equation of motion,

$v=u+at$ ...........(1)

where,

v = final velocity = 9 m/s

u = initial velocity = 0 m/s

t = time = 2 s

a = acceleration of the car = ?

Now put all the given values in the above equation 1, we get:

$9m/s=0m/s+a\times (2s)$

$a=4.5m/s^2$

The acceleration of the car is, $4.5m/s^2$

By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$ ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity = 0 m/s

t = time = 2 s

a = acceleration of the car = $4.5m/s^2$

Now put all the given values in the above equation 2, we get:

$s=(0m/s)\times (2s)+\frac{1}{2}\times (4.5m/s^2)\times (2s)^2$

By solving the term, we get:

$s=9m$

The distance covered by the car is, 9 m

3 0

3 years ago