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3 years ago

The surface below sedimentary rocks that overlie igneous or metamorphic rocks is termed a(n) ____.

Physics

1 answer:

Luden [163]3 years ago

4 0

The surface below sedimentary rocks that overlie igneous or metamorphic rocks is a(n) nonconformity.

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How is it possible for the same object to have different amounts of gravitational potential energy?

Arlecino [84]

Answer:

In Newtonian mechanics, the gravitational energy possessed by a mass, because of the gravitational field produced by a second mass.

Explanation:

8 0

3 years ago

Altitude is the angle measured above ____. North Pole horizon equator zenith

juin [17]

Altitude is the angle measured above the horizon

5 0

3 years ago

Read 2 more answers

A 25 kg child is riding on a swing. If the child travels 8.9 m/s at the bottom of their swing, how high into the air is the chil

Setler [38]

Answer:

h = 4.04 m

Explanation:

Given that,

Mass of a child, m = 25 kg

The speed of the child at the bottom of the swing is 8.9 m/s

We need to find the height in the air is the child is able to swing. Let the height is h. Using the conservation of energy such that,

$mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}$

Put all the values,

$h=\dfrac{(8.9)^2}{2\times 9.8}\\\\h=4.04\ m$

So, the child is able to go at a height of 4.04 m.

7 0

3 years ago

A 10 kg migratory swan cruises at 20m/s. A calculation that takes into ac-count the necessary forces shows that this motion requ

IgorC [24]

Answer:

Part A:

Distance=864000 m=864 km

Part B:

Energy Used=ΔE=8638000 Joules

Part C:

$\frac{\triangle m}{m}=0.004998=0.49985\%$

Explanation:

Given Data:

v=20m/s

Time =t=12 hours

In Secs:

Time=12*60*60=43200 secs

Solution:

Part A:

Distance = Speed**Time

Distance=v*t

Distance= 20*43200

Distance=864000 m=864 km

Part B:

Energy Used=ΔE= Energy Required-Kinetic Energy of swans

Energy Required to move= Power Required*time

Energy Required to move=200*43200=8640000 Joules

Kinetic Energy= $\frac{1}{2}mv^2$

$K.E\ of\ Swans=\frac{1}{2} *10*(20)^2=2000\ Joules$

Energy Used=ΔE=8640000 -2000

Energy Used=ΔE=8638000 Joules

Part C:

Fraction of Mass used=Δm/m

For This first calculate fraction of energy used:

Fraction of energy=ΔE/Energy required to move

ΔE is calculated in part B

Fraction of energy=8638000/8640000

Fraction of energy=0.99977

Kinetic Energy= $\frac{1}{2}mv^2$

Now, the relation between energies ratio and masses is:

$\frac{\triangle E}{E}=\frac{\triangle m}{2m}v^2$

$\frac{\triangle m}{m}=\frac{2}{v^2} *\frac{\triangle E}{E}\\\frac{\triangle m}{m}=\frac{2}{20^2} *0.99977$

$\frac{\triangle m}{m}=0.004998=0.49985\%$

3 0

3 years ago

If the pitched ball was traveling 77 mph before stanton's bat hit it and 120 mph after his bat hit it, by what amount did the sp

Mice21 [21]

here we will use the concept of Newton's III law

as per Newton's III law the impulse given to the ball is same as the impulse lost by the bat

So here we will say

impulse gain by the ball = impulse lost by the bat

$m_1(v_f - v_i) = m_2(\Delta v)$

given that

$m_1 = 5 ounce$

$m_2 = 32 ounce$

For ball the change in speed will be

$v_f - v_i = (120 - 77)mph$

now from above equation

$5\times (120 - 77) = 32 \times \Delta v$

$\Delta v = 6.72 mph$

so speed of bat will decrease by 6.72 mph

3 0

3 years ago