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aleksklad [387]
3 years ago
5

What happens when sediment builds up over time

Physics
2 answers:
Morgarella [4.7K]3 years ago
6 0
When sediment has built up over time layers of rock start to form, starting with sedimentry rocks, then metamorphic rocks
Arlecino [84]3 years ago
6 0
When sediments build up layers form, like the Grand canyon. over time sediments built up to form the grand canyon
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A truck is traveling at 74.5 kilometers per hour away from you. The driver is blowing the horn which has a frequency of 415 Hz.
ArbitrLikvidat [17]
It would be a high pitched sound. Hope this helps.
8 0
3 years ago
Par 1/2
BaLLatris [955]

part 1

mass = ρ x V

mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg

PE (potential energy)= mgh

PE = 6608.2 kg x 9.81 x 403

PE = 2.61 x 10⁷ J

part 2

megaton of TNT (Mt) =4.2 x 10¹⁵ J

convert PE to Mt:

2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt

4 0
2 years ago
A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A
galben [10]

Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

8 0
2 years ago
An aerobatic airplane pilot experiences weightlessness as she passes over the top of a loop-the-loop maneuver. The acceleration
Debora [2.8K]

Answer:

The radius of the loop is  20.66 km

Explanation:

let the radius of the loop be r

mass of airplane is m

At the top, the pilot experiences two radial forces, which are

1) Gravitational force is  mg

2) Centrifugal forces mv²/r out of the center

When the pilot experiences no weight,

then, mg = mv²/r

r = v² / g

 = 450² / 9.8

 = 20.66 x 10³3

 = 20.66 km

3 0
3 years ago
during a test crash an airbag inflates to stop a dummies forward motion. the dummies mass is 75 kg. if the net force on the dumm
den301095 [7]
Use Force=Mass x Acceleration (newtons second law states force is directly proportional to the acceleration) so you can say that the force is negative and solve for Acceleration.
5 0
3 years ago
Read 2 more answers
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