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2 years ago

In order for rejuvenation to take place in a stream, what must occur first?

1 answer:

7 0

The land over the stream flows must uplift - Rejuvenation is the process by which a stream r<span>enews its activity, erosive power, etc. by the uplift or removal of a barrier in the stream's bed. This lowers the base level of the stream and it so that it can begin its downward erosion again.</span>

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A fairgrounds ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follo

cestrela7 [59]

Answer:

13.37 rev/min

Explanation:

acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration ( $a_c$ ) = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².

r = 9 m

Centripetal acceleration ( $a_c$ ) is given by:

$a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s$

The velocity (v) is given by:

v = ωr; where ω is the angular velocity

Hence:

ω = v/r = 12.6 / 9

ω = 1.4 rad/s

ω = 2πN

N = ω/2π = 1.4 / 2π

N = 0.2228 rev/s

N = 13.37 rev/min

4 0

2 years ago

A tiny boat is floating in a metal bucket. If you build a fire under the bucket ,how will most of the heat be transferred from t

Dimas [21]

The molecules will heat up and move faster, some evaporating and turning to gas, the toy boat will heat up if made of conducting materials but otherwise unchanged. The water will also start to boil.

7 0

2 years ago

What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from

Bas_tet [7]

Answer:

check photo for solve

Explanation:

7 0

2 years ago

ball A of mass 5.0 kg moving at 20 meters per second collides with ball B of unknown mass moving at 10. meters per second in the

antiseptic1488 [7]

Answer:The mass of ball B is 10 kg.

Explanation;

Mass of ball A = $M_A=5 kg$

Velocity of the ball A before collision: $U_A=20 m/s$

Velocity of ball A after collision= $V_A=10 m/s$

Mass of ball B= $M_B$

Velocity of the ball B before collision: $U_B=10 m/s$

Velocity of ball B after collision= $V_B=15 m/s$

$M_AV_A+M_BV_B=M_AU_A+M_BU_B$

$5 kg\times 10 m/s+M_B\times 15=5 kg\times 20m/s+M_B\times 10m/s$

$M_B=10kg$

The mass of ball B is 10 kg.

8 0

2 years ago

Froghopper insects have a typical mass of around 11.3 mg and can jump to a height of 58.8 cm. The takeoff velocity is achieved a

allochka39001 [22]

Answer:

2874.33 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s$

Now H-h = 0.588 - 0.002 = 0.586 m

The final velocity will be the initial velocity

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.586\\\Rightarrow a=-\frac{0.586\times -9.81}{0.002}\\\Rightarrow a=2874.33\ m/s^2$

Acceleration of the frog is 2874.33 m/s²

6 0

3 years ago