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3 years ago

Check all that are true about waterfalls

1 answer:

4 0

Waterfalls are created when a river flows following a descending rapid slope. The waterfall, then, flows from the source (where it starts) to the mouth (where it ends).
Waterfalls are created when the erosion of the rocks at the bottom of the slope is more powerful than the erosion of the rocks on the top.
After many years the water is able to erode the rocks on the top as well, and the waterfall slowly disappears.

Therefore the options that apply are:
b) waterfalls move towards their mouth;
c) the top or cap rock is resistant to erosion;
<span>f) waterfalls indicate a youthful river </span>

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The normal force which the path exerts on a particle is always perpendicular to the _________________ tangent to the path. trans

Marianna [84]

Tangent to the pathh.

7 0

3 years ago

An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of 25.0 with the horizontal. The coeffici

uranmaximum [27]

Answer:

V = 10.88 m/s

Explanation:

V_i =initial velocity = 0m/s

a= acceleration= gsinθ- $\mu_k$ cosθ

putting values we get

a= 9.8sin25-0.2cos25= 2.4 m/s^2

v_f= final velocity and d= displacement along the inclined plane = 10.4 m

using the equation

$v^2_f=v^2_i-2as$

$v^2_f=0^2-2(2.4)(10.4)$

v_f= 7.04 m/s

let the speed just before she lands be "V"

using conservation of energy

KE + PE at the edge of cliff = KE at bottom of cliff

(0.5) m V_f^2 + mgh = (0.5) m V^2

V^2 = V_f^2 + 2gh

V^2 = 7.04^2 + 2 x 9.8 x 3.5

V = 10.88 m/s

6 0

3 years ago

Can someone help me?

Gwar [14]

24- series
25- parallel
26- no, because they’re connected in series
27- yes, because they’re connected in parallel

7 0

3 years ago

what is the electric potential at point A in the electric field created by a point charge of 5.5 • 10^-12 C? estimate k as 9.00

Hitman42 [59]

The electric potential at point A in the electric field= 0.099 x 10 ⁻¹v

<u>Explanation</u>:

Given data,

charge = 5.5 x 10¹² C

k =9.00 x 10⁹

The electric potential V of a point charge can found by,

V= kQ / r

Assuming, r=5.00×10⁻² m

V= 5.5 x 10⁻¹²C x 9.00 x 10⁹ / 5.00×10⁻² m

V= 49.5 x 10⁻³/ 5.00×10⁻²

Electric potential V= 0.099 x 10⁻¹v

3 0

2 years ago

If a student flicks a stationary 0.1 kg ball with 5N of force for 0.1 seconds. What is its final

Alisiya [41]

5m/s

Explanation:

Given parameters:

Mass of ball = 0.1kg

Force on the ball = 5N

time taken = 0.1s

Unknown:

final speed of the ball = ?

Solution:

According to newton's second law "the net force on a body is the product of its mass and acceleration".

Force = mass x acceleration equation 1

Acceleration =

V is the final velocity

U is the initial velocity

T is the time taken

U = O since it is a stationary body;

a = $\frac{V}{T}$

Input "a" into equation 1

F = m x $\frac{V}{T}$

5 = 0.1 x $\frac{V}{0.1}$

V = 5m/s

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

3 0

3 years ago