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Rudiy27
3 years ago
5

Water flowing through a watershed is effected greatly by which force?

Physics
2 answers:
WITCHER [35]3 years ago
8 0
Hydrotropism is the Wright answer I am misspelling and making this longer for no reason
Ket [755]3 years ago
3 0
Hydroprism is the correct answer.

Explanation:
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A 100 kg bag of sand has a weight on 100 N. When dropped its acceleration is what?
Lyrx [107]
100N describes the weight of the sandbag, while 100kg is the mass of the sandbag.

To calculate acceleration, divide your weight by the mass, thus the accleration is:

100N/100kg = 1(m/s^2)
7 0
3 years ago
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A ball is released from the top of a hill. How fast is the ball going when it reaches the base of the hill? Approximate g as 10
irina [24]
First I’ll show you this standard derivation using conservation of energy:
Pi=Kf,
mgh = 1/2 m v^2,
V = sqrt(2gh)
P is initial potential energy, K is final kinetic, m is mass of object, h is height from stopping point, v is final velocity.
In this case the height difference for the hill is 2-0.5=1.5 m. Thus the ball is moving at sqrt(2(10)(1.5))=
5.477 m/s.
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3 years ago
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What are the physical characteristics of sound waves
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6 0
3 years ago
A sports car is advertised to be able to stop in a distance of 50.0 m from a speed of 80 km. What is its acceleration and how ma
Flauer [41]

Explanation:

Given that,

Initial speed of the sports car, u = 80 km/h = 22.22 m/s

Final speed of the runner, v = 0

Distance covered by the sports car, d = 80 km = 80000 m

Let a is the acceleration of the sports car.  It can be calculated using third equation of motion as :

v^2-u^2=2ad

a=\dfrac{v^2-u^2}{2d}

a=\dfrac{0-(22.22)^2}{2\times 80000}

a=-0.00308\ m/s^2

Value of g, g=9.8\ m/s^2

a=\dfrac{-0.00308}{9.8}\ m/s^2

a=(-0.000314)\ g\ m/s^2

Hence, this is required solution.

8 0
3 years ago
What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resti
Nataliya [291]

Archimedes principle states that

 

F1 / A1 = F2 / A2

F2 = (A2 / A1) * F1

 

Also, formula for the force is F = mg. Formula for the area of the cylinder is A = πr^2, therefore we get

 

F2 = (πr2^2 / πr1^2) * mg

 

Since the diameter of the cylinders are 2 cm and 24 cm, r1 = 12 and r2 = 1.

 

Substituting the values to the derived equation, we get

 

F2 = (π 1^2 / π 12^2) * 2400 * 9.8

F2 = 163.3333 N

 

 

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6 0
3 years ago
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