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sammy [17]
2 years ago
13

A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed of 10 m/s. If air

resistance is negligible, the distance from the base of the cliff to where the rock hits the level ground below the cliff is most nearly
a.5m
b.10m
c.20m
d.40m
e.200m
Physics
1 answer:
scZoUnD [109]2 years ago
5 0

Answer:

c.20

Explanation:

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The fastest winds and heaviest rain is found in the eye of the hurricane.
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If a rock climber accidentally drops a 52.5-g piton from a height of 325 meters, what would its speed be just before striking th
alex41 [277]
Ignoring air resistance, the Kinetic energy before hitting the ground will be equal to the potential energy of the Piton at the top of the rock.  
So we have 1/2 MV^2 = MGH 
V^2 = 2GH 
V = âš2GH 
V = âš( 2 * 9.8 * 325)  
V = âš 6370
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6 0
3 years ago
Water of density 1000 kg/m3 falls without splashing at a rate of 0.373 L/s from a height of 40.5 m into a 0.64 kg bucket on a sc
Sphinxa [80]

Answer:

       F_scale = 20.18 N

Explanation:

The scale reading corresponds to two factors, the first the weight of the water in the container and the second the force of the liquid that is falling at the moment of reading.

* Let's find the amount of liquid in the container for a time of t = 2.93 s

Let's use a direct proportion rule. If 0.373 l falls in one second at t = 2.93 s, how many liters are there

        V_{water} = 2.93 s (0.373 l / 1s) = 1.09 l

        V_{water} = 1.09 10⁻³ m³

the amount of water is

       ρ = m / V

       m = ρ V

       m = 1000 1.09 10⁻³

       m = 1.09 kg

so the weight of the liquid in the container for this time is

       W = mg

       W = 1.09 9.8

       W = 10.68 N

* Let's look for the force of the falling jet

Let's use Bernoulli's equation, where the subscript 1 is for the container and the subscript 2 is for the water at a height h

        P₁ + 1/2 ρ g v₁² + ρ g y₁ = P₂ + 1/2  ρ g v₂² + ρ g y₂

In this case, the water falls freely, so the external pressure is atmospheric.

         P₂ = P_{atm}

since they indicate that the water falls, we assume that its initial velocity is zero v₂ = 0

let's use kinematics to find the speed of a drop when it reaches the container y = 0

         v² = v₀² - 2 g (y-y₀)

         v = \sqrt{0 -2 g ( 0-y_o)}

let's calculate

         v = √(2 9.8 40.5)

         v = 28.17 m / s

this is the speed in the container v₁ = 28.17 m / s

the height from where it falls is y₂ = 40.5 and reaches the container y₁ = 0

we substitute in Bernoulli's equation

         P₁ +1/2 ρ g v₁² + 0 = P_{atm} + 0 + ρ g y₂

         P₁ + ½ ρ g v₁² = P_{atm} + ρ g y₂

         P₁ = P_{atm} + ρ g y₂ - ½ ρ g v₁²

         P₁ = 1 10⁵ + 1000 9.8 40.5 - ½ 1000 28.17²

         P₁ = 1 10⁵ + 3.97 10⁵ - 3.69 10⁵

         P₁ = 1.28 10⁵ Pa

The definition of Pressure is

         P = F / A

         F = P A

We must suppose a time to carry out the reading suppose an average time of the modern equipment t = 0.1 s, in this time how much is now arriving

          m₂ = 0.373 0.2 = 0.0746 l = 0.0746 10⁻³ m³

the volume is V = A l

if the length of l = 1 m

A = 0.0746 10⁻³ m³ = 7.45 10⁻⁵ m²

the force of this jet is

            F = P A

            F = 1.28 10⁵  7.46 10⁻⁵

            F = 9.5 N

with these data let's use the equilibrium equation

           F_ scale -W - F = 0

           F_scale = W + F

           F_scale = 10.682 + 9.5

           F_scale = 20.18 N

4 0
2 years ago
Hey solution that contains all of the salute it can normally hold at a given temperature is
ankoles [38]

Answer:

A saturated solution

Explanation:

A saturated solution is one that contains the most amount of solute that can be dissolved in it at a given temperature

An example of a saturated solution is carbonated water, which readily gives off bubbles of carbon dioxide gas from areas within the solution to the region above the top surface of the gas in liquid solution

A saturation solution of salt in water can be created by continuing to dissolve salt in a given amount of water until it can no longer dissolve any more salt. However, heating the saturated salt solution, increases the amount of salt that can be dissolved.

Therefore, a solution that contains all of the solute it can normally hold at a given temperature is <u>a saturated solution</u>

8 0
2 years ago
Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long
Ksivusya [100]

Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

<u>A) Determine how long the clock run on the battery. use the relation below</u>

q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence  t = 2 * 10^7 secs

hence the time = 231.48  days

<u>B) Determine how many electrons per second flowed </u>

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

q = 0.5 * 10^-3 coulomb ( charge flowing per electron )

back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

hence : n = 3.125 * 10^15

8 0
2 years ago
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