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wolverine [178]
3 years ago
15

A boy rides a sled down a steep, snow.covered hill. Which of

Physics
2 answers:
shtirl [24]3 years ago
5 0

Answer:

Potential energy decreases and kinetic energy increases.

Explanation:

That was the answer on my CK12

Travka [436]3 years ago
4 0

Answer:

a. Potential energy decreases and Kinetic energy increases

Explanation:

Because as he comes down due to its steepness the speed of the boy or you can say his KE increases and since he comes from a high position (hill) to the lower ground his potential energy decreases simultaneously

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Hunk travelled 540 km at an average speed of 90 km/h. If Tim completed the same amount of distance with 1 hour and 30 min less t
lina2011 [118]

Answer:120

Explanation:it just is trust me

8 0
3 years ago
The roller coaster from problem #1 then tops a second hill at 15.0 m/s, how high is the second hill?
navik [9.2K]

Answer:

68.5

Explanation:

4 0
2 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
An undersea research chamber is spherical with an external diameter of 5.50 m . The mass of the chamber, when occupied, is 87600
dmitriy555 [2]

Answer:

the buoyant force on the chamber is F = 7000460 N

Explanation:

the buoyant force on the chamber is equal to the weight of the displaced volume of sea water due to the presence of the chamber.

Since the chamber is completely covered by water, it displaces a volume equal to its spherical volume

mass of water displaced = density of seawater * volume displaced

m= d * V , V = 4/3π* Rext³

the buoyant force is the weight of this volume of seawater

F = m * g = d * 4/3π* Rext³ * g

replacing values

F = 1025 kg/m³ * 4/3π * (5.5m)³ * 9.8m/s² = 7000460 N

Note:

when occupied the tension force on the cable is

T = F buoyant - F weight of chamber  = 7000460 N - 87600 kg*9.8 m/s² = 6141980 N

4 0
3 years ago
A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp
NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

f=12GHz=12\cdot 10^9 Hz is the frequency

c=3.0\cdot 10^8 m/s is the speed of light

So the wavelength of the beam is

\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

y=\frac{m\lambda D}{a}

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m

and so, the diameter is

d=2y = 750 m

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

r=\frac{750 m}{2}=375 m

So the area is

A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2

And since the power is

P=100 kW = 1\cdot 10^5 W

The average intensity is

I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2

4 0
3 years ago
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