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Lina20 [59]
3 years ago
10

To clean the outside of your house you rent a small high-pressure water sprayer. The sprayer’s pump delivers slow-moving water a

t a pressure of 9000000.0 Pa (about 90.0 atmospheres). How fast can this water move if all of its pressure potential energy becomes kinetic energy as it flows through the nozzle of the sprayer?
Physics
1 answer:
Zigmanuir [339]3 years ago
5 0

Answer:

Velocity is 134.16 m/s

Explanation:

Slow moving water pressure is the dynamic pressure. Dynamic pressure is actually the kinetic energy per unit volume of a fluid particle. It is equal to the difference between stagnation pressure and static pressure and also can be derived from Bernoulli's equation.

Formula for Dynamic Pressure (P) is

               P = 0.5 d v²

Where d is density of fluid and v is the velocity.

Density of water = 1000 kg/m³

P = 9000000 Pa

V = ?

                      P = 0.5 d v²  

         9000000 = 0.5 × 1000 × v²

                       v = √(9000000 ÷ 500)

                       v = 134.16 m/s

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If the loop the car is currently on has a radius of 26.0 m , find the minimum height h so that the car will not fall off the tra
ki77a [65]

The minimum height h is 65m so that the car will not fall off the track at the top of the circular part of the loop.

<h3>What is mechanical energy?</h3>

Potential energy plus kinetic energy are combined to form mechanical energy. According to the principle of mechanical energy conservation, mechanical energy is constant in an isolated system when only conservative forces are acting on it. Potential energy increases when an object moves in the opposite direction of a conservative net force. Kinetic energy also changes as an object's speed, not velocity, changes. However, nonconservative forces, such as frictional forces, will always be present in real systems; however, if these forces are of minimal magnitude, mechanical energy changes little, making the idea of its conservation a reasonable approximation.

For completing the vertical circle the minimum speed at the bottom must be \sqrt{5gR}

so conserving mechanical energy

mgh=\frac{1}{2} m (v_{bottom})^{2}

gh=\frac{1}{2} 5gR

⇒ h= \frac{5}{2} \times 26

h = 65m

To learn more about mechanical energy, visit:

brainly.com/question/24443465

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5 0
2 years ago
Young's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a s
PSYCHO15rus [73]

Answer:

Explanation:

wave length of light λ = 502 nm

screen distance D = 1.2 m

width of one fringe = 10.2 mm / 20

= .51 mm

fringe width = λ D / a  , a is separation of slits

Puting the values given

.51 x 10⁻³ =  502 x 10⁻⁹ x 1.2 / a

a = 502 x 10⁻⁹ x 1.2 / .51 x 10⁻³

= 1181.17 x 10⁻⁶ m

1.18 x 10⁻³ m

= 1.18 mm .

8 0
3 years ago
A carnival game consists of a two masses on a curved frictionless track, as pictured below. The player pushes the larger object
Harman [31]

Answer:

v₁₀ = 1.90 m / s

Explanation:

In this exercise we are given the maximum height data, with energy we can know how fast the body came out

Final mechanical energy, maximum height

    Em_{f} = U = m g h

Initial mechanical energy, in the lower part of the track

    Em₀ = K = ½ m v²

    Em=   Em_{f}

    ½ m v² = m g h

    v = √ 2gh

Now we can use the moment to find the speed with which objects collide

The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and  final velocity v

Initial before the crash

    p₀ = M v₁₀ + 0

Final after the crash

      p_{f} = M v1f + m v

   p₀ =   p_{f}

   M v₁₀ = M v_{1f}+ m v

As the shock is elastic the kinetic energy is conserved

     K₀ = K_{f}

    ½ M v₁₀² = ½ M v_{1f}² + ½ m v²

Let's write the system of equations

    M v₁₀ = M  v_{1f} + m v

    M v1₁₀² = M v_{1f}² + m v²

We cleared v1f in the first we replaced in the second

   v_{1f} = (M v₁₀ - mv) / M

    M v₁₀² = M (M v₁₀ - mv)² / M² + m v²

    M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²

     v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0

     2 m v₁₀ - v (m + 1) m/ M = 0

     v₁₀ = v (m +1) / (2M)

Let's substitute the value of v

     v1₁₀= √ (2gh) (m +1) / (2M)

Let's calculate

    v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2  5.41)

    V₁₀ = 7.668 (2.68) / 10.82

   v₁₀ = 1.90 m / s

5 0
3 years ago
A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. A horizontal force of 2.5 N is applied o
Masja [62]

Answers is  F=7.84 N

Friction force resists the effect of horizontal force and trying to approch to a limiting force.

we have formula for limiting friction force between block and floor

               F=Ц N

where N=mg

putting values we get answer.

Download pdf
3 0
3 years ago
Two objects of masses m1 = 0.56 kg and m2 = 0.88 kg are placed on a horizontal
ziro4ka [17]

Explanation:

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3 0
3 years ago
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