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qaws [65]
3 years ago
15

According to Bernoulli's fluid formula a An increase in the speed will lower the internal pressure b An increase in the speed wi

ll increase the internal pressure c The internal pressure doesn't change with the speed
Physics
1 answer:
lorasvet [3.4K]3 years ago
5 0

Answer:

a An increase in the speed will lower the internal pressure

Explanation:

Bernoulli's fluid formula

P_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_1+\frac{1}{2}\rho v_2^2+\rho gh_2

where

P = Pressure

ρ = Density of fluid

g = Acceleration due to gravity

h = Height

v = Velocity of fluid

If there is no change in height then we get

P_1+\frac{1}{2}\rho v_1^2=P_1+\frac{1}{2}\rho v_2^2\\\Rightarrow P+\frac{1}{2}\rho v^2=constant

According to the Bernoulli's principle when the speed of the fluid is larger in a region of streamline flow the pressure is smaller in that region. From the above equation it can be seen that increase in speed should simultaneously reduce pressure in order for their sum to be constant.

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1. A change in an object's speed has a(n) _________ effect on its kinetic energy than a change in its mass.
vredina [299]
A change in an object's speed has a(n) _________ effect on its kinetic energy than a change in its mass = <span>A greater effect.</span>
4 0
3 years ago
Read 2 more answers
A car is driving towards an intersection when the light turns red. The brakes apply a constant force of 1,398 newtons to bring t
dmitriy555 [2]

Answer:

the initial velocity of the car is 12.04 m/s

Explanation:

Given;

force applied by the break, f = 1,398 N

distance moved by the car before stopping, d = 25 m

weight of the car, W = 4,729 N

The mass of the car is calculated as;

W = mg

m = W/g

m = (4,729) / (9.81)

m = 482.06 kg

The deceleration of the car when the force was applied;

-F = ma

a = -F/m

a = -1,398 / 482.06

a = -2.9 m/s²

The initial velocity of the car is calculated as;

v² = u² + 2ad

where;

v is the final velocity of the car at the point it stops = 0

u is the initial velocity of the car before the break was applied

0 = u² + 2(-a)d

0 = u² - 2ad

u² = 2ad

u = √2ad

u = √(2 x 2.9 x 25)

u =√(145)

u = 12.04 m/s

Therefore, the initial velocity of the car is 12.04 m/s

7 0
3 years ago
The brakes on your automobile are capable of creating a deceleration of 5.0 m/s2. If you are going 135 km/h and suddenly see a s
poizon [28]

Answer:

3.33 seconds

Explanation:

We can use the velocity formula [ v = u + at ] to solve.

Find the value "u".

135km/h -> 135km*1000m/3600s -> 37.5m/s

Find the value "v".

75km/h -> 75km*1000m/3600s -> 20.83m/s

Keep in mind we are dealing with "deceleration" so when we input 5.0m/s into the formula, it will be a negative value.

Now, find "t" which is the value we aren't given with the values we're given in the question.

20.83 = 37.5 - 5t

-16.67 = -5t

3.33 = t

Best of luck!

8 0
3 years ago
Free electromagnetic oscillations occur in an electrical circuit. Knowing the characteristic dimensions of the elements, L=20mH
kirill115 [55]

Answer:

a. 0.199 ms b. 5.03 kHz c. 0.1 mJ

Explanation:

a. The period of oscillation of an L-C circuit is T = 2π√(LC) where L = inductance = 20 mH = 20 × 10⁻³ H and C = capacitance = 0.005 mF = 5 × 10⁻⁶ F.

So, T = 2π√(LC)

= 2π√(20 × 10⁻³ H × 5 × 10⁻⁶ F)

= 2π√(100 × 10⁻¹¹)

= 2π√(10 × 10⁻¹⁰)

= 2π(3.16 × 10⁻⁵)

= 19.87 × 10⁻⁵

= 1.987 × 10⁻⁴ s

= 1.99 × 10⁻⁴ s

= 0.199 × 10⁻³ s

= 0.199 ms

b. frequency , f = 1/T where T = period = 0.199 × 10⁻³ s.

So, f = 1/0.199 × 10⁻³ s

= 5.03 × 10³ Hz

= 5.03 kHz

c. The electromagnetic energy E = 1/2Li² where L = inductance = 20 × 10⁻³ H and i = current = 100 mA = 0.1 A

So, E = 1/2Li²

= 1/2 × 20 × 10⁻³ H × (0.1 A)²

= 0.1 × 10⁻³ J

= 0.1 mJ

3 0
3 years ago
a boat is moving 1.3m/s N while the steam current is pushing the boat 2.2m/s What is the resultant velocity( I also need the giv
lubasha [3.4K]

Answer:

6) 2.6 m/s, 31°

7) 9.2 m/s

8) 1.2 s

Explanation:

I'll do #6, #7, and #8 as examples.  You can solve #9 using the equation from #7, and #10 using the equation from #8.

6) Take north to be +y and east to be +x.

Given:

vₓ = 2.2 m/s

vᵧ =  1.3 m/s

Find: v

v² = vₓ² + vᵧ²

v² = (2.2 m/s)² + (1.3 m/s)²

v ≈ 2.6 m/s

θ = atan(vᵧ / vₓ)

θ = atan(1.3 / 2.2)

θ ≈ 31°

7) Given:

Δy = -4.3 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-4.3 m)

v ≈ 9.2 m/s

8) Given:

Δy = -6.7 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

-6.7 m = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.2 s

8 0
3 years ago
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