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3 years ago

What would we need to know to calculate work and power

Physics

1 answer:

san4es73 [151]3 years ago

7 0

To calculate Work we need Force and Distance or Total Change in Energy and Heat.

To calculate Power we need Current and Voltage or Voltage and Resistance.

You might be interested in

A 15kg projectile is moving with a velocity of 25m/s calculate its momentum

aliina [53]

The momentum of an object is given by:

p = mv

m is the object's mass and v is its velocity.

Given values:

m = 15kg

v = 25m/s

Plug in the values and solve for p:

p = 15*25

p = 375kg×m/s

8 0

4 years ago

The aorta is approximately 25 mm in diameter. The mean pressure there is about 100 mmHg and the blood flows through the aorta at

Ksenya-84 [330]

Answer:

Explanation:

25 mm diameter

r₁ = 12.5 x 10⁻³ m radius.

cross sectional area = a₁

Pressure P₁ = 100 x 10⁻³ x 13.6 x 9.8 Pa

a )

velocity of blood v₁ = .6 m /s

Cross sectional area at blockade = 3/4 a₁

Velocity at blockade area = v₂

As liquid is in-compressible

a₁v₁ = a₂v₂

a₁ x .6 m /s = 3/4 a₁ v₂

v₂ = .8m/s

b )

Applying Bernauli's theorem formula

P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²

100 x 10⁻³ x 13.6 x10³x 9.8 + 1/2 X 1060 x .6² = P₂ + 1/2x 1060 x .8²

13328 +190.8 = P₂ + 339.2

P₂ = 13179.6 Pa

= 13179 / 13.6 x 10³ x 9.8 m of Hg

P₂ = .09888 m of Hg

98.88 mm of Hg

8 0

4 years ago

You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius

Gennadij [26K]

Answer:

t = 0.0735 m

Explanation:

Angular acceleration of the flywheel is given as

$\alpha = 3 rad/s^2$

now after t = 8 s the speed of the flywheel is given as

$\omega = \alpha t$

$\omega = 3 \times 8$

$\omega = 24 rad/s$

now rotational kinetic energy of the wheel is given as

$K = \frac{1}{2}I\omega^2$

$K = \frac{1}{2}(\frac{1}{2}mR^2)(24^2)$

$800 = \frac{1}{4}m(0.23)^2(24^2)$

$m = 105 kg$

now we have

$m = \rho (\pi R^2) t$

$105 = 8600(\pi \times 0.23^2) t$

$t = 0.0735 m$

4 0

3 years ago

Why do people delte your answer or the question on here i is new so idk can someone plz help?

natita [175]

they are prob deleting you answers/questions because you are violating the Terms of Use or the Community Guidelines.

if you want to know how to get around Brainly and know the rules you can read the

Terms or Use: brainly.com/pages/terms_of_use

and the

Community Guidelines: brainly.com/pages/community_guidelines

5 0

3 years ago

At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial

elena55 [62]

Answer : The partial pressure of $SO_3$ is, 67.009 atm

Solution : Given,

Partial pressure of $SO_2$ at equilibrium = 30.6 atm

Partial pressure of $O_2$ at equilibrium = 13.9 atm

Equilibrium constant = $K_p=0.345$

The given balanced equilibrium reaction is,

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}$

Now put all the values of partial pressure, we get

$0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}$

$p_{SO_3}=67.009atm$

Therefore, the partial pressure of $SO_3$ is, 67.009 atm

6 0

3 years ago