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3 years ago

Factors that affect pressure in liquids

Physics

2 answers:

Alona [7]3 years ago

7 0

Two factors influence the pressure of fluids. They are the depth of the fluid and its density.

Maksim231197 [3]3 years ago

6 0

Answer:

Depth and density of the liquid determine the amount of pressure in a liquid.

Explanation:

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LASIK eye surgery uses pulses of laser light to shave off tissue from the cornea, reshaping it. A typical LASIK laser emits a 1.

Fittoniya [83]

Answer:

143 kW

Explanation:

Given that

Diameter of the beam, d = 1 mm

Wavelength of the beam, λ = 193 nm

Time used by the pulse, t = 14 ns

Energy of the pulse, U = 2 mJ

Recall that Power can be mathematically calculated using the relation,

Power = Work Done / Time,

To solve this, we apply the formula

P = U / Δt

P = 2*10^-3 J / 14*10^-9 s

P = 142857 W

P = 143 kW

3 0

3 years ago

You have a source of energy containing 21 gj of energy at 600k how much this energy can be converted to work when rejecting heat

sweet [91]

Answer:

Available energy = 35 x 10⁶ J

Explanation:

Given:

Amount of energy (Q) = 21 gj = 21 x 10⁹ J

Temperature T1 = 600 k

Temperature T0 = 27 + 273 = 300k

Find:

Available energy

Computation:

Available energy = Q[1/T0 - 1/T1]

Available energy = 21 x 10⁹ J[1/300 - 1/600]

Available energy = 35 x 10⁶ J

4 0

2 years ago

A block is released from rest, at a height h, and allowed to slide down an inclined plane. There is friction on the plane. At th

GREYUIT [131]

Here the block has two work done on it

1. Work done by gravity

2. Work done by friction force

So here it start from height "h" and then again raise to height hA after compressing the spring

So work done by the gravity is given as

$W_g = m_A g(h - h_A)$

Now work done by the friction force is to be calculated by finding total path length because friction force is a non conservative force and its work depends on total path

$W_f = -(\mu m_A g cos\theta)(\frac{h}{sin\theta} + \frac{h_A}{sin\theta})$

$W_f = -\mu m_A g cot\theta(h + h_A)$

Total work done on it

$W = m_A g(h - h_A) - \mu m_A g cot\theta(h + h_A)$

So answer will be

None of these

7 0

3 years ago

Read 2 more answers

Blocks A (mass 3.50 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block

Vedmedyk [2.9K]

Answer:

(a) V (A) = 0.7 m/s,

(b) V (A) = 0.7 m/s,

(d) u= - 0.60 m/s

(e) v = 0.75 m/s

Explanation:

Given:

M(A) =3.50 Kg, M(B)=6.50 Kg, V(A) = 2.00 m/s, V(B) = 0 m/s

Sol:

a) law of conservation of momentum

M(a) x V(A) + M(B) x V(B) = ( M(a) + M(B) ) V (let V is Common Velocity of Both block)

so 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = (3.50 Kg + 6.50 Kg ) V

after solving V = 0.7 m/s

After the collision the velocities of the both block will be as the the spring is compressed maximum.

V (A) = 0.7 m/s

b) V(A) = 0.7 m/s ( Part (a) and Part (a) are repeated )

c) as stated above the in the Part (a)

V(B) = 0.7 m/s

d) When the both blocks moved apart after the collision:

Let u=velocity of block A after the collision.

and v = velocity of block B after the collision.

then conservation of momentum

M(a) x V(A) + M(B) x V(B) = M(a) x v + M(B) x u

⇒ 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = 3.50 Kg x u + 6.50 Kg x v

⇒ 2.00 m/s = u + 1.86 v -----eqn (1) ( dividing both side by 3.50 Kg)

For elastic collision

the velocity relative approach = velocity relative separation

so 2.00 m/s = v-u ----- eqn (2)

⇒v = u + 2.00 m/s

putting this value in eqn (1) we get

2.00 m/s = u + 1.86 (v + 2.00 m/s)

u= - 0.60 m/s

e) putting v= 2.00 m/s in eqn (1)

2.00 m/s = - 2.32 m/s + 1.86 v

v = 0.75 m/s

5 0

3 years ago

a space probe with a mass of 4000 kg expels 3,500 of its mass at a velocity of 2000 m/s. what is the velocity of the remaining 5

Sonja [21]

Answer:

4.16×103 m/s

Explanation:

5 0

3 years ago