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3 years ago

Hii can someone pls help me??

Physics

2 answers:

Alexus [3.1K]3 years ago

7 0

F=51N....hope this helped out .........

ipn [44]3 years ago

3 0

F = ma

F = 8.5 × 6

F = 51 N

................

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A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is

valina [46]

Answer:

4.75 m/s

Explanation:

The computation of the velocity of the existing water is shown below:

Data provided in the question

Tall = 2 m

Inside diameter tank = 2m

Hole opened = 10 cm

Bottom of the tank = 0.75 m

Based on the above information, first we have to determine the height which is

= 2 - 0.75 - 0.10

= 2 - 0.85

= 1.15 m

We assume the following things

1. Compressible flow

2. Stream line followed

Now applied the Bernoulli equation to section 1 and 2

So we get

$\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2$

where,

P_1 = P_2 = hydrostatic

z_1 = 0

z_2 = h

Now

$\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\ \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}$

= 4.7476 m/sec

= 4.75 m/s

6 0

3 years ago

Which one of the following statements concerning superconductors is false? A constant current can be maintained in a superconduc

julsineya [31]

Answer:

All materials are superconducting at temperatures near absolute zero kelvin.

Explanation:

All materials are superconducting at temperatures near absolute zero kelvin is false concerning superconductors.

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3 years ago

If the spring constant is doubled , what value does the period have for a mass on a spring?

zhuklara [117]

Answer:

D. The period would decrease by sqrt (2)

Explanation:

The period of a mass-spring system is given by:

$T=2\pi \sqrt{\frac{m}{k}}$

where

m is the mass

k is the spring constant of the spring

If the spring constant is doubled,

k' = 2k

So the new period will be

$T'=2\pi \sqrt{\frac{m}{(2k)}}=\frac{1}{\sqrt{2}}(2\pi \sqrt{\frac{m}{k}})=\frac{T}{\sqrt{2}}$

So the correct answer is

D. The period would decrease by sqrt (2)

8 0

3 years ago

Read 2 more answers

7. Which of the following is NOT always true about a synthesis reaction!

Rainbow [258]

Answer:

Explanation:

the formula is a + b = ab

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4 years ago

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

4 years ago