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lisabon 2012 [21]
1 year ago
11

Hold a spoon such that the water will hit the back of the spoon. The handle of the spoon will be horizontal. Turn on the water,

but not full strength though.
Physics
1 answer:
antoniya [11.8K]1 year ago
6 0

This is an experiment that explains the Bernoulli principle. According to this principle, at the bottom of the spoon, there will be a reduction in pressure.

<h3>Explanation of the Bernoulli principle</h3>

So the pressure at the bottom of the spoon reduces because the velocity of the water is larger and the height of the column of water is smaller at this part.

So the spoon is attracted to the water at the point of lowest pressure because objects gravitate to low-pressure points.

Learn more about the Bernoulli Principle at:

brainly.com/question/24623919

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Newton’s law of cooling states that dx dt = −k(x−A) where x is the temperature,t is time, A is the ambient temperature, and k &g
lys-0071 [83]

Answer:

a)X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}

b)Does not affect the long term.

Explanation:

Given that

\dfrac{dx}{dt}=-k(x-A)

A = A0 cos(ωt)

\dfrac{dx}{dt}=-k(x-A_o cos(\omega t))

\dfrac{dx}{dt}+kx=kA_o cos(\omega t)

This is linear equation so integration factor ,I

I=e^{\int kdt}

I=e^{kt}

Now by using linear equation property

e^{kt} X=\int e^{kt} kA_o cos(\omega t) dt +C

e^{kt} X= kA_o \dfrac{e^{kt}}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+C

X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}

b)

at t= 0

X(0)=\dfrac{k^2A_o}{\omega^2+k^2}+C

X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+e^{-kt}\times \left ( X(0)-\dfrac{k^2A_o}{\omega^2+k^2} \right )

So the initial condition does not affect the long term.

5 0
3 years ago
(1) A Newton is 1 kgm/s2 . Is this a fundamental unit or a derived unit?
Mrac [35]
A Newton is a derived unit.
Hope that helped:)
8 0
2 years ago
During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapu
ella [17]

Answer:

The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules

Explanation:

The given spring constant of the of the spring, k = 88.0 N/m

The length by which the hose is stretched, x = 4.20 m

For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose

The elastic potential energy, P.E., of a compressed spring is given as follows;

P.E. = 1/2·k·x²

∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²

1/2 × 88.0 N/m × (4.20 m)² = 776.16 J

The work done on the hose = The potential energy given to hose, P.E. = 776.16 J

5 0
3 years ago
A 2-kg ball is thrown at 3 m/s. What is the ball's momentum? *
crimeas [40]

Answer:

Given

mass (m) =2kg

velocity (v) =3m/s

momentum (p) =?

Form

p=mv

2kgx3m/s

p=6kg.m/s

the momentum of ball's =6kg.m/s

8 0
3 years ago
6.6 kg block initially at rest is pulled to theright along a horizontal, frictionless surfaceby a constant, horizontal force of
neonofarm [45]

Answer:

v = 3.04 m/s

Explanation:

given,

mass of the block, M = 6.6 Kg

horizontal force, F = 12.2 N

distance, L = 2.5 m

initial speed  = 0 m/s

speed of the block,v = ?

we now

Work done is equal to change in Kinetic energy.

Work done = Force x displacement

W = Δ K E

Δ K E = Force x displacement

\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2= F .s

\dfrac{1}{2}\times 6.6 \times v^2 - 0= 12.2\times 2.5

 3.3 v² = 30.5

     v² = 9.242

      v = 3.04 m/s

speed of the block is equal to 3.04 m/s

4 0
2 years ago
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