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weqwewe [10]
3 years ago
5

The water is reflecting light, Is this specular or diffuse reflection? explain your answer​

Physics
1 answer:
FinnZ [79.3K]3 years ago
8 0
Yes the answer is true
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A vertical spring (spring constant =160 N/m) is mounted on the floor. A 0.340-kg block is placed on top of the spring and pushed
AleksandrR [38]

(a) 3.5 Hz

The angular frequency in a spring-mass system is given by

\omega=\sqrt{\frac{k}{m}}

where

k is the spring constant

m is the mass

Here in this problem we have

k = 160 N/m

m = 0.340 kg

So the angular frequency is

\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s

And the frequency of the motion instead is given by:

f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz

(b) 0.021 m

The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at

x = A

where A is the amplitude of the motion.

The maximum displacement is given by Hook's law:

F=kA

where

F is the force applied initially to the spring, so it is equal to the weight of the block:

F=mg=(0.340 kg)(9.81 m/s^2)=3.34 N

k = 160 N/m is the spring constant

Solving for A, we find

A=\frac{F}{k}=\frac{3.34 N}{160 N/m}=0.021 m

3 0
3 years ago
While chatting with a friend you place your book bag on a nearby slide in the playground at school. The bag remains stationary.
alukav5142 [94]

Answer:

3. fs < μmg

4. fs = mg sinθ

Explanation:

For any object placed on a slide, there are 3 external forces acting on it:

  • Fg = m*g (always downward)
  • N (normal force, always perpendicular to the surface of the slide. going upward)
  • Fs (Friction Force, always opposite to the movement of the object, parallel to the slide)

As we have only one force with components along the normal and parallel to the slide directions (gravity force), it is advisable to find the components of  this force, along these directions.

If θ is the angle of the slide above the horizontal, we have the following components of Fg:

Fgn = m*g*cosθ

Fgp = m*g*sin θ

We can apply Newton's 2nd Law to these perpendicular directions:

Fp = m*g*sin θ - Fs

Fn = N -m*g*cosθ = 0 (as the object has no movement in the direction perpendicular to the slide) (1)

Looking at the equation for the parallel direction, we have two forces, the component of Fg along the slide (which tries to accelerate the object towards the bottom of the slide), and the friction force.

While the object remains stationary, the equation for Newton's 2nd Law along this direction is as follows:

m*g*sin θ - fs =0 ⇒ fs = m*g*sinθ  (4.)

This force can take any value (depending on the angle θ) to equilibrate the component of Fg along the slide, up to a limit value, which  is given by the following expression:

fsmax = μN (2)

From (1), N= m*g*cos θ

Replacing in (2):

fsmax = μ*m*g*cos θ

While the bag remains at rest, we can say:

fs < μ*m*g*cosθ < μ*m*g (as in the limit cosθ =1)

So, the following is always true:

fs < μmg (3.)

6 0
3 years ago
If a rainstorm drops 5 cm of rain over an area of 13 km2 in the period of 3 hours, what is the momentum (in kg · m/s) of the rai
stiks02 [169]

To solve the problem it is necessary to apply the concepts related to Conservation of linear Moment.

The expression that defines the linear momentum is expressed as

P=mv

Where,

m=mass

v= velocity

According to our data we have to

v=10m/s

d=0.05m

A=13*10^6m^2

Volume (V) = A*d = (15*10^6)(0.03) = 3.9*10^5m^3

t = 3hours=10800s

\rho = 1000kg/m^3

From the given data we can calculate the volume of rain for 5 seconds

V' = \frac{V}{t}*\Delta t_{total}

Where,

\Delta t_{total} It is the period of time we want to calculate total rainfall, that is

V' = \frac{3.9*10^5}{10800}*5

V' = 1.805*10^2m^3

Through water density we can now calculate the mass that fell during the 5 seconds:

m' = V'*\rho

m' = 1.805*10^2*1000

m' = 1.805*10^5m^2

Now applying the prevailing equation given we have to

P=m'v

P = (1.805*10^5)(10)

P = 1.805*10^6 Kg.m/s

Therefore the momentum of the rain that falls in five seconds is 1.805*10^6 Kg.m/s

4 0
3 years ago
29:45
Yakvenalex [24]

Answer:

it opens the circuit so that electric charges do not flow to the timier

Explanation: a closed circut will allow electrisity to flow through.

6 0
3 years ago
Read 2 more answers
A 120-V motor has mechanical power output of 2.20 hp. It is 91.0% efficient in converting power that it takes in by electrical t
Mandarinka [93]

Answer:

a) i = 15.033\,A, b) \dot E_{in} = 1.804\,kW, c) C = 0.496\,USD

Explanation:

a) Electric power transmitted into the motor is:

\dot E _{in} = \frac{(2.20\,hp)\cdot \left(0.746\,\frac{kW}{hp} \right)}{0.91}

\dot E_{in} = 1.804\,kW

The current in the motor is:

i = \frac{1804\,W}{120\,V}

i = 15.033\,A

b) The energy delivered to the motor is:

\dot E_{in} = 1.804\,kW

c) The cost of running the motor for 2.50 hours is:

C = (1.804\,kW)\cdot (2.5\,h)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{kWh}{kJ}  \right)\cdot \left(0.110\,\frac{USD}{kWh} \right)

C = 0.496\,USD

4 0
4 years ago
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