What is the meaning of physics

A 34.00 kg sample of mud has a specific heat of 2,512 J/kg °C.

Elena-2011 [213]

Answer:

B ig lol

Explanation:

6 0

3 years ago

An ambulance is traveling east at 61.9 m/s. Behind it there is a car traveling along the same direction at 28.5 m/s. The ambulan

inessss [21]

Answer: 0.4 m

Explanation:

Given

Speed of ambulance, vs = 61.9 m/s

Speed of car = 28.5 m/s

Frequency of ambulance siren, f = 694 Hz

Velocity of sound in air, v = 343 m/s

With speed of ambulance being (61.9 m/s) -> We solve using

fd = f(v + vr) / (v - vs), where vr = 0

fd = 694 * (343 + 0) / (343 - 61.9)

fd = 694 * (343 / 281.1)

fd = 694 * 1.22

fd = 847 Hz

Recall,

λ = v/f

λ = 343/847

λ = 0.4 m

Therefore, the wavelength of the sound of the ambulance’s siren if you are standing at the position of the car is 0.4 m

7 0

2 years ago

How far away is the moon?

Fed [463]

The moon is 230,100 miles from planet earth.

5 0

2 years ago

Read 2 more answers

Kinetic vs potential energy

antiseptic1488 [7]

Answer:

Kinetic energy is the energy obtained from movement while potential energy is the energy in something when it is at a height or has been manipulated in some way; it has work done to it.

Explanation:

Kinetic energy is, like I said, energy gained through movement or action. When an object stops moving, it converts into a different type of energy. Energy cannot be destroyed or created. An object with potential energy has a reference. A reference determines how high it is. The higher and heavier an object is, the more potential energy it has.

Hope this helps!

5 0

2 years ago

P7.16 A thin flat plate 55 by 110 cm is immersed in a 6-m/s stream of SAE 10 oil at 20C. Compute the total friction drag if the

timofeeve [1]

Answer:

a

The total friction drag for the long side of the plate is 107 N

b

The total friction drag for the long side of the plate is 151.4 N

Explanation:

The first question is to obtain the friction drag when the fluid i parallel to the long side of the plate

The block representation of the this problem is shown on the first uploaded image

Where the U is the initial velocity = 6 m/s

So the equation we will be working with is

$F = \frac{1}{2} \rho C_fAU^2$

Where $\rho$ is the density of SAE 10W = $870\ kg/m^3$ This is obtained from the table of density at 20° C

$C_f$ is the friction drag coefficient

This coefficient is dependent on the Reynolds number if the Reynolds number is less than $5*10^5$ then the flow is of laminar type and

$C_f$ = $\frac{1.328}{\sqrt{Re} }$

But if the Reynolds number is greater than $5*10^5$ the flow would be of Turbulent type and

$C_f = \frac{0.074}{Re_E^{0.2}}$

Where Re is the Reynolds number

To obtain the Reynolds number

$Re = \frac{\rho UL}{\mu}$

where L is the length of the long side = 110 cm = 1.1 m

and $\mu$ is the Dynamic viscosity of SAE 10W oil $= 1.04*10^{-1} kg /m.s$

This is gotten from the table of Dynamic viscosity of oil

So

$Re = \frac{870 *6*1.1}{1.04*10^{-1}}$

$= 55211.54$

Since 55211.54 < $5.0*10^5$

Hence

$C_f = \frac{1.328}{\sqrt{55211.54} }$

$= 0.00565$

$A$ is the area of the plate = $\frac{ (110cm)(55cm)}{10000}$ = $0.55m^2$

Since the area is immersed totally it should be multiplied by 2 i.e the bottom face and the top face are both immersed in the fluid

$F = \frac{1}{2} \rho C_f(2A)U^2$

$F =\frac{1}{2} *870 *0.00565*(2*0.55)*6^2$

$F = 107N$

Considering the short side

To obtain the Reynolds number

$Re = \frac{\rho U b}{\mu}$

Here b is the short side

$Re =\frac{870*6*0,55}{1.04*10^{-1}}$

$=27606$

Since the value obtained is not greater than $5*10^5$ then the flow is laminar

And

$C_f = \frac{1.328}{\sqrt{Re} }$

$= \frac{1.328}{\sqrt{27606} }$

$= 0.00799$

The next thing to do is to obtain the total friction drag

$F = \frac{1}{2} \rho C_f(2A)U^2$

Substituting values

$F = \frac{1}{2} * 870 * 0.00799 * 2( 0.55) * 6^2$

$= 151.4 N$

4 0

2 years ago

What is the meaning of physics​

What is the meaning of physics