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Rama09 [41]
2 years ago
10

Dadas dos sustancias con distinta densidad, las cuales se mezclan, formarán una nueva sustancia con una densidad.....

Physics
1 answer:
zhannawk [14.2K]2 years ago
8 0

Answer:

Explanation:

Equal to the sum of the densities of both substances

Equal to the sum of the densities of both substances

Equal to the difference in the densities of both substances

Equal to the difference in the densities of both substances

of which I cannot predict anything about its value

of which I cannot predict anything about its value

of an intermediate value to both substances which will depend on which substance is in greater quantity

of an intermediate value to both substances which will depend on which substance is in greater quantity

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If a 50 kg student is standing on the edge of a cliff. Find the student’s gravitational potential energy if the cliff is 40 m hi
Oduvanchick [21]

you can find it using the equation: potential energy=mass*gravitational acceleration*height.


energy=50kg*9.8N/kg*40m=19600Nm=19600J or 19.6kJ


Sometimes they use 10 instead of 9.8 for the g constant. 


Rember to make me Brainliest!!!

3 0
3 years ago
What is the speed of sound at 33 °C (m/s)? For a frequency of 5 kHz, how large do you expect the wavelength to be (m)?
Vedmedyk [2.9K]

Answer:

-The speed of sound at 33°C is 362.8 m/s.

-The wavelength at a frequency at 5 kHz is 0.07256 m .

Explanation:

let v = 343 m/s be the speed of sound.

let T be the temperature.

then the speed of sound V, at 33°C is given by:

V = v + 0.6×T

   = 343 + 0.6×33

   = 362.8 m/s

Therefore, the speed of sound at 33°C is 362.8 m/s.

the wavelength at a frequency of f = 5kHz = 5000 Hz is given by:

λ = V/f

  = (362.8)/(5000)

  = 0.07256 m

Therefore, the wavelength at a frequency at 5 kHz is 0.07256 m .

7 0
3 years ago
Two particles, one with charge −7.97×10−6 C and the other with charge 6.91×10−6 C, are 0.0359 m apart. What is the magnitude of
kolbaska11 [484]

Answer:

-384.22N

Explanation:

From Coulomb's law;

F= Kq1q2/r^2

Where;

K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2

q1 and q2 = magnitudes of the both charges

r= distance of separation

F= 9 ×10^9 × −7.97×10^−6 × 6.91×10^−6/(0.0359)^2

F= -495.65 × 10^-3/ 1.29 × 10^-3

F= -384.22N

7 0
2 years ago
An airplane flies with a constant speed of 840km/h. how far can it travel in 1 hour?
arlik [135]
Distance = speed X time

In this example, the speed of the airplane = 840km. The time (that the question is asking)is how far can it travel in 1 hour.

So just plug in your numbers.

Distance = 840km X 1 hour = 840km/hour or 840km for short.
3 0
2 years ago
A torsion-bar spring consists of a prismatic bar, usually of round cross section, that is twisted at one end and held fast at th
ra1l [238]

Answer:

d₁ = 0.29 in

d₂ = 0.505 in

Explanation:

Given:

T = 1500 lbf in

L = 10 in

x = 0.5 L = 5 in

T_{1} =\frac{T(L-x)}{L} =\frac{1500*(10-5)}{10} =750lbfin

First case: T = T₁ + T₂

T₂ = T - T₁ = 1500 - 750 = 750 lbf in

If the shafts are in series:

θ = θ₁ + θ₂

θ = ((T₁ * L₁)/GJ) + ((T₂ * L₂)/GJ)

Second case: If d₁ ≠ d₂

θ = ((T₁ * L₁)/GJ₁) + ((T₂ * L₂)/GJ₂) = 0 (eq. 1)

t₁ = t₂

\frac{16T_{1} }{\pi d_{1}^{3}  } =\frac{16T_{2} }{\pi d_{2}^{3}  } (eq. 2)

T₁ + T₂ = 1500 (eq. 3)

θ₁ first case = θ₁ second case

Replacing:

\frac{750*5}{G(\frac{\pi }{32})*0.5^{4}  } =\frac{T_{1}*3.7 }{G(\frac{\pi }{32})*d_{1} ^{4}  }\\T_{1} =16216d_{1} ^{4}

The same way to θ₂:

\frac{750*5}{G(\frac{\pi }{32})*0.5^{4}  } =\frac{T_{2}*6.3 }{G(\frac{\pi }{32})*d_{2} ^{4}  } \\T_{2} =9523.8d_{2} ^{4}

From equation 2, we have:

d₁ = 0.587 * d₂

From equation 3, we have:

d₂ = 0.505 in

d₁ = 0.29 in

7 0
3 years ago
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