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sergij07 [2.7K]
3 years ago
12

Definición de variable en física

Physics
1 answer:
Dominik [7]3 years ago
5 0

Answer:

una variable es una cantidad cuyo valor puede cambiar ♥️

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A student wants to determine the speed of sound at an elevation of one mile. To do this the student performs an experiment to de
Lorico [155]

Answer:

The correct answer is a

Explanation:

The speed of a sound wave depends on the square root of the modulus of compressibility and the density of the medium.

For the same medium, the speed of sound depends on the temperature of the fora

           v = v_o \ \sqrt{1 + \frac{T}{273} }

Therefore, the different results that are obtained are due to changes in temperature. The correct answer is a

since this way it has the values ​​of the speed of sound for each temperature, for which it can compare with the results obtained from the trip.

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3 years ago
While excavating the tomb of Tutankhamen (d. 1325 BC), archeologists found a sling made of linen. The sling could hold a stone i
AleksAgata [21]

Answer:

L = 2.83 J.s

Explanation:

The formula for the angular momentum of the stone is given as follows:

L = mvr

where,

L = angular momentum of the stone  = ?

m = mass of the stone = 0.1 kg

v = linear velocity of the stone = rω

r = radius of circular path = 1.5 m

ω = angular speed of the stone = (2 rev/s)(2π rad/1 rev) = 4π rad/s

Therefore,

L = mvr = m(rω)r

L = mr²ω

using values, we get:

L = (0.1 kg)(1.5 m)²(4π rad/s)

<u>L = 2.83 J.s</u>

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3 years ago
A 16-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 24 n. starting from rest, the sle
harina [27]
The abweens is 20 because yes
7 0
3 years ago
Question 2 10
ss7ja [257]
B Ik it all just know who they’ll you
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3 years ago
How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C?
Svet_ta [14]

Answer:

229,098.96 J

Explanation:

mass of water (m) = 456 g = 0.456 kg

initial temperature (T) = 25 degrees

final temperature (t) = - 10 degrees

specific heat of ice = 2090 J/kg

latent heat of fusion =33.5 x 10^(4) J/kg

specific heat of water = 4186 J/kg

for the water to be converted to ice it must undergo three stages:

  • the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp

        Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J

  • the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp

         Q = 0.456 x 33.5 x 10^(4) = 152760 J

  • the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp

        Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J

The quantity of heat removed from all three stages would be added to get the total heat removed.

Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J

6 0
3 years ago
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