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hichkok12 [17]
3 years ago
6

Which tools would Sara use to find and irregularly shaped objects mass in volume?

Physics
1 answer:
pogonyaev3 years ago
8 0
To find the mass and volume of irregularly shaped object. One can use the balance to find the weight, and to find the volume of the object water should be poured in the beaker and note the level of water.
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Which processes cause rocks to be exposed<br> at Earth's surface? SC.7..6.2
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weathering

Explanation:

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1 . What happens when you drop the stone?

Depending on the weight from which the stone was dropped, the glass might well break

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explain why a high-speed collision between two cars would cause more damage than a low-speed collision between the same two cars
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The force equation can easily prove this.  F=ma.  This states that the force on an object is equal to mass times acceleration.  If the mass stays the same and the velocity of the cars increases than that means there is a larger force.  This is because in both cases the cars are stopping in almost an instant and the times of the crashes are theoretically identical.  Acceleration is the change in velocity over time.  If the velocity is higher with the same amount of time than that means there is a higher acceleration.  If you plug a higher acceleration into the force equation then you wind up with a higher force and in turn a more damaging collision.
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The frequency of a sound wave is 457
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The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho= rho0[1+α(T−T0)] , where
Stolb23 [73]

Answer:

At 81. 52 Deg C its resistance will be 0.31 Ω.

Explanation:

The resistance of wire =R_T =\frac{\rho_T \ l}{A}

Where R_T =Resistance of wire at Temperature T

\rho_T = Resistivity at temperature T =\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]

Where T_0 =20\ Deg\ C , \  \rho_0 = Constant,  \alpha =3.9 \times 10^-^3 DegC^-1 \ (Given)

l=Length of the wire

& A = Area of cross section of wire

For long and thin wire the resistance & resistivity relation will be as follows

\frac{R_T_1}{R_T_2}=\frac{\rho_0(1+\alpha \cdot(T_1-2 0 )}{\rho_0(1+\alpha \cdot (T_2 -20 )}

\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}

1.24=1+\alpha (T-20)

0.24=\alpha(\ T -20 )

Putting\ the\ value\ of \alpha = 3.9 \times 10^-^3 DegC^-1

T = 81.52 Deg C

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2 years ago
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