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Delicious77 [7]
3 years ago
10

If you drop an object from rest, the distance it falls is given by (1/2)at2, where a is the acceleration of the object and t is

the amount of time fallen. Suppose you drop an object and measure the distance it falls. Suppose you drop it again, but it falls for three times the amount of time as before. How far did it fall on the second drop?
Select one:
a. It will fall twice the distance compared to the first time dropped.
b. It will fall three times the distance compared to the first time dropped.
c. It will fall nine times the distance compared to the first time dropped.
d. It depends on the planet you are on when you drop it.
Physics
1 answer:
klio [65]3 years ago
7 0

If,

d \:  =  \:   \frac{1}{2} a {t}^{2}

then, with 3x time t, (suppose, new distance is h)

h \:  =  \:   \frac{1}{2} a {(3t)}^{2}

=  \frac{1}{2} a9 {t}^{2}

= 9 \:  \frac{1}{2} a{t}^{2}

= 9d

Therefore, new distance h will be 9 times bigger than distance d.

answer: c

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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)

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& 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 )}

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T = 81.52 Deg C

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