3 years ago

HOW MANY PIES DOSE IT TAKE TO GET TO THE MOON

Physics

1 answer:

Schach [20]3 years ago

8 0

i do not have an answer because it depends on the size and the distance lol

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James rode his bike 0.65 hours and traveled 8.45 km. What was his speed?

maria [59]

Speed = (distance covered) / (time to cover the distance)

= ( 8.45 km) / (0.65 hr)

= (8.45 / 0.65) km/hr

= 13 km/hr

5 0

3 years ago

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

4 0

3 years ago

iron ball weight 400 gram inside water when it is completely impressed in water 53 gram water is displaced what will be the weig

Alika [10]

Answer:

453 gm

Explanation:

<u>Immersed </u>objects are buoyed up by force equal to mass of displaced liquid

400 + 53 = 453 gm in air

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2 years ago

What is the surface area of an object weighing 100M which exerts a pressure of 20nm-2

vovangra [49]

lovely question hope 7 solve it

Explanation:

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3 years ago

A 3kg object has an initial velocity (6i - 2j) m/s (a ) what is its kinetic energy at this time? (b) Find total work done on the

guapka [62]

Answer:

$K.E = \frac{1}{2} m {v}^{2} \\ {v}^{2}_i = {v}^{2} _x + {v}^{2} _y \\ = {(6)}^{2} + {( - 2)}^{2} = 36 + 4 = 40m. {s}^{ - 1} \\ K.E_i = \frac{1}{2} (3) (40) = 60J \\ \\ {v}^{2}_f = {v}^{2} _x + {v}^{2} _y \\ = {(8)}^{2} + {(4)}^{2} = 80m. {s}^{ - 1} \\ K.E_i = \frac{1}{2} (3) (80) = 120J \\ W_{net}=K.E_f-K.E_i \\ = 120J - 60J \\ = 60J$

3 0

3 years ago