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Olegator [25]
2 years ago
13

A robin flies a distance of 45963 cm. How far has it flown in kilometers?

Physics
1 answer:
alina1380 [7]2 years ago
6 0

Answer:

0.46km

Explanation:

45963cm/100cm=459.63m/1000m=0.45963 or 0.46km

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The resistivity of gold is at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a current of 940 m
IgorC [24]

Answer:

0.0360531138247 V/m

Explanation:

\rho = Resistivity of gold = 2.44\times 10^{-8}\ \Omega .m (General value)

I = Current = 940 mA

d = Diameter = 0.9 mm

A = Area = \dfrac{\pi}{4}d^2

E = Electric field

Resistivity is given by

\rho=\dfrac{EA}{I}\\\Rightarrow E=\dfrac{\rho I}{A}\\\Rightarrow E=\dfrac{2.44\times 10^{-8}\times 940\times 10^{-3}}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}\\\Rightarrow E=0.0360531138247\ V/m

The  electric field in the wire is 0.0360531138247 V/m

6 0
3 years ago
Read 2 more answers
What is Kepler’s first law?
BartSMP [9]

Answer:

Kepler's First Law: each planet's orbit about the Sun is an ellipse. The Sun's center is always located at one focus of the orbital ellipse. The Sun is at one focus. The planet follows the ellipse in its orbit, meaning that the planet to Sun distance is constantly changing as the planet goes around its orbit.

5 0
2 years ago
Read 2 more answers
The pressure exerted by 15m of liquid is 1500pa.The acceleration due to gravity g=10m/s^2.Calculate the density liquid.​
Natali [406]

Answer:

1500 divided by 150(15m x 10m/s^2) = 10

8 0
3 years ago
How much force is needed to accelerate a toy car of mass 8 kg at 2m/s
atroni [7]

Answer:

4N

Explanation:

f = ma

8kg x 2m/s = 4N

4 0
2 years ago
The chain of length L and mass per unit length rho is released from rest on the smooth horizontal surface with a negligibly smal
devlian [24]

Answer:

Part a)

a = \frac{x}{L} g

Part b)

T = \rho x g(1 - \frac{x}{L})

Part c)

v = \sqrt{gL}

Explanation:

Part a)

Net pulling force on the chain is due to weight of the part of the chain which is over hanging

So we know that mass of overhanging part of chain is given as

m = \rho x

now net pulling force on the chain is given as

F = \rho x g

now acceleration is given as

F = Ma

\rho x g = \rho L a

a = \frac{x}{L} g

Part b)

Tension force in the part of the chain is given as

mg - T = ma

\rho x g - T = \rho x a

\rho x(g - a) = T

\rho x (g - \frac{x}{L} g) = T

T = \rho x g(1 - \frac{x}{L})

Part c)

velocity of the last link of the chain is given as

a = \frac{x}{L} g

v\frac{dv}{dx} = \frac{x}{L} g

now integrate both sides

\int v dv = \frac{g}{L} \int x dx

\frac{v^2}{2} = \frac{gL}{2}

v = \sqrt{gL}

3 0
3 years ago
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