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3 years ago

Light from the sun reaches Earth in 8.3min. The speed of light is 3.00 x 10^8 m/s. How far is the Earth from the sun?

Physics

2 answers:

ruslelena [56]3 years ago

8 0

Speed= distance/ time
distance= speed×time
distance= 3×10^8 × 8.3× 60
1494×10^8

strojnjashka [21]3 years ago

5 0

The answer is this, but i don't know how to simplify it. 3x^100000000<span />

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A television has a mass of 19 kg. What is the weight of the television?

soldi70 [24.7K]

That depends on how far it is from the nearest planet. If it's on the surface of Earth, it weighs (19 kg) x (9.8 m/s^2) = 186.2 newtons.

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3 years ago

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Yashoda prepares some lime juice on a hot day. She adds 80 g of ice at a temperature of 0°C to 0.32 kg of lime juice. The temper

Vikentia [17]

Answer:

Explanation:

a )

hear energy required to melt 1 g of ice = 340 J ,

hear energy required to melt 80 g of ice = 340 x 80 J = 27220 J .

b ) energy gained by the melted ice ( water at O°C ) = m ct

where m is mass of water , s is specific heat and t is rise in temperature

= 80 x 4.2 x ( 8°C - 0°C)

= 2688 J .

c )

energy lost by lime juice = energy gained by ice and water

= 27220 J + 2688 J .

= 29908 J .

d )

Let specific heat required be S

Heat lost by lime juice = M S T

M is mass of lime juice , S is specific heat , T is decrease in temperature

= 320 g x S x ( 29 - 8 )°C

= 6720 S

For equilibrium

Heat lost = heat gained

6720 S = 29908 J

S = 4.45 J /g °C .

4 0

3 years ago

A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles

Dimas [21]

A) The total energy of the system is sum of kinetic energy and elastic potential energy:
$E=K+U= \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
$x=A=4.00 cm = 0.04 m$
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
$E=U= \frac{1}{2}kA^2 = \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J$

b) When the position of the object is
$x=1.00 cm = 0.01 m$
the potential energy of the system is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J$
and so the kinetic energy is
$K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J$
since the mass is $m=50.0 g=0.05 kg$ , and the kinetic energy is given by
$K= \frac{1}{2}mv^2$
we can re-arrange the formula to find the speed of the object:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s$

c) The potential energy when the object is at
$x=3.00 cm=0.03 m$
is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$
Therefore the kinetic energy is
$K=E-U=0.028 J-0.016 J = 0.012 J$

d) We already found the potential energy at point c, and it is given by
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$

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3 years ago

How does a parallel circuit differ from a series circuit

Lapatulllka [165]

<span>A series circuit has one path for electrons, but a parallel circuit has more than one path.</span>

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3 years ago

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Natalka [10]

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5 0

3 years ago