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2 years ago

Answer these questions about a light year.?

Physics

2 answers:

aliina [53]2 years ago

4 0

These are five questions and five answers:

a. State the speed in m/sec using scientific notation

i) Find the place value of the digit with the highest place value (the digit most to the left) in the number 300,000,000: hundred millions.

ii) Determine the number of zeros corresponding to that place value: 9

ii) Write the digit with the highest place value, mulitplied by the power of ten with exponent equal to the number of zeros corresponding to the place of such digit less 1 (9 - 1 = 8): 3 × 10⁸

iii) Complete with the units: 3 × 10⁸ m/s ← answer

b. State the speed in km/sec using scientific notation

i) State the conversion factor: 1 km = 1000 m = 10³ m

⇒ 1 = 1 km / 10³ m

ii) Multiply by the conversion factor:

3 × 10⁸ m/s × 1 km / (10³ m)

iii) Simplify (cancel the units that are common in the numerator and denominator)

3 × (10⁸ / 10³) km/s = 3 × 10⁵ km/s ← answer

c. State the number of seconds in one year using scientific notation

i) Find the place value of the digit with the highest place value (the digit most to the left) in the number 32,000,000: ten millions.

ii) Determine the number of zeros corresponding to that place value: 8

ii) Write the digit with the highest place value, as integer, add the other significant figures as decimals, mulitplied by the power of ten with exponent equal to the number of zeros corresponding to the place of the digit with highest place value less 1 (8 - 1 = 7): 3.2 × 10⁷

iii) Complete with the units: 3.2 × 10⁷ s ← answer

d. Calculate the distance in meters of one light year

i) distance formula: distance = speed × time

ii) Replace with the numbers in scientific notation:

distance = 3 × 10⁸ m/s × 3.2 × 10⁷ s

iii) Simplify (due the operations and cancel the units that are common in the numerator and denominator)

3 × 10⁸ m/s × 3.2 × 10⁷ s = 9.6 × 10 ¹⁵ m ← answer

e. State this distance in centimeters

i) State the conversion factor: 1m = 100 cm = 10² cm

⇒ 1 = 10² cm / 1 m

ii) Multiply by the conversion factor: 9.6 × 10 ¹⁵ m × 10² cm / 1 m

iii) Simplily: 9.6 × 10⁷ cm ← answer

FrozenT [24]2 years ago

4 0

A. state the speed in m/sec using scientific notation

3.0 x 10^8 m/sec

b. state the speed in km/sec using scientific notation

3.0 x 10^5 km/sec

c. state the number of seconds in one year using scientific notation

 3.2 x 10^7 sec

d. calculate the distance in meters of one light year

1 light year = 9.4605284 x 10^15 meters

e. state this distance in centimeters

9.4605284 x 10^12 centimeters

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A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left

Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

Length of the massless beam; $L = 4.00m$

Distance of support from the left end; $x = 3.00m$

First mass; $m1 = 31.3 kg$

Distance of beam from the left end( m₁ is attached to ); $x_1 = ?$

Second mass; $m_2 = 61.7 kg$

Distance of beam from the right of the support( m₂ is attached to ); $x_1 = 0.273m$

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, $m_1g( x-x_1) = m_2gx_2$

we divide both sides by $g$

$m_1( x-x_1) = m_2x_2$

Next, we make $x_1$ , the subject of the formula

$x_1 = x - [ \frac{m_2x_2}{m_1} ]$

We substitute in our given values

$x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]$

$x_1 = 3.00m - 0.538m$

$x_1 = 2.46m$

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

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2 years ago

Solution A has a speciﬁc heat of 2.0 J/g◦C. Solution B has a speciﬁc heat of 3.8 J/g◦C. If equal masses of both solutions start

fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

We have a formula for such condition,

$Q=m.c.\Delta T$ .....................................(1)

where:

$\Delta T$ = temperature difference

Q= heat energy

m= mass of the body

c= specific heat of the body

Proving mathematically:

According to the given conditions

we have equal masses of two solutions A & B, i.e. $m_A=m_B$

equal heat is supplied to both the solutions, i.e. $Q_A=Q_B$

specific heat of solution A, $c_{A}=2.0 J.g^{-1} .\degree C^{-1}$

specific heat of solution B, $c_{B}=3.8 J.g^{-1} .\degree C^{-1}$

$\Delta T_A$ & $\Delta T_B$ are the change in temperatures of the respective solutions.

Now, putting the above values

$Q_A=Q_B$

$m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1$

Which proves that solution A attains a higher temperature than solution B.

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hodyreva [135]

Answer:

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Explanation:

W = F × s

F = 10 N,

t = 3min = 180sec

s = 0( no change in postion)

W = 10 ×0

W = 0

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