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1 year ago

In a game of Tug-of-War,

1 answer:

8 0

Eventually one of the teams will became tired and its pull will be smaller than that of the other team. When the force acting on the robe are unbalanced the rope will start to move in the direction of the great force.

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Answer these questions about a light year.?

aliina [53]

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

4 0

2 years ago

Read 2 more answers

A motorcycle has a mass of 250 kg and a velocity of 68 m/s, what is it’s momentum?

Gekata [30.6K]

Since momentum is equal to mass multiplied by velocity, you would mulitply 250 grams by 68 m/s.
The answer is 17,000 kg*m/s

3 0

2 years ago

A car on a roller coaster starts at zero speed at an elevation above the ground of 26 m. It coasts down a slope, and then climbs

nirvana33 [79]

The speed of the car at the top of the hill is 14m/s

Explanation:

given that

Initial velocity u of the car=0 m/s

The distance can be determined by finding out the difference between the elevation of the first slope and second slope.

elevation of the first slope=26 m

elevation of second slope=16m

distance s=26-16=10 m

acceleration due to gravity g=9.8 m/s2

speed of the car at the top of the hill can be determined by using the equation

$v^2=u^2+2as\\v^2=0^2+2\times 9.8\times 10\\=196\\v=\sqrt{196} =14m/s$

speed of the car at the top of the hill is 14m/s

5 0

3 years ago

A sealed cubical container 10.0 cm on a side contains three times Avogadro's number of molecules at a temperature of 24.0°C. Fin

rodikova [14]

This question involves the concepts of general gas equation and pressure.

The force exerted by the gas on one of the walls of the container is "74.08 KN".

First, we will use the general gas equation to find out the pressure of the gas:

$PV = nRT$

where,

P = Pressure of the gas = ?

V = Volume of cube = (side length)³ = (10 cm)³ = (0.1 m)³ = 0.001 m³

n = no. of moles = 3 (since molecules equal to avogadro's number make up 1 mole)

R = general gas constant = 8.314 J/mol.K

T = Absolute Temperature = 24°C + 273 = 297 K

Therefore,

$P = \frac{(3)(8.314\ J/mol.k)(297\ K)}{0.001\ m^3}$

P = 7407.78 KPa

Now, the force on one wall can be given as follows:

$P =\frac{F}{A}\\\\F=PA$

where,

A = area of one wall = (side length)² = (0.1 m)² = 0.01 m²

Therefore,

$F=(7407.78\ KPa)(0.01\ m^2)\\$

F = 74.08 KN

Learn more about the general gas equation here:

brainly.com/question/24645007?referrer=searchResults

5 0

2 years ago

A 1.51 kg ball and a 1.97 kg ball are connected by a 1.63 m long rigid, massless rod. The rod is rotating clockwise about its ce

Eddi Din [679]

Answer:

$T = 1.205\,N\cdot m$

Explanation:

Needed torque can be estimated by means of the Theorem of Angular Momentum Conservation and Impact Theorem. The center of mass of the system is:

$\bar r = \frac{(0\,m)\cdot (1.51\,kg)+(1.63\,kg)\cdot (1.97\,kg)}{1.51\,kg+1.97\,kg}$

$\bar r = 0.923\,m$

Let assume that both masses can be modelled as particles, then:

$[(1.51\,kg)\cdot (0.923\,m)^{2} + (1.97\,kg)\cdot (0.707\,m)^{2}]\cdot (38\,\frac{rev}{min} )\cdot (\frac{2\pi\,rad}{1\,rev} )\cdot (\frac{1\,min}{60\,s} ) -T\cdot (7.5\,s) = 0\,\frac{kg\cdot m^{2}}{s}$

The torque needed to stop the system is:

$T = 1.205\,N\cdot m$

5 0

3 years ago