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serg [7]
3 years ago
10

What’s 3 times 10 to the 8th power divided by 2.45 times 10 to the 9th power

Physics
1 answer:
Virty [35]3 years ago
5 0

0.12244898 is the value obtained when solving the given.

<u>Explanation:</u>

Given:

3 times 10 to the 8th power can be expressed in equation format as 3 \times 10^{8} (3 times symbolizes ‘multiplication’ then it is to the tenth power of 8)

2.45 times 10 to the 9th power can be expressed in equation format as 2.45 \times 10^{9} (2.45 times symbolizes ‘multiplication’, then it is to the tenth power of 9)

Asked to find the solution when dividing the above,

                 \frac{3 \times 10^{8}}{2.45 \times 10^{9}}=\frac{3}{2.45} \times 10^{8-9}

When the tenth power of any value goes from denominator to numerator and vice-versa, presents in opposite sign to that it possess (like when ‘10^{9} ' goes to numerator changed as ' 10^{-9} ').

Now by solving the above equation, we get

                \frac{3}{2.45} \times 10^{8-9}=1.22 \times 10^{-1}=\frac{1.22}{10}=0.12244898

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A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin return
tensa zangetsu [6.8K]

Answer:

1.68 s

Explanation:

From newton's equation of motion,

a = (v-u)/t.................................. Equation 1

Making t the subject of the equation

t =(v-u)g............................. Equation 2

Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.

Note: Taking upward to be negative and down ward to be positive,

Given: v = 0 m/s ( at the maximum height), u = 8.20 m/s, g = -9.8 m/s²

t = (0-8.20)/-9.8

t = -8.20/-9.8

t = 0.84 s.

But,

T = 2t

Where T = time taken for the bowling pin to return to the juggler's hand.

T = 2(0.84)

T = 1.68 s.

T = 1.68 s

7 0
3 years ago
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prohojiy [21]

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6 0
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Magnesium is a shiny, flexible metal that can be burned in the presence of air.
alukav5142 [94]

Answer:

A new substance was formed

Explanation:

According to this question, a shiny and flexible metal called Magnesium (Mg) is burnt in air to produce a white powder that has no shiny or flexible properties, however, has more weight than the magnesium metal itself.

This is possible because a CHEMICAL CHANGE has occured, hence, a new substance has been formed. The formation of a new substance during the burning process (chemical reaction), induced the increase in mass.

8 0
2 years ago
Compare the catching of two different water balloons.
Stels [109]

Answer:

a. The volume V₁ and V₂

b. The case that involves the greatest momentum change = Case B

c. The case that involves the greatest impulse = Case B

d. b. The case that involves the greatest force = Case B

Explanation:

Here we have

Case A: V₁ = 150-mL, v₁ = 8 m/s

Case B: V₂ = 600-mL, v₁ = 8 m/s

a. The variable that is different for the two cases is the volume V₁ and V₂

b. The momentum change is by the following relation;

ΔM₁ = Mass, m × Δv₁

The mass of the balloon are;

Δv₁ = Change in velocity = Final velocity - Initial velocity

Mass = Density × Volume

Density of water = 0.997 g/mL

Case A, mass = 150 × 0.997 = 149.55 g

Case B, mass = 600 × 0.997 = 598.2 g

The momentum change is;

Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s

Case B:  Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s

Therefore Case B has the greatest momentum change

The case that has the gretest momentum change = Case B

c. The momentum change = impulse therefore Case B involves the greatest impulse

d. Here we have;

Impulse = Momentum change = F_{average} × Δt = mΔV

∴ F_{average} = m·ΔV/Δt

∴ For Case A F_{average} = 149.55×8/Δt =  1196.4/Δt N

For Case B  F_{average} = 598.2×8/Δt =  4785.6/Δt

Where Δt is the same for Case A and Case B,  F_{average}  for Case B >>  F_{average}  for Case B

Therefore, Case B involves the greatest force.

4 0
3 years ago
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babunello [35]

Answer:

The forms of energy involved are

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Explanation:

The system consists of a ball initially at rest. The ball is pulled down from its equilibrium position (this builds up its potential energy) and then released. The released ball oscillates due to a continuous transition between kinetic and potential energy.

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