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Nadusha1986 [10]
3 years ago
5

In the sum of 54.34 and 45.66, the number of significant figure for theanswer is​

Physics
1 answer:
Natali5045456 [20]3 years ago
4 0

Answer:

5

Explanation:

54.34+45.66=100.00 (you have to use the .00 because when you add to numbers you keep the number of decimals)

so you get 100.00

all numbers that are not 0 are sig figs so 1 is a sig fig

If a number ends with a 0 after a decimal place that 0 is a sig fig

all numbers between two sig figs are sig figs so that would make all of the numbers sig figs

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Layer 1, Rock 2, Rock 1, Fault

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Does anyone taste the difference between left and right twix?
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1. A car travels 36 miles North and then 45 miles East. How far does it
salantis [7]

Answer:

Explanation:

Displacement is the shortest distance or path between two points.

1) Displacement = √(36² + 45²) = 57.63 miles

2) Displacement = √(100² + 500²) = 509.9 meters

3) Displacement = √(60² + 40²) = 72.11 miles

4) Displacement = √(700² + 500²) = 860.23 miles

5) Displacement = 300 - 300 = 0 miles

6) Displacement = 200 + 100 = 300 miles

7) Displacement = √(650² + 650²) = 919.24 miles

8) Yes, since a distance is moved in a direction

6 0
4 years ago
Tom has two pendulums with him. Pendulum 1 has a ball of mass 0.2 kg attached to it and has a length of 5 m. Pendulum 2 has a ba
mars1129 [50]

Given Information:

Pendulum 1 mass = m₁ = 0.2 kg

Pendulum 2 mass = m₂ = 0.6 kg

Pendulum 1 length = L₁ = 5 m

Pendulum 2 length = L₂ = 1 m

Required Information:

Affect of mass on the frequency of the pendulum = ?

Answer:

The mass of the ball will not affect the frequency of the pendulum.

Explanation:

The relation between period and frequency of pendulum is given by

f = 1/T

The period of pendulum is given by

T = 2π√(L/g)

Where g is the acceleration due to gravity and L is the length of the string

As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.

Bonus:

Pendulum 1:

T₁ = 2π√(L₁/g)

T₁ = 2π√(5/9.8)

T₁ = 4.49 s

f₁ = 1/T₁

f₁ = 1/4.49

f₁ = 0.22 Hz

Pendulum 2:

T₂ = 2π√(L₂/g)

T₂ = 2π√(1/9.8)

T₂ = 2.0 s

f₂ = 1/T₂

f₂ = 1/2.0

f₂ = 0.5 Hz

So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.

3 0
3 years ago
Read 2 more answers
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