Carry out the following conversions using the factor-label method. Show all your work.

1 answer:

7 0

1) 3600 sec in an hour (3600 x 24 x 365) = 31,536,000 sec
2) 100 cm in 1 m (.14 x 100) = 14 cm
3) 1000 g in 1 kg (20/1000) = .02 kg
4) 10^-9 m = 1 nm (637 x 10^-9) = 6.37 x 10 ^-7m
5) 1000 m in 1 km (1000 x 3.2) = 3200 m
6) 10^-6 m in 1 micrometers (24 x 10^-6) = 2.4 x 10^-5

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An attractive force of 7.2 N occurs between two point charges that are 0.10 m apart. If one charge is -4.0 µC, what is the other

Umnica [9.8K]

The answer is c. +2.0 µC

To calculate this, we will use Coulomb's Law:

F = k*Q1*Q2/r²

where F is force, k is constant, Q is a charge, r is a distance between charges.

k = 9.0 × 10⁹ N*m/C²

It is given:

F = 7.2 N

d = 0.1 m = 10⁻¹ m

Q1 = -4.0 µC = 4 * 1.0 × 10⁻⁶ = 4.0 × 10⁻⁶

Q2 = ?

Thus, let's replace this in the formula for the force:

7.2 = 9.0 × 10⁹ * 4.0 × 10⁻⁶ * Q2/(10⁻¹)²

7.2 = 9 * 4 * 10⁹⁻⁶ * Q2/10⁻¹°²

7.2 = 36 × 10³ * Q2 / 10⁻²

Multiply both sides of the equation by 10⁻²:

7.2 × 10⁻² = 36 × 10³ * Q2

⇒ Q2 = 7.2 × 10⁻² / 36 × 10³ = 7.2/36 × 10⁻²⁻³ = 0.2 × 10⁻⁵ = 2 × 10⁻⁶

Since µC = 1.0 × 10^-6:

Q2 = 2 * 1.0 × 10^-6 = 2 µC

5 0

3 years ago

(II) How much work did the movers do (horizontally) pushing a 46.0-kg crate 10.3 m across a rough floor without acceleration, if

Gwar [14]

Answer:

2324 J

Explanation:

The formula for work is:

$W=F*d$

where $F$ is the force applied, and $d$ is the distance moved, in this case $d=10.3m$

and we need to find $F.$

Since the crate is not moving up or down, we conclude that the <u>normal force must be equal to the weight </u>of the object:

$N=w$

where $N$ is the normal force and $w$ is the weight, which is: $w=mg$ , where g is the gravitational acceleration $g=9,81m/s^2$ and $m$ is the mass $m=46kg$ .