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2 years ago

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The weight of a person on Earth is 67.0 N. If the gravitational acceleration on the moon is 1.62 m/s2, calculate the person’s we

ight on the moon.
Group of answer choices

A) 56.0 N

B) 11.0 N

C) 108 N

D) 4.00 N

1 answer:

Rzqust [24]2 years ago

5 0

Answer:

$\large{ \boxed{ \tt{b) \: 11.0 \: N}}}$

- Please see the attached picture for full solution!:)

--------------- HappY LearninG <3 ---------

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Sunlight reflects from a concave piece of broken glass, converging to a point 34 cm from the glass. what is the radius of curvat

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The rays of light coming from the Sun are parallel to each other, so when they are reflected by the concave piece of glass (which acts as a concave mirror) they converge into the focus of the mirror, which is
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3 years ago

Calculate the acceleration using the formula acceleration = (final velocity - initial velocity)/time

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Explanation:

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4 years ago

A car accelerates uniformly from rest and

miv72 [106K]

Answer:

29.75 revolutions

Explanation:

The kinematic formula for distance, given a uniform acceleration a and an initial velocity v₀, is

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So, the car has travelled 44.62 m in 9.7 seconds - we want to know how many of the tire's <em>circumferences</em> fit into that distance, so we'll first have to calculate that circumference. The formula for the circumference of a circle given its diameter is $c=\pi{d}$ , which in this case is 47.8π cm, or, using π ≈ 3.14, 47.8(3.14) = 150.092 cm.

Before we divide the distance travelled by the circumference, we need to make sure we're using the same units. 1 m = 100 cm, so 105.092 cm ≈ 1.5 m. Dividing 44.62 m by this value, we find the number of revs is

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7 0

3 years ago

Two identical charges, each -8.00 E-5 C, are separated by a distance of 20.0 cm (100 cm = 1 m). What is the force of repulsion?

Rom4ik [11]

F = 1440 N. The repulsion force between two identical charges, each -8.00x10⁻⁵C separated by a distance of 20.0 cm is 1440 N.

The easiest way to solve this problem is using Coulomb's Law given by the equation $F=k\frac{|q_{1}*q_{2}|}{r^{2} }$ , where k is the constant of proportionality or Coulomb's constant, q₁ and q₂ are the charges magnitude, and r is the distance between them.

We have to identical charges of -8.00x10⁻⁵C, are separated by a distance of 20.0 cm, and we need to know the force of repulsion between the charges.

First, we have to convert 20.0 cm to meters.

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$F = 9.00x10^{9}\frac{Nm^{2}}{C^{2}} \frac{|-8.00x10^{-5}C*-8.00x10^{-5}C|}{(0.20m)^{2} }$

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3 years ago