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evablogger [386]
3 years ago
13

If it takes Charlie .45 hrs walking at 2.5 km/hr to get to the chocolate factory and it takes Willy Wonka 25 hrs walking at 3.5

km/hr then who lives closer and by how far?​
Physics
1 answer:
Dominik [7]3 years ago
3 0

Answer:

6.4

Explanation:

trust

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The emission of light from a radioisotope occurs during ? decay.
jeka94

Answer:

Gamma

Explanation:

7 0
3 years ago
Read 2 more answers
Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the
Svet_ta [14]

Answer:

b)  x_{cm} = 4.88 cm , c) x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃) and d)

x_{cm}’= 1.88 cm

Explanation:

The definition of mass center is

    x_{cm} = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

    D₂ = d₁ + d₂

    D₂ = 1.1 + 1.9

    D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

    D₃ = D₂ + d₂

    D₃ = 3.0 + 3.9

    D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

    M = m₁ + m₂ + m₃

    M = 24 + 12 + 56

    M = 92 g

    x_{cm} = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

    x_{cm} = 1/92 (448.8)

    x_{cm} = 4,878 cm

    x_{cm} = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at   d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

    x_{cm} ’= 1/M  (m₁ d₁ + m₂ d₂ + m₃ d₃)

   

   x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃)

Part d) Let's calculate the value of the center of mass

    x_{cm}’= 1/92 ((24 (-1.9) +56 3.9)

    x_{cm}’= 1/92 (172.8)

    x_{cm}’= 1.88 cm

This distance is to the right of the central marble

3 0
3 years ago
PLEASE HELP WITH THIS!
jok3333 [9.3K]

Answer: A

Explanation:

Parallel circuit. The source current is three times the current of a single bulb

7 0
3 years ago
Formula for finding displacement
FrozenT [24]
Displacement = (final position) - (initial position) = change in position.

HOPE IT HELPS YOU
5 0
4 years ago
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Hey besties, im having a mental breakdown on this question. pls help :D
forsale [732]

With "toward the wall" designated as the positive direction, the average acceleration felt the ball in the 0.0158 s of contact with the wall is

a_{\rm ave} = \dfrac{-26.412\frac{\rm m}{\rm s} - 28.4\frac{\rm m}{\rm s}}{0.0158\,\mathrm s} ≈ \boxed{-3470 \dfrac{\rm m}{\mathrm s^2}}

8 0
3 years ago
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