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lozanna [386]
3 years ago
12

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. the heavie

r fragment slides 6.70 m before stopping. how far does the lighter fragment slide?
Physics
1 answer:
Doss [256]3 years ago
6 0
By conservation of momentum,
Distance lighter fragment slide= 7*6.7=46.9m
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A 74.1 kg high jumper leaves the ground with
pishuonlain [190]

Answer:

4.80 m

Explanation:

We are given the mass of the high jumper, its initial velocity, and the acceleration of gravity. We are trying to find the vertical displacement of the high jumper.

Let's set the upwards direction to be positive and the downwards direction to be negative.

List out the relevant known variables.

  1. v₀ = 9.7 m/s
  2. a = -9.8 m/s²
  3. Δx = ?

We still need one more variable in order to use the constant acceleration equations. Since we are trying to find the max height of the jumper, we can use the fact that at the top of its trajectory, its final velocity will be 0 m/s.

    4. v = 0 m/s

Using these four variables, let's find the constant acceleration equation that contains these variables:

  • v² = v₀² + 2aΔx

Substitute the known values into the equation and solve for Δx.

  • (0)² = (9.7)² + 2(-9.8)Δx
  • 0 = 94.09 + (-19.6)Δx
  • -94.09 = -19.6Δx
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The high jumper can jump to a max height of 4.80 m.

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2 years ago
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Answer: 1.1 x 10^-10 C

Explanation:

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The answer is:
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What is the differece between force and motion?​
Jobisdone [24]

Answer:

---

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A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg
alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

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Solution:

Magnification of lens, m = -\frac{v}{u}

where

u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

\frac{1}{16.5} = \frac{ 0.90}{1.90u}

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

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