What does the author explain in this passage?

on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker

Aleksandr-060686 [28]

B4 the tackle:

The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north 

and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east 

After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle 

The vector triangle is right angled: 

magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s 

so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] 

v(f) = 5.6 m/s (to 2 sig figs) 

direction of v(f) is the same as the direction of the final momentum 

so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) 

so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E 

btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it

4 0

2 years ago

Why electrons flow in a wire when connected to a battery?

vesna_86 [32]

Answer:

The metal atoms in the wire can't move, but their outer electrons can. The force pushes those electrons and they move to further parts of the wire, trying to reach the other end. As the electrons move away, new electrons flow into the wire through the battery to take their place.

Explanation:

6 0

2 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

2 years ago

How do the processes of weathering, erosion, and deposition affect the ecoregions of Texas? Give a specific example of how each

NISA [10]

Answer:

Weathering, erosion, and deposition from the terrestrial surface topography and soil characteristics. These processes, for example, have formed a variety of landforms in Texas like beaches, plateaus, mountains, and canyons as well as soil types like fertile soil, clay-rich soil, and sandy soil. The combination of topography, soil, and climatic conditions in an area defines the types of habitats that the area can support this is crucial to ecoregion classification. Ten separate ecoregions occur in Texas including 1) East Texas Pineywoods, 2) Gulf Coast Prairies and Marshes, 3) Oak Woods and Prairies, 4) Blackland Prairie, 5) cross timbers and prairies (6) Rolling Plains, (7) High Plains, (8) TransPecos, (9) South Texas Plains, (Brush Country), and (10) Edwards Plateau. Such ecoregions are named for the major types of habitats topographical features (e.g. Edwards Plateau) present in their areas. The weathering, erosion, and deposition of each of these ecoregions have an important influence.

5 0

2 years ago

Four balls have the same temperature. Which ball has the most thermal

puteri [66]

Answer:

a bowling ball because it has the most mass.

7 0

2 years ago

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