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katovenus [111]

3 years ago

9

Which is a resource that the National Park Service manages?

2 answers:

gizmo_the_mogwai [7]3 years ago

7 0

Answer:

wildlife refuges

Explanation:

Rasek [7]3 years ago

3 0

Answer:

Which is a resource that the National Park Service manages?

A. memorials

B. wildlife refuges

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4 years ago

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An atom is the most __________ unit of living and nonliving things.

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Basic is the answer. (i think so)

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4 years ago

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statuscvo [17]

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7 0

2 years ago

A 91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

matrenka [14]

Answer:

The velocity of the players will be <u>2.88 m/s</u> in the <u>east</u> direction.

Explanation:

Let 'v' be the velocity of the players after collision.

Consider the east direction as positive direction.

Given:

Mass of the first player is, $m_1 = 91.5$ kg

Initial velocity of the first player is, $u_1 = 2.73$ m/s

Mass of the second player is, $m_2 = 63.5$ kg

Initial velocity of the second player is, $u_2 = 3.09$ m/s

In order to solve this problem we use the law of conservation of momentum which says that the total momentum must be conserved before and after the collision. So we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$

Solving for v, we get:

$v = \frac{m_1 u_1+m_2 u_2}{m_1+m_2}=\frac{(91.5)(2.73)+(63.5)(3.09)}{91.5+63.5}=2.88\ m/s$

Therefore, their velocity after the collision is 2.88 m/s.

The sign of the velocity after collision is positive. So, the players will move in the east direction only after collision.

3 0

3 years ago

An electron is moving east in a uniform electric field of 1.47 {\rm N/C} directed to the west. At point A, the velocity of the e

zimovet [89]

Answer:

a) $v = 6.25\times 10^{5} m/s$

b) $v = 1.73\times 10^{4} m/s$

Explanation:

Given data:

Electric field = 1.47 N/C

velocity of electron is $4.55\times 10^5 m/s$

distance of point b from point A is 0.55 m

we know that acceleration of particle is given as

a) for electron

$a =\frac{q E}{m}$

$a = \frac{1.6\times 10^{-19} \times 1.47}{9.1\times 10^{-31}}$

$a = 2.58\times 10^{11} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 20.7025 \times 10^{10} + 2\times 2.58\times 10^{11} \times 0.355$

$v = 6.25\times 10^{5} m/s$

b) for proton

$a = \frac{1.6\times 10^{-19} \times -1.47}{1.6\times 10^{-27}}$

$a = -1.41\times 10^{8} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 3.8025 \times 10^{8} - 2\times 1.41\times 10^{8} \times 0.355$

$v = 1.73\times 10^{4} m/s$

3 0

3 years ago