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bixtya [17]
3 years ago
12

5.0 kg, with a bullet of mass 0.1 kg. The target was mounted on

Physics
1 answer:
kari74 [83]3 years ago
8 0

Answer:

306 m/s

Explanation:

Law of conservation of momentum

m1v1 + m2v2 = (m1+m2)vf

m1 is the bullet's mass so it is 0.1 kg

v1 is what we're trying to solve

m2 is the target's mass so it is 5.0 kg

v2 is the targets velocity, and since it was stationary, its velocity is zero

vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s

plugging in, we get

(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)

(0.1)(v1) + 0 = 30.6

(0.1)(v1) = 30.6

v1 = 306 m/s

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The element Ne, neon, has ____ (number) complete shells having its ____ electrons in second _____.
KatRina [158]

Answer:

2, 8 and shell

Explanation:

Neon as atomic number 10. Since for each shell, electrons equal 2n².

When n = 1, 2n² = 2(1)² = 2

When n = 2, 2n² = 2(2)² = 8

So it fills both the first and second shell with 2 and 8 electrons respectively to achieve its stable atomic state. The rest of the 8 electrons go into the second shell because the first shell has achieved its stable dual configuration of two electrons. The next shell requires a maximum of 8 electrons to achieve stability so, the remaining electrons fill it up to achieve the stable octet configuration.

8 0
3 years ago
The Indianapolis speedway consists of a 2.5 mile track having four turns, each 0.25 mile long and banked at 9 12'
blsea [12.9K]

Answer: Your question is missing below is the question

Question : What is the no-friction needed speed (in m/s ) for these turns?

answer:

20.1 m/s

Explanation:

2.5 mile track

number of turns = 4

length of each turn = 0.25 mile

banked at 9 12'

<u>Determine the no-friction needed speed </u>

First step : calculate the value of R

2πR / 4 =  πR / 2

note : πR / 2 = 0.25 mile

∴ R = ( 0.25 * 2 ) / π

      = 0.159 mile ≈ 256 m

Finally no-friction needed speed

tan θ = v^2 / gR

∴ v^2 = gR * tan θ

 v = √9.81 * 256 * tan(9.2°)  = 20.1 m/s

8 0
3 years ago
While playing catch with my dog I bounced the ball off the ground at a 30-degree angle. It had a range of 6 meters.
baherus [9]

The initial velocity of the ball is 8.2 m/s

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A uniform motion along the horizontal direction, at constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (g=9.8 m/s^2, acceleration of gravity)

The range of a projectile can be derived by the equation of motions along the two directions, and it is found to be:

d=\frac{v^2 sin 2\theta}{g}

where

v is the initial velocity of the projectile

\theta is the angle of projection

g is the acceleration of gravity

For the ball in this problem, we have

\theta=30^{\circ}

d = 6 m is the range

Solving for v, we find the initial velocity:

v=\sqrt{\frac{gd}{sin 2\theta}}=\sqrt{\frac{(9.8)(6)}{sin (2\cdot 30^{\circ})}}=8.2 m/s

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0
3 years ago
1 point
Nuetrik [128]

Answer:

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Explanation:

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7 0
3 years ago
Would a pencil rolling off a desk be kinetic energy?
Anit [1.1K]
Yes. A pencil sitting on the desk in potential energy. <span />
5 0
3 years ago
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