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wel
3 years ago
6

An atomic nucleus is composed of A) protons. B) protons and neutrons. C) protons and electrons. D) protons, neutrons, and electr

ons.
Physics
1 answer:
In-s [12.5K]3 years ago
7 0

Answer:

B) protons and neutrons.

Explanation:

The protons and neutrons are located in the nucleus of the atom and represent most of the 'mass' of the atom, that's their count that determine the 'mass' of an atom (like 12 for Carbon).

The electrons rotate around the nucleus and have a negligible mass.

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Electric charges can only be transferred through a process called conduction.true or false
umka21 [38]

I'm going to say false hope that helped

5 0
3 years ago
Read 2 more answers
Read the list of phrases from the article.
ICE Princess25 [194]

Answer:

(D)

to establish an understanding of key concepts relating to population biology

Explanation:

Thats what I would go with but I didn't read the article so I don't know what context was used. Good luck! :)

8 0
3 years ago
An astronaut has a mass of 75 kg and is floating in space 500 m from his 125,000 kg spacecraft. What will be the force of gravit
nikdorinn [45]

Answer:

1. 2.5×10¯⁹ N

2. 3.33×10¯¹¹ m/s²

Explanation:

1. Determination of the force of attraction.

Mass of astronaut (M₁) = 75 Kg

Mass of spacecraft (M₂) = 125000 Kg

Distance apart (r) = 500 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

The force of attraction between the astronaut and his spacecraft can be obtained as follow:

F = GM₁M₂ /r²

F = 6.67×10¯¹¹ × 75 × 125000 / 500²

F = 2.5×10¯⁹ N

Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N

2. Determination of the acceleration of the astronaut.

Mass of astronaut (m) = 75 Kg

Force (F) = 2.5×10¯⁹ N

Acceleration (a) of astronaut =?

The acceleration of the astronaut can be obtained as follow:

F = ma

2.5×10¯⁹ N = 75 × a

Divide both side by 75

a = 2.5×10¯⁹ / 75

a = 3.33×10¯¹¹ m/s²

Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²

5 0
3 years ago
arzan, who weighs 700 N, swings from a cliff at the end of a convenient vine thatis 20 m long. From the top of the cliff to the
vekshin1

Answer:

= 7.07 m

Explanation:

The Tarzan reaches bottom of swing after descending 2.5 m,

change in his potential energy equals his kinetic energy at bottom of swing

m g h = (1/2) m v²   ,  

hence speed v of Tarzan at bottom of swing is given as  

v = ( 2 g h )1/2

= ( 2 × 9.8 × 2.5 )1/2

= 7 m/s

At the bottom of swing, if the vine breaks, then he is moving with horizontal velocity 7 m/s in gravitational field.  

If vertical distance from ground to bottom of swing is 5 m, then time t for Tarzan to reach ground is given by

S = (1/2)g t2 or   t = (2S/g)1/2

= ( 2 × 5 / 9.8 )1/2

= 1.01 s

Horizontal distance traveled by Tarzan = 1.01 × 7

= 7.07 m

7 0
3 years ago
Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as
viva [34]

Answer:

a) L=0. b) L = 262 k ^   Kg m²/s and c)  L = 1020.7 k^   kg m²/s

Explanation:

It is angular momentum given by

      L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]

Let's apply this relationship to our case

Let's start by breaking down the speed

      v₀ₓ = v₀ cosn 45

      voy =v₀ sin 45

      v₀ₓ = 9 cos 45

      voy = 9 without 45

      v₀ₓ = 6.36 m / s

      voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

   vfy² = voy²- 2 g y

   y = voy² / 2g

   y = (6.36)²/2 9.8

   y = 2.06 m

Let's calculate the angular momentum

L= \left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^   Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

     R = vo² sin 2θ / g

     R = 9² sin (2 45) /9.8

     R = 8.26 m

L = \left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^   kg m² /s

5 0
3 years ago
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