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2 years ago

Which statement accurately describes the charge of the nucleus of an atom?

Physics

1 answer:

Aleksandr [31]2 years ago

3 0

Answer:

A. The nucleus can be either positively charged or neutral.

Explanation:

The nucleus of an atom contains protons and neutrons only. Whereas electrons revolve around the nucleus. Protons are positively charged, neutrons have no charge and electrons are negatively charged.

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4) The components of Vector A are given as follows Ax = -2.4 Ay = + 3.8 What is the magnitude of the A, and what is the angle th

Monica [59]

Answer:

A = 4.49

α = 57.72°

Explanation:

Knowing the magnitude of x & y of a vector we can determine the total magnitude of a vector.

$A=\sqrt{A_{x}^{2} +A_{y}^{2} } \\A=\sqrt{(2.4)^{2} +(3.8)^{2} }\\A=4.49$

The angle tangent can be used to determine the angle.

$tan(\alpha )=\frac{3.8}{2.4}\\tan(\alpha ) =1.5833\\\alpha =tan^{-1}(1.5833) \\\alpha = 57.72 (deg)$

5 0

2 years ago

A cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5 m/s2 for 8.0

Katarina [22]

Answer:

V = 15m/s

Explanation:

Given the following data;

Initial velocity = 3m/s

Time = 8secs

Acceleration = 1.5m/s²

To find the final velocity, we would use the first equation of motion;

V = U + at

Substituting into the equation, we have

V = 3 + 1.5*8

V = 3 + 12

V = 15m/s

6 0

2 years ago

HELP!!!! WITH THIS PLSSS!!!

AlekseyPX

Answer:

i cant do your whole assignment for you man

Explanation:

5 0

2 years ago

Let’s look at a radio-controlled model car. Suppose that at time t1=2.0st1=2.0s the car has components of velocity vx=1.0m/svx=1

Nonamiya [84]

Answer:

$a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

Magnitude of accleration is $6\ \text{m/s}^2$ and the direction is $0^{\circ}$

Explanation:

$t_1=2\ \text{s}$

$v_x=1\ \text{m/s}$

$v_y=3\ \text{m/s}$

$t_2=2.5\ \text{s}$

$v_x=4\ \text{m/s}$

$v_y=3\ \text{m/s}$

Average acceleration in the different axes

$a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2$

$a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2$

The components of the acceleration is $a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

The magnitude of acceleration

$a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2$

Direction

$\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}$

The magnitude of accleration is $6\ \text{m/s}^2$ and the direction is $0^{\circ}$ .

7 0

2 years ago

The nuclear particle carrying a charge of +1e is the

babunello [35]

The nuclear particle is a proton

6 0

2 years ago

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