Where were the girls heading when the car broke down? in the movie hidden figures

Physics

1 answer:

Ilia_Sergeevich [38]2 years ago

5 0

Explanation:

Katherine Johnson, NASA Mathematician Featured in 'Hidden Figures,' Dies at 101

Feb 25, 2020 — Their story was told in the 2016 Hollywood film “Hidden Figures,” based on Margot Lee Shetterly's nonfiction book of the same title , ...

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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

Ball A 1.55kg moving right at 8.76 m/s makes a head-on collision with ball B (0.752 kg) moving left at 11.4 m/s. After, ball B m

Illusion [34]

1.15 m/s to the left (3 sig. fig.).

<h3>Explanation</h3>

Momentum is conserved between the two balls if they are not in contact with any other object. In other words,

$p_{\text{A,initial}} + p_{\text{B,initial}}=p_{\text{A,final}} + p_{\text{B,final}}$

$m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}}$ , where

$m$ stands for mass and
$v$ stands for velocity, which can take negative values.

Let the velocity of objects moving to the right be positive.

$m_\text{A} = 1.55\;\text{kg}$ ,
$m_\text{B} = 0.752\;\text{kg}$ .

Before the two balls collide:

$v_\text{A} = +8.76\;\text{m}\cdot\text{s}^{-1}$ ,
$v_\text{B} = -11.4\;\text{m}\cdot\text{s}^{-1}$ .

After the two balls collide:

$v_\text{A}$ needs to be found,
$v_\text{B} = +9.03\;\text{m}\cdot\text{s}^{-1}$ .

Again,

$m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}}$ ,

$1.55 \times (+8.76) + 0.752 \times (-11.4) = 1.55\;{\bf v_{\textbf{A,final}}} + 0.752 \times (+9.03)$ .

$v_{\text{A,final}} = \dfrac{1.55 \times (+8.76) + 0.752 \times (-11.4)-0.752 \times (+9.03)}{1.55} = -1.15\;\text{m}\cdot\text{s}^{-1}$ .

$v_{\text{A,final}}$ is negative? Don't panic. Recall that velocities to the right is considered positive. Accordingly, negative velocities are directed to the left.