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2 years ago

All of the following are contributions of the Space Center for Florida except which one?

Physics

2 answers:

jekas [21]2 years ago

6 0

What are the choices

Anna007 [38]2 years ago

4 0

Answer:

You have to give us an answer choice : )

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Calculer l intensité du courant qui le traverse

MissTica

Answer:

Le calcul du courant se fait avec deux éléments : la tension et la valeur de la résistance. Courant (A) = tension (V) / résistance (Ohm) ce qui donne la formule I = U/R.

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2 years ago

Which of the following is a type of T cell?

Bess [88]

Answer:

Explanation:

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2 years ago

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12. A frain moves from rest to a speed of 25 m/s in 30.0 seconds. What is its acceleration?

ziro4ka [17]

initial velocity=u=0m/s
Final velocity=v=25m/s
Time=t=30s

$\\ \tt\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \tt\longmapsto Acceleration=\dfrac{25-0}{30}$

$\\ \tt\longmapsto Acceleration=\dfrac{25}{30}$

$\\ \tt\longmapsto Acceleration=0.8m/s^2$

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2 years ago

An airplane is moving at 350 km/hr. If a bomb is

bogdanovich [222]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final height

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb's initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's final velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity </h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

2 years ago

A 3000-N force gives an object an acceleration of 15 m/s2. The mass of the object is

7nadin3 [17]

<h3>Answer:</h3>

200 kg

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Physics</u>

<u>Newton's Law of Motions </u>

Newton's 1st Law of Motion: An object at rest remains at rest and an object in motion stays in motion

Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration)

Newton's 3rd Law of Motion: For every action, there is an equal and opposite reaction<u> </u>

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] F = 3000 N

[Given] a = 15 m/s²

[Solve] m = <em>x</em> kg

<u>Step 2: Solve for </u><em><u>m</u></em>

Substitute in variables [Newton's Second Law of Motion]: 3000 N = m(15 m/s²)
[Mass] [Division Property of Equality] Isolate <em>m</em> [Cancel out units]: 200 kg = m
[Mass] Rewrite: m = 200 kg

5 0

2 years ago