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3 years ago

Accelerations are produced by equal forces or unequal forces?

Physics

1 answer:

sukhopar [10]3 years ago

5 0

Answer: unequal forces

Explanation: in order for something to accelerate it must speed up or slow down . When unequal forces react it casuse a change in motion which is also know as acceleration .

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It has been suggested that rotating cylinders about 9 mi long and 5.9 mi in diameter be placed in space and used as colonies. Wh

mezya [45]

Answer:

$\omega = 4.5\times 10^{-2} rad/s$

Explanation:

Given data:

Rotating cylinder length = 9 mi

diameter of cylinder is 5.9 mi

we know that linear acceleration is given as

a = r ω^2

where ω is angular velocity

so $\omega = \sqrt{\frac{a}{r}}$

$r = \frac{5.9}[2} \frac{1609 m}{1 mi} = 4.746\times 10^{3} m$

$\omega = \sqrt{\frac{9.80}{4.746\times 10^{3}}}$

$\omega = 4.5\times 10^{-2} rad/s$

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3 years ago

The electric potential at a point equidistant from two particles that have charges +Q and –Q is larger than zero. a. smaller tha

Marizza181 [45]

Answer:

idk, idk cause i'm steppin on my toes and i can't stop i make flips ou of my flops

Explanation:

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3 years ago

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

Molodets [167]

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 jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e 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 fina 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 ony 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

3 years ago

Describe the changes of motion when marble is moved from one viewpoint to another.

slavikrds [6]

Answer:

Explanation:

according to Newton First Law of Motion (Law of Inertia); An object at rest will stay at rest, forever, as long as nothing pushes or pulls on it. An object in motion will stay in motion, traveling in a straight line, forever, until something pushes or pulls on it.

the marble will move in a straight line

5 0

3 years ago

What does aerobic refer to?

Gnesinka [82]

Answer:

A) how your body uses oxygen

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3 years ago

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