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Alex17521 [72]
3 years ago
5

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

7 0
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:

8 0
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:

<h3>b) Time</h3>

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

7 0
3 years ago
Read 2 more answers
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