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1 year ago

What type of motion would produce positive acceleration in an object? What type of motion would produce negative acceleration?

Physics

1 answer:

Bumek [7]1 year ago

4 0

Answer:

positive acceleration: releasing an object from a height - acceleration due to the force of gravity, the speed of the object will increase as it falls

negative acceleration: applying breaks on the tires to slow down the car - deceleration due to frictional forces

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Which number is larger value : 5.6*10^23 or 8.9*10^6<br><br> Need Help

oee [108]

Answer:

5.6*10^23. if 10^n is greater, that means its the larger value. hope dis helps

Explanation:

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2 years ago

Sam is walking through the park. He hears a police car siren moving on the street coming toward him. What happens to the siren s

xeze [42]

The sound gets louder as it gets closer and when it passes is gets softer

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3 years ago

Read 2 more answers

A 3 Ω resistor, an 8 Ω resistor, and a 14 Ω resistor are connected in parallel across a 6 V battery. Determine the equivalent re

Schach [20]

Answer:

R=25Ω

Explanation:

Given data:

R1=3 Ω

R2=8 Ω

R3=14 Ω

V=6

Now,

To Find,

R=?

Formula

R=R1+R2+R3

put the values,

R=3+8+14

R=25Ω

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3 years ago

What is responsible for the unusual shape of a coastline

solong [7]

If a coastline has a very unusual shape it's normally caused by either a flood or dam.

hope this helps!

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3 years ago

Whenever two apollo astronauts were on the surface of the moon, a third astronaut orbited the moon. assume the orbit to be circu

almond37 [142]

Missing question:
"Determine (a) the astronaut’s orbital speed v and (b) the period of the orbit"

Solution

part a) The center of the orbit of the third astronaut is located at the center of the moon. This means that the radius of the orbit is the sum of the Moon's radius r0 and the altitude ( $h=430 km=4.3 \cdot 10^5 m$ ) of the orbit:
$r= r_0 + h=1.7 \cdot 10^6 m + 4.3 \cdot 10^5 m=2.13 \cdot 10^6 m$
This is a circular motion, where the centripetal acceleration is equal to the gravitational acceleration g at this altitude. The problem says that at this altitude, $g=1.08 m/s^2$ . So we can write
$g=a_c= \frac{v^2}{r}$
where $a_c$ is the centripetal acceleration and v is the speed of the astronaut. Re-arranging it we can find v:
$v= \sqrt{g r}= \sqrt{(1.08 m/s^2)(2.13 \cdot 10^6 m)}=1517 m/s = 1.52 km/s$

part b) The orbit has a circumference of $2 \pi r$ , and the astronaut is covering it at a speed equal to v. Therefore, the period of the orbit is
$T= \frac{2 \pi r}{v} = \frac{2\pi (2.13 \cdot 10^6 m)}{1517 m/s} =8818 s = 2.45 h$
So, the period of the orbit is 2.45 hours.

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3 years ago