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iVinArrow [24]
3 years ago
13

Help please!!!! Brainlyist to the person who answers

Physics
2 answers:
valkas [14]3 years ago
8 0

Answer:

Speed up yes it could depends on initial forces and ending forces

keep in constant motion yes

Slow down yes it could depends on initial forces and ending forces

Change directions yes it could depends on initial forces and ending forces

Could not have a net force of Zero

Explanation:

Speed up yes it could depends on initial forces and ending forces

keep in constant motion yes

Slow down yes it could depends on initial forces and ending forces

Change directions yes it could yes it could depends on initial forces and ending forces

Could not have a net force of Zero

Ksenya-84 [330]3 years ago
6 0
An unbalanced force acting on an object causes it to accelerate. ... The more mass the object has, the more inclined it is to resist any change to its motion.
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-- While it's drifting in space, it's a <em>meteoroid</em>.

-- While it's falling through the atmosphere burning up and making a bright streak of light, it's a <em>meteor</em>.

-- If it doesn't completely burn up and there's some of it left to fall on the ground, then the leftover piece on the ground is a <em>meteorite</em>.

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4 years ago
Look at the resistor illustrated in the figure above. Based on your knowledge of the resistor color code, what are the resistanc
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A because of the resistors are four in this options first option is multiplied by 4
6 0
3 years ago
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A particle moves along the x axis so that its velocity at time t is given by v(t)=
vodomira [7]

Answer:

a = 0.7267 ,  acceleration is positive therefore the speed is increasing  

Explanation:

The definition of acceleration is

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they give us the function of speed

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remember that the angles are in radians

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6 0
3 years ago
How is the acceleration of an object in newtons second law related to its mass and the net force on it?
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B) Acceleration is directly proportional to the mass of the object
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4 years ago
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What is 64 nanometers to m?
defon
\sf Hello!

\sf We\: know \:that,
\sf 1\: meter = \sf 10^{9} nm

\sf Then,

\sf Distance\: in \:m = \sf Distance\: in\: nm × \dfrac{\sf 1}{\sf 10^{9}}\: \sf m

⇒ \sf Distance\: in \:m = \sf 64 × 10^{-9} \:m

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