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Cloud [144]
3 years ago
10

A 200-kg object has a 50-Newton rightward net force being applied to it. What is the magnitude of the rightward acceleration on

the object m/s^2?
Physics
1 answer:
katovenus [111]3 years ago
8 0
The formula for acceleration is force divided by mass.

Force = 50 N

Mass = 200 kg

Acceleration = ???

\frac{50}{200} = \frac{1}{4}

\frac{1}{4} = 0.25

~~

Acceleration = 0.25 m/s²

~~

Next time use, f = ma in a triangle,

F

M A

~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey
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irina [24]
As the water plunges, its velocity increases. Its potential energy<span> becomes kinetic</span>energy<span>. The law of conservation of </span>energy<span> states that when one form of </span>energy<span> is</span>transformed<span> to another, no </span>energy<span> is destroyed in the process. ... So the total amount of </span>energy<span> is the same before and after any </span>transformation<span>.

hope it helps

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5 0
3 years ago
What is the average speed of a car that moved 40 km in 3 hours?
Naya [18.7K]

<u>A</u> would be the answer, since it would take about 3 hours to cover most of it, 10km/h would be the average speed.


really hope this helps.

3 0
3 years ago
Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m
Phoenix [80]

Answer:

a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. t_2=2.5714\ s is the time taken to stop after braking

Explanation:

Given:

  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
  • distance between the cars, \Delta s=45\ m
  • deceleration of the leading car after braking, a_1=-2\ m.s^{-2}

a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

where:

v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

7 0
3 years ago
Consider a horse pulling a buggy. Is the
Degger [83]

Yes, it's true.

But 2nd Newton Law always come to play when the horse is to move forward because obviously the forces interact antagonistically and mass has to be accounted for.

That's what I think. Hope it's right, all the best.

8 0
3 years ago
Traveling in circle requires a net force
seraphim [82]

Answer:

<u><em>Circular motion requires a net inward or "centripetal" force. Without a net centripetal force, an object cannot travel in circular motion. In fact, if the forces are balanced, then an object in motion continues in motion in a straight line at constant speed.</em></u>

Explanation:

3 0
3 years ago
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