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dangina [55]
3 years ago
5

Ramon and sally are observing a toy car speed up as it goes around a circular track. ramon says, "the car's speeding up, so ther

e must be a net force parallel to the track." "i don't think so," replies sally. "it's moving in a circle, and that requires centripetal acceleration. the net force has to point to the center of the circle."
Physics
1 answer:
Anestetic [448]3 years ago
4 0
The changing speed demonstrates the nearness of an increasing speed; this centripetal quickening is of enduring with significance and coordinated in any regard cases toward the hub of the pivot. This increasing speed is, thus, created by utilizing a centripetal anxiety that is furthermore reliable in significance and coordinated toward the hub of revolution. contained in regards to revolution around a settled hub of a firm build that is not irrelevantly little in contrast with the span of the course, each and every molecule of the constitution portrays a uniform around activity with the equivalent rakish speed, yet with speed and increasing speed different with the range with comprehend to the hub.
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A set of statements that belong together as a group and contribute to the function definition is known as a(n) ________.
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Block. hope this helps.
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3 years ago
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Feliz [49]

a. 46 m/s east

The jet here is moving with a uniform accelerated motion, so we can use the following suvat equation to find its velocity:

v=u+at

where

v is the velocity calculated at time t

u is the initial velocity

a is the acceleration

The jet in the problem has, taking east as positive direction:

u = +16 m/s  is the initial velocity

a=+3 m/s^2  is the acceleration

Substituting t = 10 s, we find the final velocity of the jet:

v=16 + (3)(10)=46 m/s

And since the result is positive, the direction is east.

b. 310 m

The displacement of the jet can be found using another suvat equation

s=ut+\frac{1}{2}at^2

where

s is the displacement

u is the initial velocity

a is the acceleration

t is the time

For the jet in this problem,

u = +16 m/s  is the initial velocity

a=+3 m/s^2  is the acceleration

t = 10 s is the time

Substituting into the equation,

s=(16)(10)+\frac{1}{2}(3)(10)^2=310 m

4 0
3 years ago
PLEASE HELP MEEEEEE!!!!!!!!!!!!!!!!!!
Simora [160]
You would need to freeze it in a freezer. Hope this helps if it does could I have brainlist thanks
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A block of 200 g is attached to a light spring with a force constant of 5 N / m and freely in a horizontal plane vibrates. The m
Finger [1]

Answer:

m  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 mm  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 m

Explanation:

That is a reason

8 0
3 years ago
Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

3 0
2 years ago
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