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makvit [3.9K]
2 years ago
6

If the average time it takes for the cart from point 1 to point 2 is 0.2 s, calculate the angle θ from the horizontal of the tra

ck. Assume the track is frictionless. Hint: use the definitions of acceleration and Newton’s second law.
Physics
1 answer:
inn [45]2 years ago
5 0

Answer:

hehe

Explanation:

I dont know because I am a noob ant study

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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
Suppose you monitor a large number (many thousands) of stars over a period of 3 years, searching for planets through the transit
nasty-shy [4]

Answer:

2. You must be able to precisely measure variations in the star's brightness with time.

5. As seen from Earth, the planet's orbit must be seen nearly edge–on (in the plane of our line-of-sight).

6. You must repeatedly obtain spectra of the star that the planet orbits.

Explanation:

The transit method is a very important and effective tool for discovering new exoplanets (the planets orbiting other stars out of the solar system). In this method the stars are observed for a long duration. When the exoplanet will cross in front of theses stars as seen from Earth, the brightness of the star will dip. To observe this dip following conditions must be met:

1. The orbit of the planet should be co-planar with the plane of our line of sight. Then only its transition can be observed.

2. The brightness of the star must be observed precisely as the period of transit can be less than a second as seen from Earth. Also the dip in brightness depends on the size of the planet. If the planet is not that big the intensity dip will be very less.

3. The spectrum of the star needs to be studied and observe during the transit and normally to find out the details about the planets.

4. Also, the orbital period should be less than the period of observation for the transit to occur at least once.

4 0
3 years ago
How much is the velocity of a body when it travels 600m in 5 minutes?​
maks197457 [2]

answer

so unit of velocity is m/s

displacement=600m

5minutes should be converted to seconds

5×60=300 seconds

so,

velocity= displacement÷time

= 600m ÷300s

=2m/s or 2ms^-1

4 0
2 years ago
Read 2 more answers
Based on the table above, _______ planets are made of the same things as _______ moons.
wolverine [178]

Answer:

b

Explanation:

3 0
3 years ago
A 0.20 kg baseball is traveling at 40 m/s toward the batter. The ball is hit by the bat with a force of 200N, and is
Paraphin [41]

Answer:

Time, t = 0.015 seconds.

Explanation:

Given the following data;

Mass, m = 0.2kg

Force, F = 200N

Initial velocity, u = 40m/s

Final velocity, v = 25m/s

To find the time;

Ft = m(v - u)

Time, t = m(v - u)/f

Substituting into the equation, we have;

Time, t = 0.2(25 - 40)/200

Time, t = 0.2(-15)/200

Time, t = 3/200

Time, t = 0.015 seconds.

Note: We ignored the negative sign because time can't be negative.

8 0
3 years ago
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