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aev [14]
3 years ago
12

Help help help help help

Physics
1 answer:
mina [271]3 years ago
7 0
Please mark me brainliest

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Two stunt drivers drive directly toward each other. At time t=0 the two cars are a distance D apart, car 1 is at rest, and car 2
lesantik [10]

Answer: Hello there!

We know this:

The distance between the cars at t= 0 is D.

car 2 has an initial velocity of v0 and no acceleration.

car 1 has no initial velocity and a acceleration of ax that starts at  t = 0

then we could obtain the acceleration of the car 1 by integrating the acceleration over the time; this is v(t) = ax*t where there is not a constant of integration because the car 1 has no initial velocity.

Because the cars are moving against each other, we want to se at what time t they meet, this is equivalent to see:  

position of car 1 + position of car 2 = D

and in this way we could ignore constants of integration :D

for the position of each car we integrate again:  

P1(t) = (1/2)ax*t^2 and P2(t) = v0t

v0t + (1/2)ax*t^2 = D

v0t + (1/2)ax*t^2  - D = 0

now we can solve it for t using the Bhaskara equation.

t = \frac{-v0 +\sqrt{v0^{2} + 4*(1/2)ax*D } }{2(1/2)ax} =\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}

that we cant solve witout knowing the values for v0, D and ax. But you could replace them in that equation and obtain the time, where you must remember that you need to choose the positive solution (because this quadratic equation has two solutions).

Now we want to know the velocity of car 1 just before the impact, this can be calculated by valuating the time in the as the time that we just found in the velocity equation for the car 1, this is:

v(\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}) = ax*\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax} = {-v0 +\sqrt{v0^{2} + 2ax*D }

where again, you need to replace the values of v0, D and ax.

7 0
3 years ago
After a while, the car started to go around a long bend, still maintaining its constant speed of 55 miles per hour. Is there a n
sukhopar [10]

Answer:

4) True. The change of direction needs an unbalanced force

Explanation:

Let us propose the resolution of the problem using Newton's second law.

    F = m a

As the car is spinning the acceleration is centripetal

    a = v2.r

   

    F = m v2 / r

We can see that as the velocity of a vector even if its module does not change, the change of direction requires an external force.

Now we can analyze the statement if they are true or false

1) and 3) False, even when the speed changes, the direction changes

2) False with the speed change can be determined

4) True. The change of direction needs an unbalanced force

5) False are different things. the direction is where it is going and the speed is the magnitude of the vector

6 0
2 years ago
A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =
Stels [109]

Answer:

628.022466 N

8.61 m/s

Explanation:

m = Mass

\mu = Coefficient of friction

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

F_f=\mu mg\\\Rightarrow F_f=0.646\times 99.1\times 9.81\\\Rightarrow F_f=628.022466\ N

Magnitude of frictional force is 628.022466 N

F=ma\\\Rightarrow a=\frac{F_f}{m}\\\Rightarrow a=\frac{628.022466}{99.1}\\\Rightarrow a=6.33726\ m/s^2

v=u+at\\\Rightarrow 0=u-6.33726\times 1.36\\\Rightarrow u=8.61\ m/s

Initial speed of the player is 8.61 m/s

4 0
3 years ago
If you were creating a presentation that would cover the different types of muscles in the body, which groups of muscle would yo
diamong [38]
It would be best to cover the cardiac, smooth, and skeletal muscles! =)
6 0
2 years ago
Read 2 more answers
A spring is 17 cm long when it is lying on a table. One end is then attached to a hook and the other end is pulled by a force th
RideAnS [48]

Answer:

im pretty sure it B but I recommend waiting for another person. I used the workdone formula (Force*Dictance*cos(theta) and got 55 Joules

Explanation:

8 0
3 years ago
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