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zhuklara [117]
2 years ago
15

A rocket is launched at an angle of = 49° above the horizontal with an initial speed vi = 54 m/s, as shown below. It moves for 2

5 s along its initial line of motion wth an acceleration of 24.4 m/s2. At this time, its engines fail and the rocket proceeds to move as a free body.

Physics
1 answer:
4vir4ik [10]2 years ago
7 0
I never seem to pretty best friends
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Transverse, surface, and longitudinal waves are all __________ waves because they __________.
algol [13]

Answer:

a. mechanical; require a medium to travel through

Explanation:

Longitudinal, transverse and surface waves are types of mechanical waves. For example, within the longitudinal waves are the sound waves, which needs a medium to propagate like the air. This is why sound does not travel in a vacuum.

And an example of a transverse wave is the waves that form in the water when a rock is thrown (ripples), these waves need a medium (the water) to propagate.

On the other hand, electromagnetic waves such as light waves do not need a medium to propagate, this is why we can see the light of distant stars because their light travels through the vacuum until it reaches us.

So, the answer is:

Transverse, surface, and longitudinal waves are all mechanical waves because they require a medium to travel through .

5 0
3 years ago
Un automóvil acelera de 0 a 140 km/h en 9.8 segundos, determina su aceleración
guajiro [1.7K]

Answer:

The acceleration is 14.28 km/h^2

Explanation:

Step one:

Given data

initial speed u= 0 km/h

final speed v= 140km/h

time t= 9.8 seconds

Required

The acceleration of the car

Step two:

From a= v-u/t

substitute

a= 140-0/9.8

a=140/9.8

a=14.28 km/h^2

6 0
3 years ago
The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T
zaharov [31]

a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than -0.25 m/s^2

f) The two trains avoid collision if the acceleration of the freight train is at least 0.35 m/s^2

Explanation:

a)

We can describe the position of the passenger train at time t with the equation

x_p(t)=u_p t + \frac{1}{2}at^2

where

u_p = 25.0 m/s is the initial velocity of the passenger train

a=-0.100 m/s^2 is the deceleration of the train

On the other hand, the position of the freight train is given by

x_f(t)=x_0 + v_f t

where

x_0=200 m is the initial position of the freight train

v_f = 15.0 m/s is the constant velocity of the train

The collision occurs if the two trains meet, so

x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

b)

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

x_f(t)=x_0 +v_ft

And substituting t = 22.5 s, we find:

x_f(22.5)=200+(15)(22.5)=537.5 m

We can verify that the passenger train is at the same position at the time of the collision:

x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m

So, the two trains collide at x = 537.5 m.

c)

In the graph in attachment, the position-time graph of each train is represented. We have:

  • The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
  • The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

d)

In order to avoid the collision, the freight train must have a initial position of

x_0'

such that the two trains never meet.

We said that the two trains meet if:

x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t

Re-arranging,

\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0

Substituting the values for the acceleration and the velocity,

0.05t^2-10t+x_0'=0

The solution of this equation is given by the formula

t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m

Therefore, the freight train must have a head start of 500 m.

e)

In this case, we want to find the acceleration a' of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t

Re-arranging

\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0

Substituting,

-0.5at^2-10t+200=0

The solution to this equation is

t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}

Again, the two trains never meet if the discriminant is negative, so

10^2-4\cdot (-0.5a') \cdot (200)

So, the deceleration must be smaller (towards negative value) than -0.25 m/s^2

f)

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2

where a_f is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2

Re-arranging,

\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0

And the solution is

t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}

Again, the two trains avoid collision if the discriminant is negative, so

10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
Water has a ph of 7. ph is defined as the negative log of the hydrogen ion concentration [h ]. explain why water is assigned a p
AfilCa [17]

Answer: Concentration of H+ =10^-7; log of 10^-7=7 so pH=7

Explanation:

4 0
1 year ago
Relate the structure of water to its ability to act as a solvent
pishuonlain [190]

Explanation:

Water (H2O) as a polar covalent molecule has its arrangement of oxygen and hydrogen atoms where one end (hydrogen) has a partially positive charge while the other side (oxygen) had a partially negative charge.

It is also capable of forming hydrogen bonds with polar molecules. Each water molecule can form two hydrogen bonds involving their hydrogen atoms and two further hydrogen bonds using the hydrogen atoms attached to neighboring water molecules.

5 0
3 years ago
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