1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
DedPeter [7]
3 years ago
11

A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is d

irection of its velocity?
FIND THE DIRECTION I WILL MARK YOU BRAINLIEST FOR THE CORRECT ANSWER

Physics
1 answer:
QveST [7]3 years ago
3 0

Answer:

  The target's velocity is about 1320 m/s in the direction 265.7°.

Explanation:

In order for there to be a collision between missile and target, we must have ...

  (target starting position) + (target movement) = (missile movement)

assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...

  (10.2 s)(1350 m/s) = 13,770 m

__

We are interested in the target movement, so we can solve for that:

  (target movement) = (missile movement) - (target starting position)

In terms of meters, this is ...

  (target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°

The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.

As a positive angle, the target's direction is ...

  -94.3° +360° = 265.7°

The direction of the target's velocity is 265.7°.

_____

If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.

Using rectangular coordinates, we have ...

  13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)

  23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)

Then the difference is ...

  (12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)

and the (3rd-quadrant) angle is ...

  target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°

__

The target's speed is found by dividing the distance it covers by the time it takes.

  √(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s

You might be interested in
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
vlada-n [284]

Answer:

\frac{1}{10}M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒I_{1} \times W_{1} =I_{2} \times W_{2}

  • I_{1} is the moment of interia of earth before impact
  • W_{1} is the angular velocity of earth about an axis passing through the center of earth before impact
  • I_{2} is moment of interia of earth and asteroid system
  • W_{2} is the angular velocity of earth and asteroid system about the same axis

let  W_{1}=W

since \text{Time period of rotation}∝\frac{1}{\text{Angular velocity}}

⇒ if time period is to increase by 25%, which is \frac{5}{4} times, the angular velocity decreases 25% which is \frac{4}{5}  times

therefore W_{1} = \frac{4}{5} \times W_{1}

I_{1}=\frac{2}{5} \times M\times R^{2}(moment of inertia of solid sphere)

where M is mass of earth

           R is radius of earth

I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where M_{1} is mass of asteroid

⇒ \frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})

\frac{1}{2} \times M\times R^{2}= (\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})

M_{1}\times R^{2}= \frac{1}{10} \times M\times R^{2}

⇒M_{1}=}\frac{1}{10} \times M

3 0
2 years ago
A force of 19 newtons is applied on a cart of 2 kilograms, and it experiences a frictional force of 1.7 newtons. What is the acc
Umnica [9.8K]
Frictional force always opposes applied force, so the net force on the cart would have to be 19N - 1.7N. The acceleration can then be solved by using the relation: F = ma. This is shown below:

Net force = 19 - 1.7 = 17.3 N

Acceleration = Force / mass
Acceleration = 17.3 / 2
Acceleration = 8.65 N/m
4 0
2 years ago
Read 2 more answers
What does it mean if the moon is waxing and waning
konstantin123 [22]
Both verbs come from Olde English. 
That's why everybody clearly understood their meaning until
a hundred years ago, but nobody understands them now.


"Waxing"  =  growing

For two weeks after the New Moon, it's growing toward Full.
First it's a waxing crescent for a week, then it's waxing gibbous.


"Waning"  =  shrinking

For two weeks after the Full Moon, it's shrinking toward New.
First it's waning gibbous for a week, then it's a waning crescent.
6 0
2 years ago
A 40-cmcm-long tube has a 40-cmcm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. A
Licemer1 [7]

Answer:

1070 Hz

Explanation:

First, I should point out there might be a typo in the question or the question has inconsistent values. If the tube is 40 cm long, standing waves cannot be produced at 42.5 cm and 58.5 cm lengths. I assume the length is more than the value in the question then. Under this assumption, we proceed as below:

The insert in the tube creates a closed pipe with one end open and the other closed. For a closed pipe, the difference between successive resonances is a half wavelength \frac{\lambda}{2}.

Hence, we have

\dfrac{\lambda}{2}=58.5-42.5=16 \text{ cm}

\lambda=32\text{ cm}=0.32 \text{ m}.

The speed of a wave is the product of its wavelength and its frequency.

v=f\lambda

f=\dfrac{v}{\lambda}

f=\dfrac{343}{0.32}=1070 \text{ Hz}

7 0
3 years ago
What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from
ElenaW [278]
When is at the end of the runway the velocity of the plane is given by the equation vf^{2}=0+2*a*s    where s=1800 m is the runway length. Thus
vf^{2}=2*5*1800=18000 (m/s)^{2}      
vf =134.164 (m/s)  

At half runway the velocity of the plane is
v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}&#10; 
v= \sqrt{9000}=94.87 ( \frac{m}{s})

Therefore at midpoint of runway the percentage of takeoff velocity is
‰P= \frac{v}{vf}=  \frac{94.87}{134.164}=0.707
6 0
3 years ago
Other questions:
  • Two blocks of masses 3.0 kg and 5.0 kg are connected by a spring and rest on a frictionless surface. They are given velocities t
    10·1 answer
  • Which objects would sink in honey, which has a density of 1.4 g/cm³? Check all that apply.
    10·2 answers
  • The ballistic pendulum is a device used to measure the speed of a projectile such as a bullet. The projectile of mass m is fired
    14·1 answer
  • Ethan made a diagram to compare examples of the first and second laws of thermodynamics. What belongs in the areas marked X and
    5·1 answer
  • a train leaves a station heading east at 50 mph from that same station a car drives north at a speed of 30mph after 3 hours how
    5·1 answer
  • Electrical systems are governed by Ohm’s law, which states that V = IR, where V is the voltage, I is the current, and R is the r
    11·1 answer
  • Ball Distance Traveled
    10·1 answer
  • The amount of matter in an object is called its
    14·2 answers
  • How do icebergs form
    5·1 answer
  • What is meant by an acceleration of negative 2metre per second square​
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!