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
Vitek1552 [10]
3 years ago
7

Two airplanes leave an airport at the same time. The velocity of the first airplane is 650 m/h at a heading of 60.5 ◦ . The velo

city of the second is 560 m/h at a heading of 85◦ . How far apart are they after 2.9 h? Answer in units of m.

Physics
1 answer:
Assoli18 [71]3 years ago
7 0
First, illustrate the problem as shown in the attached picture. Next, let's find the distance traveled by planes A and B after 2.9 h.

Distance of A: 650 m/h * 2.9 h = 1,885 m
Distance of B: 560 m/h * 2.9 h = 1,624 m

Then, we use the cosine law to determine the distance x. The angle should be: 85 - 60.5 = 24.5°

x² = 1,885² + 1,624² - 2(1,885)(1,624)(cos 24.5°)
x = √619381.3183
<em>x = 787 m</em>

You might be interested in
Hey can someone send me the answer to this
bagirrra123 [75]

Cars 'A' and 'C' look like they're moving at the same speed.  If their tracks are parallel, then they're also moving with the same velocity.

5 0
3 years ago
A rock is lifted by a machine to a height of 10m. If it has a mass of 22 kilograms
zmey [24]

Answer:

2156J

Explanation:

Given parameters:

Height of lift  = 10m

Mass  = 22kg

Unknown:

Work done by the machine  = ?

Solution:

Work done is the force applied to move a body through a certain distance.

So;

        Work done  = Force x distance

Here;

       Work done  = mass x acceleration due to gravity x height

      Work done  = 22 x 9.8 x 10  = 2156J

5 0
3 years ago
Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4
Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0
3 years ago
Consider the following FBDs.FBDsAn elevator is suspended by a cable and moves uniformly upward. Which of the above free body dia
Drupady [299]

The force of gravity F_g will act downwards.

Normal force F_N will act upwards equal to the force of gravity.

A force due to uniform acceleration F_a will act upwards to move the elevator upwards.

Thus, figure E is the correct answer.

6 0
1 year ago
how does the force of attraction between large masses compare with the force of attraction between small masses at the same dist
Veseljchak [2.6K]
The force and amount of attracton is great for larger masses
4 0
3 years ago
Other questions:
  • A 125-g metal block at a temperature of 93.2 °C was immersed in 100. g of water at 18.3 °C. Given the specific heat of the metal
    14·1 answer
  • A convex lens is placed on a flat glass plate and illuminated from above with monochromatic red light. When viewed from above, c
    5·1 answer
  • The largest number of neurons within the brain and spinal cord are responsible for the central nervous system's complex internal
    13·1 answer
  • Can someone please help me
    6·2 answers
  • A rocket with mass 5.00 X 103 kg is in a circular orbit of radius 7.20 X 106 m around the earth. The rocket’s engines fire for a
    13·2 answers
  • .. As you increase the
    8·1 answer
  • F F= {mango, apple, banana, orange)​
    12·1 answer
  • 6) Calculate the density of sulfuric acid if 35.4 mL of the acid weighs 65.14 g.
    5·2 answers
  • What type of reaction occurs when an egg cooks in a pan?
    14·2 answers
  • Can any one give me a good egg drop design that is over 500 grams?​
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!