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vaieri [72.5K]
3 years ago
8

An airplane was moving 1200 km per minute traveling in a north-to-south direction when they were spotted by air traffic control.

Because they were only 2 miles off the ground and 5 miles away the airplane was contacted for safety concerns.
Which best describes the motion of the airplane?

Its acceleration was 1200 km/s
Its speed was 1200 km/m southward
Its velocity was 20 km/s southward
Its velocity was 20 km/m
Physics
2 answers:
Feliz [49]3 years ago
7 0

Its velocity was 20 km/s southward

bulgar [2K]3 years ago
4 0

Answer:

Its velocity was 20 km/s southward

Explanation:

As we know that airplane is moving from North to South

So here we will have

v = 1200 km/min

now we know

1 min = 60 s

so we will have

v = 1200 \times \frac{1}{60}

v = 20 km/s

so the velocity of the airplane is towards South moving with speed v = 20 km/s so correct answer will be

Its velocity was 20 km/s southward

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Oksanka [162]

The work done by a constant force in a rectilinear motion is given by:

W=Fd\cos\theta

where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.

In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:

\begin{gathered} W=(17)(12)\cos0+(36)(12)\cos30 \\ W=578.123 \end{gathered}

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1 year ago
How does tightening a string on a guitar affect its natural frequency?
iragen [17]
By tightening a string you are actually putting more stress on the string you are giving it a new frequency that isn't natural.

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3 0
3 years ago
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On an ice skating rink, a girl of mass 50 kg stands stationary, face to face with a boy of mass 80 kg. The children push off of
sasho [114]

The velocity of the girl is  -4.8 m/s.

Using the principle of conservation of linear momentum, The total momentum of  bodies before and after collision is constant. Since the two objects are stationary, the initial momentum of each body is zero.

Thus;

0 = (80 × 3) + (50 × v)

0 = 240 + 50 v

-240 = 50 v

v = -240/50

v = -4.8 m/s

Note that the negative sign shows that the velocity of the girl is in opposite direction that that of the girl.

Learn more about momentum: brainly.com/question/904448

5 0
3 years ago
Helpp pls
Kipish [7]

Answer:

The intensity of the electric field is

|E|=10654.37 \:N/C

Explanation:

The electric field equation is given by:

|E|=k\frac{q}{d^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge at 0.4100 m from the balloon
  • d is the distance from the charge to the balloon

As we need to find the electric field at the location of the balloon, we just need the charge equal to 1.99*10⁻⁷ C.

Then, let's use the equation written above.

|E|=(9*10^{9})\frac{1.99*10^{-7}}{0.41^{2}}

|E|=10654.37 \:N/C

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5 0
3 years ago
We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

density of air of = 1.2 kg/m^3

V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N

B) the torque can be calculated using below formula below

Torque= (Force × distance)

= 114.1 × 7

= 798.5Nm

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