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Andrews [41]
3 years ago
15

During takeoff, an airplane climbs with a speed of 130 m/s at an angle of 47 degrees above the horizontal. The speed and directi

on of the airplane constitute a vector quantity known as the velocity. The sun is shining directly overhead. How fast is the shadow of the plane moving along the ground? (That is, what is the magnitude of the horizontal component of the plane's velocity?)
Physics
1 answer:
Reptile [31]3 years ago
5 0
130cos(47deg) = 88.6598 m/s
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The basketball coach tells his team to run sprints back and forth across the court, which is 30 m long. They start at the left e
sp2606 [1]

Distance is the total length of an object's path. Displacement is the overall change in position, ie. how far an object is from its initial position.


The court is 30 m long, so a path going back and forth once is 60 m long. Going along this path 6 times totals 360 m.


The end point is the same as the starting point, so the displacement is 0 m.

5 0
3 years ago
Heat from burning fuel warms the walls of the firebox section of the furnace in
nydimaria [60]

Heat from burning fuel warms the walls of the firebox section of the furnace in

A. a hot-water heating system.

B. a hot-air heating system.

C. a compressor compartment.

D. an evaporation system.

7 0
3 years ago
A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total accelera
AlekseyPX

Answer:

Explanation:

Given

Radius of bicycle wheel r=0.3\ m

Initial angular velocity \omega _0=0

It rotates 3 revolution in 5 s therefore

\omega =2\pi 3=\6\pi =18.85\ rad/s

using \omega =\omega _0+\alpha t

where \alpha =angular\ acceleration

\omega =Final\ angular\ velocity

t=time

\alpha =\frac{18.85}{5}=3.77 rad/s^2

Total acceleration of any point will be a vector sum of tangential acceleration and centripetal acceleration

\omega at t=1

\omega =0+3.77\times 1=3.77 rad/s

a_c=\omega ^2\cdot r

a_c=(3.77)^2\cdot 0.3=4.26 m/s^2

Tangential acceleration a_t=\alpha \times r

a_t=3.77\times 0.3=1.13 m/s^2

a_{net}=\sqrt{a_t^2+a_c^2}

a_{net}=\sqrt{(1.13)^2+(4.26)^2}

a_{net}=4.41 m/s^2

                       

7 0
3 years ago
A car travels straight for 20 miles on a road that is 30° north of east. What is the east component of the car’s displacement
Over [174]

The east component of the cars displacement is 17.3 miles.

Trigonometric ratio is used to show the relationship between the sides of a right angled triangle and its angles.

Let x represent the east component of the cars displacement.

Using trigonometric ratio:

cos(30) = x / 20

x = 20 * cos(30)

x = 17.3 miles

The east component of the cars displacement is 17.3 miles.

Find out more on Trigonometric ratio at: brainly.com/question/1201366

4 0
2 years ago
Calculate the current flowing if a charge of 36 kilocoulombs flows in 1 hour.
o-na [289]

Answer:

Current = 10 Amperes.

Explanation:

Given the following dat;

Quantity of charge, Q = 36 kilocoulombs (KC) = 36 * 1000 = 36000C

Time = 1 hour to seconds = 60*60 = 3600 seconds

To find the current;

Quantity of charge = current * time

Substituting in the equation

36000 = current * 3600

Current = 36000/3600

Current = 10 Amperes.

6 0
3 years ago
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