An airplane takes off with a velocity of 40 m/s at 15 degrees. What is the altit

Question

An airplane takes off with a velocity of 40 m/s at 15 degrees. What is the altitude (height above the ground) of the plane 2.5 min after it takes off?

Select one:

a. 25.9 m

b. 1.55 km

c. 5.80 km

d. 6.00 km

An airplane takes off from a runway at a 30 degree angle going 30 m/s. How fast must a car travel along the ground in order to stay directly under the airplane?

Select one:

a. 15 m/s

b. 26.0 m/s

c. 30 m/s

d. 60 m/s

An amateur astronomy measures 0.000412 degrees for the parallax of a star. How far is the star from our Sun if the Sun-Earth distance is 1.0 AU (astronomical unit)?

Select one:

a. 3.60 x 106 AU

b. 1.39 x 105 AU

c. 2.78 x 105 AU

d. none of the above

Two tractors are pulling on a stump. One tractor pulls with 120 lbs of force at 60 degrees north of east. The other tractor pulls with 80 lbs of force at 75 degrees north of west. What is the magnitude of the net force due to the two tractors pulling together?

Select one:

a. 47.5 lbs

b. 185 lbs

c. 198 lbs

d. 200 lbs

Explanation / Answer

a) speed of airplane = 40 m/s

angle of airplane = 15 degrees

time = 2.5 min = 2.5*60=150 seconds

distance travelled in 2.5 minutes = 40*150=6000 m

height reached abive ground = 6000*sin15=1552.9 m=1.55 km

correct option is b

speed of airplane = 30 m/s

angle of airplane = 30 degrees

for a car to be always under the airplane, the speed of car should be equal to horizontal component of airpplane = 30*cos30=25.98 m/s

correct option is (b)

parallax of star = 0.000412 degrees = 0.000412*pi/180 = 7.19*10-6 rad

Sun -Earth distnace = 1 AU

distnace of start from Sun = 1/7.19*10-6 = 1.39*105 AU

Option b is correct

compomnent of net force along East = 120cos60 - 80cos75=39.29 lb

component of net force along North = 120sin60+80sin75 = 181.2 lb

magnitude of net force = sqrt(39.292+181.22)=185.4

correct option is b

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