As per the question, the velocity of the airplane [v] = 660 miles per hour.
The total time taken by airplane [t] = 3.5 hours.
We are asked to determine the total distance travelled by the airplane during that period.
The distance covered [ S] by a body is the product of velocity with the time.
Mathematically distance covered = velocity × total time
S = v × t
= 660 miles/hour ×3.5 hours
= 2310 miles.
Hence, the total distance travelled by the airplane in 3.5 hour is 2310 miles.
The voltage from one side of the battery all the way around to the other side of the battery is 12v .
If 4 of those volts show up across the circle-thing, then the rest of the 12v ... 8v ... Must show up across the set of parallel rectangles.
To get that answer, I subtracted the 4 from the 12.
Just like it says in choice-C.
Answer:
Coefficient of friction.
Explanation:
The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :
N is normal force.
= coefficient of friction
Answer:
The launching point is at a distance D = 962.2m and H = 39.2m
Explanation:
It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.
X axis
x = Vox t
t = x / vox
t = 7.1 / 340
t = 2.09 10-2 s
In this same time the height of the window fell
Y = Voy t - ½ g t²
Let's calculate the initial vertical speed, this speed is in the window
Voy = (Y + ½ g t²) / t
Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209
Voy = 27.7 m / s
We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s
Vy = Voy - gt₂
Vy = 0 -g t₂
t₂ = Vy / g
t₂ = 27.7 / 9.8
t₂ = 2.83 s
This is the time it also takes to travel the horizontal and vertical distance
X = Vox t₂
D = 340 2.83
D = 962.2 m
Y = Voy₂– ½ g t₂²
Y = 0 - ½ g t2
H = Y = - ½ 9.8 2.83 2
H = 39.2 m
The launching point is at a distance D = 962.2m and H = 39.2m
