Answer: 17939.74 yards
Explanation:
Given , A rectangular measures 100 meters by 150 meters
To find : Area of rectangle.
Formula :
Area of rectangle = Length x width
Here, let length = 100 meters and width = 150 meters
Then, Area of rectangle = 100 meters x 150 meters = 15,000 square meters
Also , 1 meter = 1.09361 yards
Then, Area of rectangle = 15,000 x 1.09361 x 1.09361 square yards
= 17939.7424815 square yards
≈ 17939.74 yards
Hence, the area of rectangle is 17939.74 yards .
C. is correct : ) it is applied when you push and that causes the desk to move
red goes to red, black goes to white, yellow goes to green, blue goes to blue.
Answer:
a) (0, -33, 12)
b) area of the triangle : 17.55 units of area
Explanation:
<h2>
a) </h2>
We know that the cross product of linearly independent vectors 
and 
gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.
Luckily for us, we know that vectors 
and 
are living in the plane through the points P, Q, and R, and are linearly independent.
We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).
If they weren't linearly independent, we will obtain vector zero as the result of the cross product.
So, for our problem:
<h2>B)</h2>
We know that and are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:
so:
Answer:
Explanation:
Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .
velocity of approach = 1.5 - 0 = 1.5
velocity of separation = v₁ + v₂
coefficient of restitution = velocity of separation / velocity of approach
.8 = v₁ + v₂ / 1.5
v₁ + v₂ = 1.2
applying law of conservation of momentum
m x 1.5 + 0 = mv₂ - mv₁
1.5 = v₂ - v₁
adding two equation
2 v ₂= 2.7
v₂ = 1.35 m /s
v₁ = - .15 m / s
During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.
For second collision ,
coefficient of restitution = velocity of separation / velocity of approach
.5 = v₃ + v₄ / 1.35
v₃ + v₄ = .675
applying law of conservation of momentum
m x 1.35 + 0 = mv₄ - mv₃
1.35 = v₄ - v₃
adding two equation
2 v ₄= 2.025
v₄ = 1.0125 m /s
v₃ = - 0 .3375 m / s
