plane is flying at an altitude of 70 m
now if an object is dropped from it then time taken by object to drop on ground will be given as
here initial speed in vertical direction must be zero as plane is moving horizontal
given that
y = 70 m
a = 9.8 m/s^2
now since the plane is moving horizontally with speed v = 44 m/s
so the horizontal distance moved by the object will be
so the distance moved by the box is 166.3 m
Answer:
avriage force F = 2722.5 N
Explanation:
For this problem we can use Newton's second law, to calculate the average force and acceleration we can find it by kinematics.
vf² = v₀² - 2 ax
The final carriage speed is zero (vf = 0)
0 = v₀² - 2ax
a = v₀² / 2x
a = 1.1²/(2 0.200)
a = 3.025 m / s²
a = 3.0 m/s²
We calculate the average force
F = ma
F = 900 3,025
F = 2722.5 N
Answer:
D) The ball exerts a force on the wall and the wall exerts a force back.
Explanation:
Newton's third law of motion states that:
"When an object A exerts a force on another object B, then object B exerts an equal and opposite force on object A"
In this problem, we can identify (for instance) object A with tha ball and object B with the wall. Therefore, if we apply Newton's third law, we get:
The ball (object A) exerts a force on the wall (object B), therefore the wall (object B) exerts an equal and opposite force on the ball (object A). So, option D is the correct one.
1 Newton in Earth gravity is the equivalent weight of 1/9.80665 kg on Earth. This is derived using Newton's second law f=ma and assuming Earth gravity of 9.80665 m/s2. 1 N (Earth) = 0.101971621297793 kg.
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:
