Answer:
1.28 s
Explanation:
Given:
Δy = 8 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(8 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 1.28 s
The displacement is the vector quantity that defines the distance and direction between two positions. It is a change in your position.
A displacement is a vector in geometry and mechanics that has a length equal to the shortest distance between a point P's initial and final positions.
It calculates the length and angle of the net motion, or total motion, in a straight line from the starting point to the destination of the point trajectory. The translation that links the starting point and the ending point can be used to spot a displacement.
The average velocity, which is a vector for motion over a certain amount of time, is determined by dividing the displacement by the duration of the time interval.
Hence, the displacement is the vector quantity that defines the distance and direction between two positions. It is a change in your position.
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I believe you're talking about displacement. It's a directional vector that depicts the movement of a point between two instances.
Answer:
This is because of the first Newton's law:
If an object is at rest or at moving with constant velocity, it will remain doing that untill a force is applied on the object.
Now, if the car is still, and then it speeds, the box will want to keep the movement that it was doing before, so it will slide in the seat until the box gets the same speed as the car (this can happen when the box hits the back of the seat, and the back of the seat "pushes" the box)
Now, the same happens when the car stops, now the box has a velocity and when the car stops the box will keep moving forward, because there is no force that stops the box.
velocity does the bomb strike the earth? (b) How
long does it take the bomb to fall? (c) What is the
bomb's range?
Answers:
a) -171.402 m/s
b) 17.49 s
c) 1700.99 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
(3)
Where:
is the bomb's final jeight
is the bomb'e initial height
is the bomb's initial vertical velocity, since the airplane was moving horizontally
is the time
is the acceleration due gravity
is the bomb's range
is the bomb's initial horizontal velocity
is the bomb's fina velocity
Knowing this, let's begin with the answers:
With the conditions given above, equation (1) is now written as:
(4)
Isolating:
(5)
(6)
(7)
Since , equation (3) is written as:
(8)
(9)
(10) The negative sign ony indicates the direction is downwards
Substituting (7) in (2):
(11)
(12)
The bomb hits the earth at approximately 540.3 km/hr, it takes approximately 17.5 seconds for the bomb to hit the ground, and the bomb's range is approximately 1.36km.
To answer these questions, we're going to use the principles of kinematics that involve motion in a straight line with constant acceleration (that is the acceleration due to gravity in this case). (a) The velocity of the bomb when it strikes the earth can be calculated using the kinematic equation v = u + gt, where u is the initial velocity (0 m/s since the bomb was dropped, not thrown), g is the acceleration due to gravity (approximately -9.8 m/s²), and t is the time. However, we do not know the time yet, so we will use the equation v = sqrt(2×g×h), where h is the height (1500m), and this gives us approximately 540.3 km/hr. (b) The time it takes for the bomb to fall can be calculated using t = sqrt((2×h)/g) which gives approximately 17.5 seconds. (c) The range of the bomb is the horizontal distance it travels, which is the velocity of the plane multiplied by the time it takes for the bomb to hit the ground, or 350 km/hr × 17.5/3600 hr = approximately 1.36km.
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