Answer:
Explained
Explanation:
Dynamically continuous innovation:
- Falls in between continuous and discontinuous innovation.
-Changes in customer habits are not as large as in discontinuous innovation and not as negligeble as in continuous innovation.
best example can as simple as transformation in Television. New HD TVs have flat panels, wide screens and improved performance The Added features are considered dynamically improved.
Discontinuous innovation:
- discontinuous innovation comprise of new to world product only so they are discontinuous to every customer segment.
- these product are so fundamentally different from the the product that already exist that they reshape market and competition.
For example- the mobile and the internet technology are reshaping the market through regular innovation and change.
<span>a) 1960 m
b) 960 m
Assumptions.
1. Ignore air resistance.
2. Gravity is 9.80 m/s^2
For the situation where the balloon was stationary, the equation for the distance the bottle fell is
d = 1/2 AT^2
d = 1/2 9.80 m/s^2 (20s)^2
d = 4.9 m/s^2 * 400 s^2
d = 4.9 * 400 m
d = 1960 m
For situation b, the equation is quite similar except we need to account for the initial velocity of the bottle. We can either assume that the acceleration for gravity is negative, or that the initial velocity is negative. We just need to make certain that the two effects (falling due to acceleration from gravity) and (climbing due to initial acceleration) counteract each other. So the formula becomes
d = 1/2 9.80 m/s^2 (20s)^2 - 50 m/s * T
d = 1/2 9.80 m/s^2 (20s)^2 - 50m/s *20s
d = 4.9 m/s^2 * 400 s^2 - 1000 m
d = 4.9 * 400 m - 1000 m
d = 1960 m - 1000 m
d = 960 m</span>
Answer:
The chunk went as high as
2.32m above the valley floor
Explanation:
This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.
Applying the principle of energy conservation for the two ice we have based on the scenery
Momentum before impact = momentum after impact
M1U1+M2U2=(M1+M2)V
Given data
Mass of ice 1 M1= 5.20kg
Mass of ice 2 M2= 5.20kg
velocity of ice 1 before impact U1= 13.5 m/s
velocity of ice 2 before impact U2= 0m/s
Velocity of both ice after impact V=?
Inputting our data into the energy conservation formula to solve for V
5.2*13.5+5.2*0=(10.4)V
70.2+0=10.4V
V=70.2/10.4
V=6.75m/s
Therefore the common velocity of both ice is 6.75m/s
Now after impact the chunk slide up a hill to solve for the height it climbs
Let us use the equation of motion
v²=u²-2gh
The negative sign indicates that the chunk moved against gravity
And assuming g=9.81m/s
Initial velocity of the chunk u=0m/s
Substituting we have
6.75²= 0²-2*9.81*h
45.56=19.62h
h=45.56/19.62
h=2.32m
Given:
u = 0, initial velocity
v = 125 m/s, final velocity
s = 0.0800 m, distance traveled
9.20 g, the mass of the pellet
If the acceleration is a, then
0² + 2*(a m/s²)*(0.800 m) = (125 m/s)²
1.6a = 15625
a = 9765.625 m/s²
Calculate the force.
F = (9.20 x 10⁻³ kg)*(9765.625 m/s²) = 98.84 N
Answer: 98.8 N (nearest tenth)
Answer:
v = 0.33 [m/s]
Explanation:
We must remember that speed is defined as the relationship between the displacement in a given time. In this way, we can propose the following equation.
where:
v = velocity [m/s]
d = displacement = 40 [m]
t = 2 [min] = 120 [s]
Now replacing we have: