Answer:
Explanation:
Work done = ∫Fdx
= ∫(cx-3.00x²) dx
[ c x² / 2 - 3 x³ / 3 ]₀²
= change in kinetic energy
= 11-20
= - 9 J
[ c x² / 2 - x³ ]₀² = - 9
c x 2² / 2 - 2³ = -9
2c - 8 = -9
2c = -1
c = - 1/2
When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
<h3>
Frictional force between the block and the horizontal surface</h3>
The frictional force between the block and the horizontal surface is determined by applying Newton's law;
∑F = ma
F - Ff = ma
Ff = F - ma
Ff = 4 - 2(1.2)
Ff = 4 - 2.4
Ff = 1.6 N
When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;
F - Ff = ma
5 - 1.6 = 2a
3.4 = 2a
a = 3.4/2
a = 1.7 m/s²
Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
Learn more about frictional force here: brainly.com/question/4618599
Answer:
How long or wide something is
Explanation:
<span>Here I think you have to find the velocity in x and y components where x is east and y is north
So as air speed indicator shows the negative speed in y component and adding it in
air speed while multiplying with the direction component we will get the velocity as velocity is a vector quantity so direction is also required
v=-28 m/s y + 18 m/s (- x/sqrt(2) - y/sqrt(2))
solving
v= -12.7 m/s x-40.7 m/s y
if magnitude of velocity or speed is required then
speed= sqrt(12.7^2 + 40.7^2)
speed= 42.63 m/s
if angle is asked
angle = arctan (40.7/12.7)
angle = 72.67 degrees south of west</span>
