Answer:
By force and sloppy surface.
Explanation:
By applying force on the sled and slope of the path are the two ways the sled accelerate as it descends. If there is more friction between sled and the ground then force is required to push the sled to move downward while on the other hand, if the path on which sled moves is sloppy then it will move automatically without the use of force, so these two ways can accelerate the sled.
Buddy i mean gravity can be one example to the mechanical weathering
Answer:
<h2>a) t = 4</h2>
Explanation:
In projectile the time taken by an object to reach its maximum height is the maximum time (tmax).
tmax = Usin
/g
U = velocity of the ball
= angle of projection = 90° (vertical velocity)
g = acceleration due to gravity = 9.8m/s²
tmax = Usin90°/9.8
tmax = U/9.8
T the maximum height U = Uy = 39.2 m/s²
tmax = 39.2/9.8
tmax = 4secs
To calculate the time at which the ball will have an horizontal velocity, we will use the same formula but
will be 0° in this case
If Ux = Ucos
8 = Ucos0
U = 8m/s
Since t = Usin
/g
t = 8sin0/9.8
t = 0sec
evaporation (if that's to fill in the blanks question)
Answer:
a)32.34 N/m
b)10cm
c)1.6 Hz
Explanation:
Let 'k' represent spring constant
'm' mass of the object= 330g =>0.33kg
a) in order to find spring constant 'k', we apply Newton's second law to the equilibrium position 10cm below the release point.
ΣF=kx-mg=0
k=mg / x
k= (0.33 x 9.8)/ 0.1
k= 32.34 N/m
b) The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object.
Therefore, amplitude of the oscillation is 10cm
c)frequency of the oscillation can be determined by,
f= 1/2π 
f= 1/2π 
f= 1.57
f≈ 1.6 Hz
Therefore, the frequency of the oscillation is 1.6 Hz