Calcium sulfate hope this helps
Answer:
The question implies that force of friction is 2.1E4 N
F = M v^2 / R frictional force needed to supply this centripetal force
2.1E4 = M v^2 / R
v^2 = R * 2.1E4 / M
v^2 = 120 * 2.1E4 / 2.5E3 = 1200 * 2.1 / 2.5 = 1008 m^2/s^2
v = 31.7 m/s max speed of truck (about 70 MPH)
First convert grams to kilograms (so that the answer will be N) 1000g=1kg so 2.9g=.0029kg.
To solve for force we use the equation F=ma, so we need to find acceleration (a) .Next use the equation a=Δv/Δt. Δt is given as being .00043 seconds. Δv is the difference in velocities - final minus initial (this is important to discern the direction of the force vector); 0-304=-304. Now plugin and solve for a. The result is -706976.7442 m/s², or -710000 m/s² if you are using significant figures.
Now use the values for mass and acceleration to solve for force. F=ma=-2050.232558N or F=- 2100N (if you are using significant figures).
The maximum height is reached when the vertical component of the velocity is zero.
vertical direction:
acceleration: a = -g = -9.81m/s²
velocity: v = -g*t + v₀
position: y = -0.5*g*t² + v₀*t + y₀
For v= 0:
0 = -g*t + v₀ => t = v₀/g
Insert into position equation gives:
y(max) = (-0.5*v₀²/g) + (v₀²/g) + y₀ = (0.5*v₀²/g) + y₀
Answer:
m/s/s
Explanation:
Acceleration is change in velocity over time.
a = Δv / t
a = [m/s] / [s]
a = [m/s/s]