A 1500 kg car is pushing a 4000 kg truck. The car and truck are accelerating at 2.0 m/s^2. Assuming that the frictional force on
1 answer:
To solve this problem we will use the Force equation according to the definition given in Newton's second law. There we have that the Force is equal to
Where,
m = mass
a = acceleration
Our values are given as
Considering that both mass are equal to one, we have that:
Therefore the truck exert a force on the car of 11kN
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Answer:
The gas was Hexane
Explanation:
taking the diference between the mass of the flask and the final mass qe can calculate the mass of liquid injected (assuming none escaped the flask):
with the volume of the flask we can get the density of the gas at the indicated pressure and temperature:
From the ideal gases law we have that the density can be calculated as:
Where R is the ideal gases constant = , and M the molecular weight of the fluid. Solving for M:
Note that the temperature is computed in Kelvin T= 18+273=291K
The gas with the closer molar mass is Hexane
This question is incomplete, the complete question is;
In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.
Angular Velocity at time 0s = 12 rad/s
Angular Velocity at time 0.15s = 24 rad/s
a) What is the angular acceleration?
b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g
Answer:
a) the angular acceleration is 80 rad/s²
b) the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
Explanation:
Given the data in the question;
from the graph below;
Angular Velocity at time 0s = 12 rad/s
Angular Velocity at time 0.15s = 24 rad/s
a) What is the angular acceleration;
Angular acceleration ∝ = ( -
) / dt
we substitute
Angular acceleration ∝ = ( 24 - 12 ) / 0.15
Angular acceleration ∝ = 12 / 0.15
Angular acceleration ∝ = 80 rad/s²
Therefore, the angular acceleration is 80 rad/s²
b)
If the ball is 0.60 m from her shoulder, i.e s = 0.6 m
the tangential acceleration of the ball will be;
a = ∝ × s
we substitute
a = 80 × 0.6
a = 48 m/s²
a = ( 48 / 9.8 )g
a = 4.9 g
Therefore, the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
