A projectile is launched upward at an angle of 70⁰ from the horizontal and strikes the ground a certain distance down range. For
1 answer:
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Answer:
(a) vmax = 0.34m/s
(b) v = 0.13m/s
(c) v = 0.31m/s
(d) x = 0.039m
Explanation:
Given information about the spring-mass system:
m: mass of the object = 0.30kg
k: spring constant = 22.6 N/m
A: amplitude of the motion = 4.0cm = 0.04m
(a) The maximum speed of the object is given by the following formula:
(1)
w: angular frequency of the motion.
The angular frequency is calculated with the following relation:
(2)
You replace the expression (2) into the equation (1) and replace the values of the parameters:
The maximum speed of the object is 0.34 m/s
(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:
The speed is 0.13m/s
(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:
You solve the previous equation for t:
With this value of t, you can calculate the speed of the object with the following formula:
The speed of the object for x = 1.5cm is v = 0.31 m/s
(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:
The position will be:
The position of the object on which its speed is one-half its maximum velocity is 0.039
Answer:
An increase in pressure
Explanation:
The ideal gas law states that:
where
p is the gas pressure
V is the volume
n is the number of moles
R is the gas constant
T is the temperature of the gas
in the equation, n and R are constant. For a gas kept at constant volume, V is constant as well. Therefore, from the formula we see that if the temperature (T) is increase, the pressure (p) must increase as well.