<h3>
Answer: The acceleration doubles</h3>
===========================================================
Explanation:
Consider a mass of 10 kg, so m = 10
Let's say we apply a net force of 20 newtons, so F = 20
The acceleration 'a' is...
F = ma
20 = 10a
20/10 = a
2 = a
a = 2
The acceleration is 2 m/s^2. Every second, the velocity increases by 10 m/s.
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Now let's double the net force on the object
F = 20 goes to F = 40
m = 10 stays the same
F = ma
40 = 10a
10a = 40
a = 40/10
a = 4
The acceleration has also doubled since earlier it was a = 2, but now it's a = 4.
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In summary, if you double the net force applied to the object, then the acceleration doubles as well.
The frequency of oscillation on the frictionless floor is 28 Hz.
<h3>
Frequency of the simple harmonic motion</h3>
The frequency of the oscillation is calculated as follows;
f = (1/2π)(√k/m)
where;
- k is the spring constant
- m is mass of the block
f = (1/2π)(√7580/0.245)
f = 28 Hz
Thus, the frequency of oscillation on the frictionless floor is 28 Hz.
Learn more about frequency here: brainly.com/question/10728818
#SPJ1
Answer:
The value is 
Explanation:
From the question we are told that
The initial pressure is 
The initial temperature is ![T_1 = 50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283 \ K](https://tex.z-dn.net/?f=T_1%20%3D%20%2050%20%5C%20F%20%3D%20%2850%20-%2032%29%20%2A%20%5B%5Cfrac%7B5%7D%7B9%7D%20%5D%20%2B%20273%20%3D%20283%20%20%5C%20%20K)
The final temperature is ![T_2 = 320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433 \ K](https://tex.z-dn.net/?f=T_2%20%3D%20%20320%20%5C%20F%20%3D%20%28320%20-%2032%29%20%2A%20%5B%5Cfrac%7B5%7D%7B9%7D%20%5D%20%2B%20273%20%3D433%20%20%5C%20%20K)
Generally the equation for adiabatic process is mathematically represented as
=> 
Generally for a monoatomic gas 
So
![14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }](https://tex.z-dn.net/?f=14%20%2A%20283%5E%7B%5Cfrac%7B%5Cfrac%7B5%7D%7B3%7D%20%7D%7B1-%20%5B%5Cfrac%7B5%7D%7B3%7D%20%5D%7D%20%7D%20%3DP_2%20%2A%20433%5E%7B%5Cfrac%7B%5Cfrac%7B5%7D%7B3%7D%20%7D%7B1-%20%5B%5Cfrac%7B5%7D%7B3%7D%20%5D%7D%20%7D)
=> 
=>
<span>3.92 m/s^2
Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so
0.4 * 9.8 m/s^2 = 3.92 m/s^2
If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so
20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N
Multiply by the coefficient of static friction, giving
196 N * 0.4 = 78.4 N
So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass
78.4 N / 20.0 kg
= 78.4 kg*m/s^2 / 20.0 kg
= 3.92 m/s^2
And you get the same result.</span>
