To solve this question we will use ideal gas equation:
Where:
p = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
We can rearrange formula to get:
We are working woth same gas so we can write following formula. Index 1 stands for conditions before change and index 2 stands for conditions after change.
We are given:
p1=92.1kPa = 92100Pa
V1=200mL = 0.2L
T1=275K
p2= 101325Pa
T2=273K
V2=?
We start by rearranging formula for V2. After that we can insert numbers:
Nothing will change, it's still going at 9 km/h.
Hope this helps.
Answer:
The distance the mass would stretch it is 
The correct option is A
Explanation:
From the question we are told that
The period of the slit is 
The acceleration due to gravity is 
Generally the period is mathematically represented as
Whee m is the particle and k is the spring constant
making 
the subject
![\frac{m}{k} = [\frac{T}{2 \pi} ]^2](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bk%7D%20%20%3D%20%5B%5Cfrac%7BT%7D%7B2%20%5Cpi%7D%20%5D%5E2)
The weight on the particle is related to the force stretching the spring by this mathematical relation
where x is the length by which the spring is stretched
Substituting the equation above for
![[\frac{T}{2 \pi} ]^2 = \frac{x}{g}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BT%7D%7B2%20%5Cpi%7D%20%5D%5E2%20%3D%20%5Cfrac%7Bx%7D%7Bg%7D)
making x the subject
![x = g [\frac{T}{2 \pi} ]](https://tex.z-dn.net/?f=x%20%3D%20g%20%5B%5Cfrac%7BT%7D%7B2%20%5Cpi%7D%20%5D)
Substituting the value
![x = 9.8 * [\frac{0.2}{2 * 3.142} ]^2](https://tex.z-dn.net/?f=x%20%3D%209.8%20%2A%20%5B%5Cfrac%7B0.2%7D%7B2%20%2A%203.142%7D%20%5D%5E2)
10.4 N
Given
m = 1.10 kg
θ = 15.0°
g = 9.81 m/s2
Solution
Fnet, y = ΣF y = Fn − Fg, y = 0
Fn = Fg, y = Fg
cosθ = mgcosθ
Fn = (1.10 kg)(9.81 m/s2
)(cos15.0°) = 10.4 N
Answer:
14 m/s
Explanation:
We can solve the problem by using the law of conservation of energy.
At the beginning, when the ball is thrown from the ground, it has only kinetic energy, which is given by
where m = 5.9 kg is the mass of the ball and v is its initial speed.
As the ball goes up, its speed decreases, so its kinetic energy decreases and converts into gravitational potential energy. When the ball reaches its maximum height, the speed has become zero, and all the kinetic energy has been converted into gravitational potential energy, given by:
where g = 9.8 m/s^2 is the gravitational acceleration and h = 10 m is the maximum height reached by the ball.
Since we can ignore air resistance, energy must be conserved, so the initial kinetic energy must be equal to the final potential energy of the ball, so we can write:
And we can solve the equation to find v, the initial speed of the ball:
