If we assume also that the temperature of the air does not change, we can use Boyle's Law:
p₁V₁ = p₂V₂
Now, we know:
p₁ = 100kPa
V₂ = 100cm³ (the volume of the tyre)
V₁ = 120cm³ (becuse the air is contained inside the tyre AND the pump)
We can solve for p₂:
p₂ = (p₁V₁)/V₂
= (100×120)/100
= 120kPa
Therefore your answer is: 120kPa
Answer:
A: 4 times as much
B: 200 N/m
C: 5000 N
D: 84,8 J
Explanation:
A.
In the first question, we have to caculate the constant of the spring with this equation:

Getting the k:
![k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bm%2Ag%7D%7Bx%7D%20%3D%5Cfrac%7B0%2C2%5Bkg%5D%2A9%2C81%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%5D%7D%7B0%2C05%5Bm%5D%7D%20%3D39%2C24%5B%5Cfrac%7BN%7D%7Bm%7D%5D)
Then we can calculate how much the spring stretch whith the another mass of 0,2kg:
![x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bm%2Ag%7D%7Bk%7D%20%3D%5Cfrac%7B0%2C4%5Bkg%5D%2A9%2C81%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%5D%7D%7B39%2C24%5B%5Cfrac%7BN%7D%7Bm%7D%5D%7D%20%3D0%2C1%5Bm%5D%5C%5C)
The energy of a spring:

For the first case:
![E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%7D%20%2A39%2C24%5B%5Cfrac%7BN%7D%7Bm%7D%5D%2A%280%2C05%5Bm%5D%29%5E%7B2%7D%20%3D0%2C049%20%5BJ%5D)
For the second case:
![E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%7D%20%2A39%2C24%5B%5Cfrac%7BN%7D%7Bm%7D%5D%2A%280%2C1%5Bm%5D%29%5E%7B2%7D%20%3D0%2C0196%20%5BJ%5D)
If you take the relation E2/E1 = 4.
B.
We have the next facts:
x=0,005 m
E = 0,0025 J
Using the energy equation for a spring:
⇒![k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]](https://tex.z-dn.net/?f=k%3D%5Cfrac%7BE%2A2%7D%7Bx%5E%7B2%7D%20%7D%20%3D%5Cfrac%7B0%2C0025%5BJ%5D%2A2%7D%7B%280%2C005%5Bm%5D%29%5E%7B2%7D%20%7D%20%3D200%5B%5Cfrac%7BN%7D%7Bm%7D%20%5D)
C.
The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.
![E=W*h=500[N]*10[m]=5000[J]](https://tex.z-dn.net/?f=E%3DW%2Ah%3D500%5BN%5D%2A10%5Bm%5D%3D5000%5BJ%5D)
Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.
D.
The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.
The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:
W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J
<span>Here the force that is applied between the electron and proton is centripetal, so equate the two forces to determine the velocity.
We know charge of the electron which for both Q1 and Q2, e = 1.60 x 10^-19 C
The Coulombs Constant k = 9.0 x 10^9
Radius r = 0.053 x 10^-9m = 5.3 x 10^-11 m
Mass of the Electron = 9.11 x 10^-31
F = k x Q1 x Q2 / r^2 = m x v^2 / r(centripetal force)
ke^2 / r^2 = m x v^2 / r => v^2 = ke^2 / m x r
v^2 = ((1.60 x 10^-19)^2 x 9.0 x 10^9) / (9.11 x 10^-31 x 5.3 x 10^-11 )
v^2 = 4.77 x 10^12 = 2.18 x 10^6 m/s
Since one orbit is the distance,
one orbit = circumference = 2 x pi x r; distance s = v x t.
v x t = 2 x pi x r => t = (2 x 3.14 x 5.3 x 10^-11) / (2.18 x 10^6)
t = 33.3 x 10^-11 / 2.18 x 10^6 = 15.27 x 10^-17 s
Revolutions per sec = 1 / t = 1 / 15.27 x 10^-17 = 6.54 x 10^15 Hz</span>
too much sun is dangerous for humans and can cause cancer so it's important that light is reflected for example a pool reflects water back to space that is why water sometimes is cold because it reflects light
Refer to the diagram shown below.
Neglect wind resistance, and use g = 9.8 m/s².
The pole vaulter falls with an initial vertical velocity of u = 0.
If the velocity upon hitting the pad is v, then
v² = 2*(9.8 m/s²)*(4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
The pole vaulter comes to res after the pad compresses by 50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
0 = (9.037 m/s)² + 2*(a m/s²)*(0.5 m)
a = - 82.32/(2*0.5) = - 82 m/s²
Answer: - 82 m/s² (or a deceleration of 82 m/s²)