Answer:
the expected distance is 4.32 m
Explanation:
given data
half life time = 1.8 ×
s
speed = 0.8 c = 0.8 × 3 ×
to find out
expected distance over
solution
we know c is speed of light in air is 3 ×
m/s
we calculate expected distance by given formula that is
expected distance = half life time × speed .........1
put here all these value
expected distance = half life time × speed
expected distance = 1.8 ×
× 0.8 × 3 ×
expected distance = 4.32
so the expected distance is 4.32 m
Answer:
1, 3 and 4
Explanation:
Did the test and saw the answers.
Answer:
The cannonball fly horizontally before it strikes the ground, S = 323.25 m
Explanation:
Given data,
The height of the cliff, h = 80 m
The horizontal velocity of the cannonball, Vₓ = 80 m/s
The range of the cannon ball with initial vertical velocity is zero is given by the formula,


S = 323.25 m
Hence, the cannonball fly horizontally before it strikes the ground, S = 323.25 m
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
Answer:
a. FTh = 30 N
b. Fw = 30 N
c. a = 200 m/s2
Explanation:
See full explanation in the picture. Please rate as brainliest