Lets say that:
Eric is E
Sheena is S
Yael is Y
We know that S+2=Y, E= 1/S and that E+S+Y=52. So, we can insert S+2=Y and E= 1/S into our other equation, which would bring us to (1/2S)+(S+2)+S=52. This is equal to 1/2S+S+2+S=52. We then use algebra to get the answer:
Given:
S+2=Y
E=1/S
E+Y+S=52
Equation:
1/2S+S+2+S=52
1/2S+S+S=52
2.5S=50
S=50/2.5
S=20
And since Yael has 2 more points then Yael then we do 20+2 and get 22.
Hope this helps! <3
Explanation:
It is given that a particle covers 10m in first 5s and 10m in next 3s. so using the equation of motion
Case I
s=ut+
2
1
at
2
10=5u+
2
1
a(5)
2
20=10u+25a
4=2u+5a..............(1)
Case 2
In next 3s the particle covers more 10m distance. So
20=8u+
2
1
a(8)
2
5=2u+8a.........(2)
On solving equation (1) and (2)
4=2u+5a
5=2u+8a
a=
3
1
m/s
2
Put the value of a in equation (1)
u=
6
7
m/s
Now to find distance in next 10 s. total time will be 10s
s=
6
7
×10+
2
1
×
3
1
×(10)
2
s=28.33m
Distance travelled in next 2 sec
s=28.33−20=8.33m
Answer:
velocity. height. weight. possition. place. energy. force.
Explanation: 50/50 % chance they are wrong and write.
Answer:
11.962337 × 10^-4 N
Explanation:
Given the following :
Length L = 11.8
Charge = 29nC = 29 × 10^-9 C
Linear charge density λ = 1.4 × 10^-7 C/m
Radius (r) = 2cm = 2/100 = 0.02 m
Using the relation:
E = 2kλ/r ; F =qE
F = 2kλq/L × ∫dr/r
F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))
2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4
In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214
Hence,
(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N
