Answer:
29.96m/s
Explanation:
Given parameters:
Initial speed = 25.5m/s
Acceleration = 1.94m/s²
Time = 2.3s
Unknown:
Final speed of the car = ?
Solution:
To solve this problem, we are going to apply the right motion equation:
v = u + at
v is the final speed
u is the initial speed
a is the acceleration
t is the time taken
Now insert the parameters and solve;
v = 25.5 + (1.94 x 2.3) = 29.96m/s
Answer:
Charge on each is 2 x 10⁻¹⁰.
Explanation:
We know that Force between two point charges is given b the Coulomb's law as:
F = kq₁q₂/r^2
k = 9 x 10^9
r = 3.00 cm
= 0.03 m
q₁ = q₂
F = 4.00 x 10^-7
Rearranging the formula, we get:
F = k q²/r²
q² = Fr²/k
q² = 4 x 10⁻⁷ x 0.03²/(9x10⁹)
q² = 4 x 10⁻²⁰
q = 2 x 10⁻¹⁰
As there is force of repulsion between the charges, the charges must be both positive or both negative.
Complete Question
A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.
If the acceleration of the projective is : a = c/s m/s2
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?
Answer:
The value of the constant is 
Explanation:
From the question we are told that
The acceleration is 
The initial position of the projectile is s= 1.5m
The final position of the projectile is 
The velocity is 
Generally 
and acceleration is 
so

=> 

integrating both sides

Now for the limit
a = 200 m/s
b = 0 m/s
c = s= 3 m
d =
= 1.5 m
So we have

![[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bv%5E2%7D%7B2%7D%20%5D%20%5Cleft%20%7C%20200%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.%20%20%3D%20c%20%5Bln%20s%5D%5Cleft%20%7C%203%7D%20%5Catop%20%7B1.5%7D%7D%20%5Cright.)
![\frac{200^2}{2} = c ln[\frac{3}{1.5} ]](https://tex.z-dn.net/?f=%5Cfrac%7B200%5E2%7D%7B2%7D%20%20%3D%20%20c%20ln%5B%5Cfrac%7B3%7D%7B1.5%7D%20%5D)
=> 

Answer: i really don't know bu i recomened you watch josie b asmr she is the best
Explanation:
Answer:
Explanation: The equation that relates resistance of tungsten at different temperatures is as follows
R = R₀ [1 + α ∆T] , R₀ is resistance at lower temperature , R is resistance at higher temperature . α is temperature coefficient of resistivity and ∆T is rise in temperature .
Putting the values
170 = 26 [1 + .0045 ∆T]
∆T = 1230.75
lower temperature = 40◦C
higher temperature = 1230 + 40
= 1270◦C