Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.
force on it due to rest of the charges will be equal and opposite so
k3q Q / x² =k 8q Q / (L+x)²
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal
kQ . 3q / x² = k3q 8q / L²
Q = 8q (x² / L²)
so charge required = - 8q (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )
Answer:
W = 9.93 10² N
Explanation:
To solve this exercise we must use the concept of density
ρ = m / V
the tabulated density of copper is rho = 8966 kg / m³
let's find the volume of the cylindrical tube
V = A L
V = π (R_ext ² - R_int ²) L
let's calculate
V = π (4² - 2²) 10⁻⁴ 3
V = 1.13 10⁻² m³
m = ρ V
m = 8966 1.13 10⁻²
m = 1.01 10² kg
the weight of the tube
W = mg
W = 1.01 10² 9.8
W = 9.93 10² N
Answer:
The final pressure of the gas is 9.94 atm.
Explanation:
Given that,
Weight of argon = 0.16 mol
Initial volume = 70 cm³
Angle = 30°C
Final volume = 400 cm³
We need to calculate the initial pressure of gas
Using equation of ideal gas
Where, P = pressure
R = gas constant
T = temperature
Put the value in the equation
We need to calculate the final temperature
Using relation pressure and volume
Hence, The final pressure of the gas is 9.94 atm.
Answer:
<h3>The answer is 300 N</h3>
Explanation:
The force acting on an object given the mass and acceleration we use the formula
<h3>force = mass × acceleration</h3>
We have
force = 15 × 20
We have the final answer as
<h3>300 N</h3>
Hope this helps you
Answer:
Explanation:
Let v is the launch speed of the plastic ball and the angle of projection is θ.
So, in horizontal direction
v Cosθ x t = 4.8 .... (1)
In th evertical direction
1.4 = v Sin θ x t - 0.5 gt² .... (2)
As , v Sin θ x t = 3.8 .... (3) , put in equation (2)
1.4 = 3.8 - 4.9 t²
t = 0.7 s
Put in (1) and (3)
v Cosθ x 0.7 = 4.8
v Cosθ = 6.86
and v Sinθ x 0.7 = 3.8
v Sinθ = 5.43
Now
v = 8.75 m/s
