Answer:
275 kPa
Explanation:
mass of the gas=m=1.5 kg
initial volume if the gas=V₁=0.04 m³
initial pressure of the gas= P₁=550 kPa
as the condition is given final volume is double the initial volume
V₂=final volume
V₂=2 V₁
As the temperature is constant
T₁=T₂=T
=
putting the values in the equation.
=
P₂=
P₂=
P₂=275 kPa
So the final pressure of the gas is 275 kPa.
Answer:
The tension in the rod as the ball moves through the bottom circle is 9.8 N
Explanation:
When the ball is released from rest, the centripetal force equals the weight of the ball. So mv²/r = mg where m = mass of ball = 0.5 kg, v = speed of ball, r = radius of vertical circle = length of rod = 0.5 m and g = acceleration due to gravity = 9.8 m/s²
v = √gr = √9.8 m/s² × 0.5 m = √4.9 = 2.21 m/s
Now at the bottom of the circle T - mg = mv²/r where T = tension in the rod
T = m(g + v²/r)
= m(g + (√gr)²/r)
= m(g+ gr/r)
= m(g + g)
= 2mg
= 2 × 0.5 kg × 9.8 m/s²
= 9.8 N
So, the tension in the rod as the ball moves through the bottom circle is 9.8 N
Isostasy is the concept used to explain the dynamics of the earth's upper layers. It is considered that there is a gravitational equilibrium that exists between the earth's upper most layer, the crust, and the next layer, the mantle. The crust is said to "float" on the mantle.
Therefore, in this model, the block of wood represents the crust and the water represents the earth's mantle.
Answer:
Explanation:
a )
in the regions r < R₁
charge q inside sphere of radius R₁ = 0
Applying gauss's law for electric field E at distance r <R₁
electric flux through Gaussian surface of radius r = 4π r² E
4π r² E = q / ε₀ = 0 / ε₀
E = 0
Applying gauss's law for electric field E at distance R₁ < r < R₂ .
charge q inside sphere of radius R₁ = q₁
Applying gauss's law for electric field E at distance R₁ < r < R₂
electric flux through Gaussian surface of radius r = 4π r² E
4π r² E = q₁ / ε₀
E = q₁ / 4πε₀
in the regions r> R₂
charge q inside sphere of radius R₂ = (q₁ + q₂)
Applying gauss's law for electric field E at distance r > R₂
electric flux through Gaussian surface of radius r = 4π r² E
4π r² E = (q₁ + q₂) / ε₀
E = (q₁ + q₂) /4π ε₀
b )
For electric flux to be zero at r > R₂
(q₁ + q₂) /4π ε₀ = 0
q₁ + q₂ = 0
q₁ / q₂ = - 1 .
Answer:
a)
4 times
b)
2 times
c)
0.5 times
d)
0.25 times
Explanation:
= Applied force by the person = F
= distance from the hinge = R
Torque is given as
(a)
= Applied force by the person = 2F
= distance from the hinge = 2R
New Torque is given as
So Torque becomes 4 times
b)
= Applied force by the person = F
= distance from the hinge = 2R
New Torque is given as
So Torque becomes 2 times
c)
= Applied force by the person = (0.5)F
= distance from the hinge = R
New Torque is given as
So Torque becomes 0.5 times
d)
= Applied force by the person = (0.5)F
= distance from the hinge = (0.5)R
New Torque is given as
So Torque becomes 0.25 times