Answer:
its acceleration decrease because of the force acted upon by the viscous liquid
The volume of the balloon is given by:
V = 4πr³/3
V = volume, r = radius
Differentiate both sides with respect to time t:
dV/dt = 4πr²(dr/dt)
Isolate dr/dt:
dr/dt = (dV/dt)/(4πr²)
Given values:
dV/dt = 72ft³/min
r = 3ft
Plug in and solve for dr/dt:
dr/dt = 72/(4π(3)²)
dr/dt = 0.64ft/min
The radius is increasing at a rate of 0.64ft/min
The surface area of the balloon is given by:
A = 4πr²
A = surface area, r = radius
Differentiate both sides with respect to time t:
dA/dt = 8πr(dr/dt)
Given values:
r = 3ft
dr/dt = 0.64ft/min
Plug in and solve for dA/dt:
dA/dt = 8π(3)(0.64)
dA/dt = 48.25ft²/min
The surface area is changing at a rate of 48.25ft²/min
Explanation:
It is given that,
The horizontal and vertical component of velocity of an electron is:
The magnetic field acting there is given by :
(a) The magnitude of the magnetic force on the electron is given by :
q = e
(b) We know that the charge on proton is :
The magnetic force as same as for electron but the direction is opposite i.e.
Hence, this is the required solution.
Answer:
322 kJ
Explanation:
The work is the energy that a force produces when realizes a displacement. So, for a gas, it occurs when it expands or when it compress.
When the gas expands it realizes work, so the work is positive, when it compress, it's suffering work, so the work is negative.
For a constant pressure, the work can be calcutated by:
W = pxΔV, where W is the work, p is the pressure, and ΔV is the volume variation. To find the work in Joules, the pressure must be in Pascal (1 atm = 101325 Pa), and the volume in m³ (1 L = 0.001 m³), so:
p = 60 atm = 6.08x10⁶ Pa
ΔV = 82.0 - 29.0 = 53 L = 0.053 m³
W = 6.08x10⁶x0.053
W = 322x10³ J
W = 322 kJ
