Answer:

Explanation:
Given:
Solute Diffusion rate = 4.0 × 10⁻¹¹ kg/s
Area of cross-section = 0.50 cm²
Length of channel =0.25 cm
Now for the new channel
Area of cross-section = 0.30 cm²
Length of channel =0.10 cm
let the Solute Diffusion rate of new channel = s
now equating the diffusion rate per unit volume for both the channels

thus,

Answer:
Q = 5267J
Explanation:
Specific heat capacity of copper (S) = 0.377 J/g·°C.
Q = MSΔT
ΔT = T2 - T1
ΔT=49.8 - 22.3 = 27.5C
Q = change in energy = ?
M = mass of substance =508g
Q = (508g) * (0.377 J/g·°C) * (27.5C)
Q= 5266.69J
Approximately, Q = 5267J
Answer:
bus momentum
p_bus= m_bus x v_bus
=18,200 x 16.5
basball momentum
pball=mball x vball
=0.142 x v
p_bus = pball
18200 x 16.5 = 0.142 x v
v=(18200 x 16.5)/0.142
v is the answer for baseball
Explanation:
⚠️not my answer tryna be honest here⚠️
Answer:
a) ![(Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]](https://tex.z-dn.net/?f=%28Qa%2Ag%2AVb%29-%28Qh%2AVb%2Ag%29%3D%28Qh%2AVb%2Aa%29%5C%5Cwhere%20%5C%5Cg%3Dgravity%20%5Bm%2Fs%5E2%5D%5C%5Ca%3Dacceleration%20%5Bm%2Fs%5E2%5D)
b) a = 19.61[m/s^2]
Explanation:
The total mass of the balloon is:
![massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\](https://tex.z-dn.net/?f=massball%3Ddensityheli%2Avolumeheli%5C%5C%5C%5Cmassball%3D0.41%20%5Bkg%2Fm%5E3%5D%2A0.048%5Bm%5E3%5D%5C%5Cmassball%3D0.01968%5Bkg%5D%5C%5C%5C%5C)
The buoyancy force acting on the balloon is:
![Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]](https://tex.z-dn.net/?f=Fb%3Ddensityair%2Agravity%2Avolumeball%5C%5CFb%3D1.23%5Bkg%2Fm%5E3%5D%2A9.81%5Bm%2Fs%5E2%5D%2A0.048%5Bm%5E3%5D%5C%5CFb%3D0.579%5BN%5D)
Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.
In the attached image we can see the free body diagram and the equation deducted by Newton's second law