The solution for the problem is:
1 Watt = 1 Joule per second
1 Watt*second = 1 Joule
a Kilowatt is 1,000 Watts
an hour is 60 seconds times 60 minutes or 3,600 seconds
a Kilowatt * hour is 1,000 Watts in 3,600 seconds
15 W*h = 15,000 Watt*hour = 15,000 Watt * 3,600 seconds = 54,000,000
Watt*second
54,000,000 Watt*second = ? Joules
54,000,000 Joules / second = 54,000,000 Watts
Answer:
Option A. 1 bar = 1 atm
Explanation:
Pressure has various units of measurement. Each unit of measurement can be converted to other units of measurement. For example:
1 atm = 1 bar
1 atm = 760 mmHg
1 atm = 760 torr
1 atm = 1×10⁵ N/m²
1 atm = 1×10⁵ Pa
With the above conversion scale we can convert from one unit to the other.
Considering the question given above, it is evident from the coversion scale illustrated above that only option A is correct.
Thus,
1 bar = 1 atm
Answer:
25.6 m/s
Explanation:
Draw a free body diagram of the sled. There are two forces acting on the sled:
Normal force pushing perpendicular to the hill
Weight force pulling straight down
Take sum of the forces parallel to the hill:
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.8 m/s²) (sin 38.0°)
a = 6.03 m/s²
Given:
v₀ = 0 m/s
a = 6.03 m/s²
t = 4.24 s
Find: v
v = at + v₀
v = (6.03 m/s²) (4.24 s) + (0 m/s)
v = 25.6 m/s
Answer:
T_ww = 43,23°C
Explanation:
To solve this question, we use energy balance and we state that the energy that enters the systems equals the energy that leaves the system plus losses. Mathematically, we will have that:
E_in=E_out+E_loss
The energy associated to a current of fluid can be defined as:
E=m*C_p*T_f
So, applying the energy balance to the system described:
m_CW*C_p*T_CW+m_HW*C_p*T_HW=m_WW*C_p*T_WW+E_loss
Replacing the values given on the statement, we have:
1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8 kg/s*4,18 kJ/(kg°C)*75°C=1.8 kg/s*4,18 kJ/(kg°C)*T_WW+30 kJ/s
Solving for the temperature Tww, we have:
(1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8 kg/s*4,18 kJ/(kg°C)*75°C-30 kJ/s)/(1.8 kg/s*4,18 kJ/(kg°C))=T_WW
T_WW=43,23 °C
Have a nice day! :D
Answer:
W=561.41 J
Explanation:
Given that
m = 51 kg
μk = 0.12
θ = 36.9∘
Lets F is the force applied by man
Given that block is moving at constant speed it mans that acceleration is zero.
Horizontal force = F cos θ
Vertical force = F sinθ
Friction force Fr= μk N
N + F sinθ = m g
N = m g - F sinθ
Fr = μk (m g - F sinθ)
For equilibrium
F cos θ = μk (m g - F sinθ)
F ( cos θ +μk sinθ) = μk (m g
Now by putting the values
F ( cos 36.9∘ + 0.12 x sin36.9∘)=0.12 x 51 x 10
F= 70.2 N
We know that Work
W= F cos θ .d
W= 70.2 x cos 36.9∘ x 10
W=561.41 J
