Answer:
Energy consumed by the electric kettle in 9.5 min =Pt=(2.5×10
3
)×(9.5×60)=14.25×10
5
J
Energy usefully consumed =msΔT=3×(4.2×10
3
)×(100−15)=10.71×10
5
where s=4.2J/g
o
C= specific heat of water and boiling point temp=100
o
C
Heat lost =14.25×10
5
−10.71×10
5
=3.54×10
5
Answer:
u wanna do my edge bro the answer is b
Explanation:
Answer: D. 5cm
Explanation:
Given the following :
Focal length (f) = - 6.0 cm
Height of object = 15.0cm
Distance of object from mirror (u) = 12.0cm
Height of image produced by the mirror =?
Firstly, we calculate the distance of the image from the mirror.
Using the mirror formula
1/f = 1/u + 1/v
1/v = 1/f - 1/u
1/v = 1/-6 - 1/12
1/v = - 1/6 - 1/12
1/v = (- 2 - 1) / 12
1/v = - 3 / 12
v = 12 / - 3
v = - 4
Using the relation :
(Image height / object height) = (- image distance / object distance)
Image height / 15 = - (-4) / 12
Image height / 15 = 4 / 12
Image height = (15 × 4) / 12
Image height = 60 / 12
Image height = 5cm
Answer:
Δp = 0.05 p
Explanation:
The moment is
p = m v
The uncertainty of the moment is
Δp = dp/dm Δm + dp/dv Δv
Like the uncertainty in the mass is zero
Δm = 0
ΔP = m Δv
We divide for the moment
Δp / p = Δv / v
They do not indicate that Δv / v = 0.05
Δp / p = 0.05
Δp = 0.05 p
In the case of a system consisting of two cars
p = m₁ v₁ + m₂ v₂
Δp = dp / dv₁ Δv₁ + dp / dv₂ Δv₂
Δp = m₁ Δv₁ + m₂ Δv₂
Δv₁ / v₁ = 0.05
Δv₁ = 0.05 v₁
Δv₂ / v₂ = 0.05
Δv₂ = 0.05 v₂
We replace
Δp = m₁ 0.05 v₁ + m₂ 0.05 v₂
Δp = 0.05 p
According to the continuity equation, the rate at which mass enters the system equals the rate at which mass exits in any steady state process.
An equation that explains the movement of a particular quantity is a continuity equation, also known as a transport equation. Although it can be applied generally to any significant quantity, it is extremely simple and useful when used with preserved quantities.
The radius is seven centimeters, and the mass flow rate is 0 to 5 kg/s. Find the mass flow rate at a point with a 3.5 cm radius. We can consequently deduce that based on the equation. As we all know, the mass flow rate is constant.
If the rate of mass entering and leaving the system is equal, the rate of mass leaving the system should be processed.
The mass flow rate air section A and the mass flow rated section B are equivalent, according to the continuity equation. Mass flow rate in section B is therefore 0.02, or five kilograms per second.
To know more about continuity equation, click on the link below:
brainly.com/question/19566865
#SPJ4
