Answer:
40479.6 J
Explanation:
Applying,
q = cm(t₂-t₁).................... Equation 1
Where q = change in heat content of the system, c = specific heat capacity of the system, m = mass of the system, t₁ = initial temperature, t₂ = final temperature.
From the question,
Given: m = 79 g = 0.079 kg, t₁ = 21°C, t₂ = 143°C
Constant: c = 4200 J/kg.°C
Substitute these values into equation 1
q = 4200(0.079)(143-21)
q = 331.8(122)
q = 40479.6 J
Answer:
162500000.
Explanation:
Given that
Diameter of the wire , d= 1.8 mm
The length of the wire ,L = 15 cm
Current ,I = 260 m A
The charge on the electron ,e= 1.6 x 10⁻¹⁹ C
We know that Current I is given as

I=Current
q=Charge
t=time
q= I t
q= 260 m t
The total number of electron = n
q= n e

n=162500000 t

The number of electron passe per second will be 162500000.
Answer: Heterogeneous mixture - the parts are not uniformly mixed.
A mixture contains components having distinct chemical properties. There are two types of mixtures: homogeneous and heterogeneous. In a homogeneous mixture there is uniform distribution of components. we cannot distinguish one portion of the mixture from another. for example salt mixed in water. In heterogeneous mixture, the components are not uniformly mixed. hence, we are able to distinguish different parts of a mixture, like the mixture of iron, sand and salt given in this question.
A futuristic design for a car is to have a large solid disk-shaped flywheel within the car storing kinetic energy. The uniform flywheel has mass 370 kg with a radius of 0.500 m and can rotate up to 320 rev/s. Assuming all of this stored kinetic energy could be transferred to the linear velocity of the 3500-kg car, find the maximum attainable speed of the car.
Use the Inverse square law, Intensity (I)<span> of a light </span>is inversely proportional to the square of the distance(d).
I=1/(d*d)
Let Intensity for lamp 1 is L1 distance be D1 so on, L2 D2 for Intensity for lamp 2 and its distance.
L1/L2=(D2*D2)/(D1*D1)
L1/15=(200*200)/(400*400)
L1=15*0.25
L1=3.75 <span>candela</span>