Answer:
40 J
Explanation:
From the question given above, the following data were obtained:
Force (F) = 10 N
Distance (s) = 4 m
Workdone (Wd) =?
Work done is simply defined as the product of force and distance moved in the direction of the force. Mathematically, we can express the Workdone as:
Workdone = force × distance
Wd = F × s
With the above formula, we can obtain the workdone as follow:
Force (F) = 10 N
Distance (s) = 4 m
Workdone (Wd) =?
Wd = F × s
Wd = 10 × 4
Wd = 40 J
Thus, 40 J of work was done.
Answer:
Induced emf in the coil, E = 0.157 volts
Explanation:
It is given that,
Number of turns, N = 100
Diameter of the coil, d = 3 cm = 0.03 m
Radius of the coil, r = 0.015 m
A uniform magnetic field increases from 0.5 T to 2.5 T in 0.9 s.
Due to this change in magnetic field, an emf is induced in the coil which is given by :


E = -0.157 volts
Minus sign shows the direction of induced emf in the coil. Hence, the induced emf in the coil is 0.157 volts.
Density: g/mL, kg/cubic meter
Volume: L, teaspoon
Mass: g, MeV/sq. C
Answer: The potential difference between the plates = 0.4061V
Explanation:
Given that the
Electric field strength E = 155 N/C
Distance d = 0.00262 m
From the definition of electric field strength, is the ratio of potential difference V to the distance between the plates. That is
E = V/d
Substitute E and d into the above formula
155 = V/0.00262
Cross multiply
V = 155 × 0.00262
V = 0.4061 V
The potential difference between the plates is 0.4061 V
Answer:
v = 24 cm and inverted image
Explanation:
Given that,
The focal length of the object, f = +8 cm
Object distance, u = -12 cm
We need to find the position &nature of the image. Let v be the image distance. Using lens formula to find it :

Put all the values,

So, the image distance from the lens is 24 cm.
Magnification,

The negative sign of magnification shows that the formed image is inverted.