Answer:
All statement are correct.
Explanation:
1. Electric field lines are the same thing as electric field vectors, electric field are mathematically vectors quantity. These vectors point in the direction in which a positive test charge would move.
2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space. Yes it is correct tangent drawn at any point on these lines gives the direction of electric filed at that point.
3. The number of electric field lines that start or end at a charged particle is proportional to the magnitude of charge on the particle, is a correct statement.
4.The electric field is strongest where the electric field lines are close together, again a correct statement as relative closeness of field lines indicate a stronger strength of electric field.
Hence we can say that all the statement are correct.
Answer:
29.38 seconds
Explanation:
Half life, T = 22.07 s
No = 1293
Let N be the number of atoms left after time t
N = 1293 - 779 = 514
By the use of law of radioactivity
Where, λ is the decay constant
λ = 0.6931 / T = 0.6931 / 22.07 = 0.0314 decay per second
so,
take natural log on both the sides
0.9225 = 0.0314 t
t = 29.38 seconds
Answer:
109.385m
Explanation:
In 1 day, the hour hand travels 2 circles, or 4π rad in angular. The distance it travels is its angle times the radius
7.4 * 4π = 93 mm
In 1 day, the minute hand travels 24*60 = 1440 circles, or 1440 * 2π = 2880π rad in angular. The distance it travels is
12.1 * 2880π = 109478 mm
So the distance traveled by the tip of the minute hand that exceed the distance traveled by the tip of the hour hand is
109478 - 93 = 109385 mm or 109.385 m
Answer:
the change in thermal energy of the projectile is 43.8 kJ
Explanation:
Given;
mass of the object, m = 5kg
initial velocity of the projectile, v₁ = 200 m/s
final velocity of the projectile, v₂ = 150 m/s
To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.
Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²
KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)
KE = ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ
Therefore, the change in thermal energy of the projectile is 43.8 kJ
Answer:
<h2>Rate of heat transfer from the body of the person is given as</h2><h2>
</h2>
Explanation:
As we know that the rate of heat transfer due to thermal conduction is given as
so we have
k = 0.04 W/m C
