Answer:
Final speed of the car, v = 24.49 m/s
Explanation:
It is given that,
Initial velocity of the car, u = 0
Acceleration, 
Time taken, t = 7.9 s
We need to find the final velocity of the car. Let it is given by v. It can be calculated using first equation of motion as :
v = 24.49 m/s
So, the final speed of the car is 24.49 m/s. Hence, this is the required solution.
Answer ) Sound level equation
The intensity of a sound wave is related to its amplitude squared by the following relationship: I=(Δp)22ρvw I = ( Δ p ) 2 2 ρ v w . Here Δp is the pressure variation or pressure amplitude (half the difference between the maximum and minimum pressure in the sound wave) in units of pascals (Pa) or N/m2.
Answer:
proportional to the current in the wire and inversely proportional to the distance from the wire.
Explanation:
The magnetic field produced by a long, straight current-carrying wire is given by:
where
is the vacuum permeability
I is the current intensity in the wire
r is the distance from the wire
From the formula, we notice that:
- The magnitude of the magnetic field is directly proportional to I, the current
- The magnitude of the magnetic field is inversely proportional to the distance from the wire, r
Therefore, correct option is
proportional to the current in the wire and inversely proportional to the distance from the wire.
Answer:
Explanation:
Given that,
Mass of ball m = 2kg
Ball traveling a radius of r1= 1m.
Speed of ball is Vb = 2m/s
Attached cord pulled down at a speed of Vr = 0.5m/s
Final speed V = 4m/s
Let find the transverse component of the final speed using
V² = Vr²+ Vθ²
4² = 0.5² + Vθ²
Vθ² = 4²—0.5²
Vθ² = 15.75
Vθ =√15.75
Vθ = 3.97 m/s.
Using the conservation of angular momentum,
(HA)1 = (HA)2
Mb • Vb • r1 = Mb • Vθ • r2
Mb cancels out
Vb • r1 = Vθ • r2
2 × 1 = 3.97 × r2
r2 = 2/3.97
r2 = 0.504m
The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m
The required time,
Using equation of motion
V = ∆r/t
Then,
t = ∆r/Vr
t = (r1—r2) / Vr
t = (1—0.504) / 0.5
t = 0.496/0.5
t = 0.992 second
Answer:
I. a, c, f and h
II. e
III. b, d, g and i
IV. i
Explanation:
I. Chemical symbols are simple abbreviations used to represent various elements or compound. They consist entire of alphabet.
For the diagram given above, the labelled parts which represent chemical symbol are: a, c, f and h
II. Coefficients are numbers written before the chemical symbol of elements or compound.
For the diagram given above, the labelled part which represent Coefficient is: e
III. Number of atoms of element present in a compound is simply obtained by taking note of the numbers written as subscript in the chemical formula of the compound.
For the diagram given above, the labelled part which represent the number of atoms of the element are: b, d, g and i
IV. When no number is written as subscript in the formula of the element in the compound, it means the element has just 1 atom in the compound.
For the diagram given above, the labelled part which indicates that only 1 atom of the element is present is: i
