Answer:
the frequency heard by the observer is equal to 2677 Hz
Explanation:
given,
velocity of the observer = 17 m/s
speed of the sound = 343 m/s
velocity of the source = 0 m/s
frequency emitted from the source = 2550 Hz
velocity of observer is negative as it is approaching the source. f = 2676.38 Hz ≈ 2677 Hz
hence, the frequency heard by the observer is equal to 2677 Hz
Answer:
2.65m/s
Explanation:
Using the equation of motion:
v² = u²+2a∆S where
v is the final velocity
u is the initial velocity
∆S is the change in distance
a is the acceleration
Given
u = 0m/s
a = 9.8m/s²
∆S = 1.3-0.943
∆S = 0.357m
Substituting the given parameters into the formula
v² = 0²+2(9.8)(0.357)
v² = 0+6.9972
v² = 6.9972
v=√6.9972
v = 2.65m/s
Hence the velocity at which it hit the ground is 2.65m/s
Answer:
Explanation:
We put the charges in the ascending order as follows
1.53 P
3.26 P
4.66 P
5.09 P
6.39 P
where P is equal to 10⁻¹⁹
we round off given charges as follows
1.53 P → 1.6 P
3.26 P → 3.2 P
4.66 P → 4.8 P
5.09 P → 4.8 P
6.39 P → 6.4 P
We see that 2 nd to 4 th charges are integral multiples of first charge . That means these charges are supposed to be made of combination of first charge . So first charge appears to be minimum possible charge .
Hence this charge may exist on single electron.
Answer:
Explanation:
work done=force*displacement
=350N*15m
=5250 joule
The resultant force on the positive charge is mathematically given as
X=40N
<h3>What is the magnitude of the electrostatic force on the negative charge?</h3>
Question Parameters:
Three-point charges, two positive and one negative, each having a magnitude of 20
Generally, the -ve charge is mathematically given as
Q+=X
Therefore
X=40N
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