We have: F = m×a
Here, m = 90 Kg
a = 15 m/s²
Substitute their values into the expression:
F = 90 × 15
F = 1350 N
In short, Your Answer would be Option D
Hope this helps!
Answer:
1. 8437500 N
2. The force between the two charges is attractive.
Explanation:
1. Determination of the force between the two charges.
Charge 1 (q₁) = –2.0 C
Charge 2 (q₂) = 3.0 C
Distance apart (r) = 80 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
F = Kq₁q₂ / r²
F = 9×10⁹ × 2 × 3 / 80²
F = 5.4×10¹⁰ / 6400
F = 8437500 N
Thus, the force of attraction between the two charges is 8437500 N
2. From the question given, the charges are:
Charge 1 (q₁) = –2.0 C
Charge 2 (q₂) = 3.0 C
We understood that like charges repels while unlike charges attract. Since the two charges (i.e –2 C and 3 C) has opposite signs, it means they will attract each other.
Thus the force between them is attractive.
Answer:
C
Explanation:
Im not sure but I did somthing simalier
Answer
given,
Length of the string, L = 2 m
speed of the wave , v = 50 m/s
string is stretched between two string
For the waves the nodes must be between the strings
the wavelength is given by
where n is the number of antinodes; n = 1,2,3,...
the frequency expression is given by
now, wavelength calculation
n = 1
λ₁ = 4 m
n = 2
λ₂ = 2 m
n =3
λ₃ = 1.333 m
now, frequency calculation
n = 1
f₁ = 12.5 Hz
n = 2
f₂= 25 Hz
n = 3
f₃ = 37.5 Hz
