Answer:
Answer:
Speed of the wave in the string will be 3.2 m/sec
Explanation:
We have given frequency in the string fixed at both ends is 80 Hz
Distance between adjacent antipodes is 20 cm
We know that distance between two adjacent anti nodes is equal to half of the wavelength
So \frac{\lambda }{2}=20cm
2
λ
=20cm
\lambda =40cmλ=40cm
We have to find the speed of the wave in the string
Speed is equal to v=\lambda f=0.04\times 80=3.2m/secv=λf=0.04×80=3.2m/sec
So speed of the wave in the string will be 3.2 m/sec
Answer:
new force is 6 times of the initial force.
Explanation:
Let the charges on two objects is q₁ and q₂. The electric force between charges is given by :
Objects 1 and 2 attract each other with a electrostatic force of 18.0 units
If the charge of Object 1 is doubled and the charge of object 2 is tripled, it means, and . New force is given by :
So, the new electrostatic force between objects will become 6 times of the initial force.
Answer: The result of "the upper bound of the density" does not go on the denominator.
So simplified, no. The answer is no.
Al(OH)3 = 26.98 + [(16×3) + (1.01×3)] = 26.98 + 51.03 = 78.01 and the unit will be g/mol
<h3>
<em>Al(OH)3 = 78.01 g/mol</em></h3>
Answer:
the energy of the spring at the start is 400 J.
Explanation:
Given;
mass of the box, m = 8.0 kg
final speed of the box, v = 10 m/s
Apply the principle of conservation of energy to determine the energy of the spring at the start;
Final Kinetic energy of the box = initial elastic potential energy of the spring
K.E = Ux
¹/₂mv² = Ux
¹/₂ x 8 x 10² = Ux
400 J = Ux
Therefore, the energy of the spring at the start is 400 J.