With acceleration

and initial velocity

the velocity at time <em>t</em> (b) is given by




We can get the position at time <em>t</em> (a) by integrating the velocity:

The particle starts at the origin, so
.



Get the coordinates at <em>t</em> = 8.00 s by evaluating
at this time:


so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).
Get the speed at <em>t</em> = 8.00 s by evaluating
at the same time:


This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

Answer:
The answer to your question is: 20
Explanation:
Atomic number is the number of proton and atom has. Each element has a specific number of protons, if the number of protons change, then this is a new element.
Mass number is the number of protons and neutrons and atom has.
Mass number = protons + neutrons
Data
Number of protons = ?
Atomic number = 20
Then,
atomic number = number of protons = 20
Answer:
0.035 N
Explanation:
Parameters given:
Charge q1 = -3.31x10^(-7) C
Charge q2 = -5.7x10^(-7) C.
Distance between them, R = 22 cm = 0.22 m
Electrostatic force between to particles is given as:
F = (k* q1 * q2) / R²
F = (9 * 10^9 * -3.31 * 10^(-7) * -5.7 * 10^(-7)) / 0.22²
F = 0.035 N
Answer:
The distance is 5 m.
Explanation:
Given that,
Horizontal velocity = 20.0 m/s
Time t = 0.250 s
We need to calculate the horizontal distance
The distance traveled by the baseball with horizontal velocity in time t is given by
Using formula for horizontal distance



Hence, The distance is 5 m.
Answer:
11.07Hz
Explanation:
Check the attachment for diagram of the standing wave in question.
Formula for calculating the fundamental frequency Fo in strings is V/2L where;
V is the velocity of the wave in string
L is the length of the string which is expressed as a function of its wavelength.
The wavelength of the string given is 1.5λ(one loop is equivalent to 0.5 wavelength)
Therefore L = 1.5λ
If L = 3.0m
1.5λ = 3.0m
λ = 3/1.5
λ = 2m
Also;
V = √T/m where;
T is the tension = 0.98N
m is the mass per unit length = 2.0g = 0.002kg
V = √0.98/0.002
V = √490
V = 22.14m/s
Fo = V/2L (for string)
Fo = 22.14/2(3)
Fo = 22.14/6
Fo = 3.69Hz
Harmonics are multiple integrals of the fundamental frequency. The string in question resonates in 2nd harmonics F2 = 3Fo
Frequency of the wave = 3×3.69
Frequency of the wave = 11.07Hz