Answer:
The velocity of the mailbag just before it hits the ground is 14.57 m/s.
Explanation:
Given that,
Acceleration = 2 m/s
Time = 3 sec
We need to calculate the velocity of mailbag
Using equation of motion
Put the value into the formula
We need to calculate the height at which the mailbag dropped
Using equation of motion
Put the value into the formula
We need to calculate the velocity of the mailbag just before it hits the ground
Using equation of motion
Put the value into the formula
Hence, The velocity of the mailbag just before it hits the ground is 14.57 m/s.
A. wavelength
the formula for frequency is frequency= wavelength x velocity
Answer:
16613 m/s
Explanation:
Given that
mass of the fly, m = 0.55 g = 0.55*10^-3 kg
Kinetic Energy of the fly, E = 7.6*10^4 J
Speed of the fly, v = ? m/s
We know that the Kinetic Energy is that energy that an object, in this case, the fly, possesses due to its motion.
The Kinetic Energy, KE of any object is represented by the formula
KE = 1/2 * m * v²
If we substitute the values in the relation, we have,
7.6*10^4 = 1/2 * 0.55*10^-3 * v²
v² = (15.2*10^4) / 0.55*10^-3
v² = 2.76*10^8
v = √2.76*10^8
v = 16613 m/s
Thus, the fly would need a speed of 16.6 km/s in order to have a Kinetic Energy of 7.6*10^4 J
On connecting a 9V battery to a Capacitor consisting of two circular plates of radius 0.066 m separated by an air gap of 2. 0 mm. The charge on the positive plate is 544.7 ×10⁻¹² C.
Capacitance of the capacitor is determined by the area of the plate of Capacitor and distance between the plates of capacitor.
Let the area of the Capacitor be A , radius of circular plates be r and distance between the plates of capacitor be d.
Given, Voltage, V = 9V
Radius, r = 0.066m
Distance, d = 2mm = 0.002m
Area of the Capacitor, A = πr²
A = π(0.066)²
A = 0.013m²
Capacitance, C = ε₀A / d
C = 8.85×10⁻¹²×0.013/0.002
C = 60.5 ×10⁻¹² F
C = 60.5 pF
We know that Q = CV where Q is the charge on capacitor.
Q = 60.5 ×10⁻¹² × 9
Q= 544.7 ×10⁻¹² C
Since, both plates of a capacitor acquire equal and opposite charge.
Hence the charge on the positive plate of the capacitor is 544.7 ×10⁻¹² C.
Learn more about Capacitance here, brainly.com/question/14746225
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