Answer:
Buoyancy causes things to float in water or liquid which appears to defeat the force of gravity. Hope this helps.
Answer:
10N
Explanation:
Equation: ΣF = ma
Fapp = ma
Fapp = (2kg)(5m/s^2) (im guessing you mean 5.00 m/s^2 not m/s)
Fapp = 10*kg*m/s^2
Fapp = 10N
Answer:
The question is wrong Since if you apply Force on 0.0m²It would mean That the pressure exerted=F/A=F/0
An since we can't divide a number by 0, the question is wrong
Answer:
f = 931.1 Hz
Explanation:
Given,
Mass of the wire, m = 0.325 g
Length of the stretch, L = 57.7 cm = 0.577 m
Tension in the wire, T = 650 N
Frequency for the first harmonic = ?
we know,
![v =\sqrt{\dfrac{T}{\mu}}](https://tex.z-dn.net/?f=v%20%3D%5Csqrt%7B%5Cdfrac%7BT%7D%7B%5Cmu%7D%7D)
μ is the mass per unit length
μ = 0.325 x 10⁻³/ 0.577
μ = 0.563 x 10⁻³ Kg/m
now,
![v =\sqrt{\dfrac{650}{0.563\times 10^{-3}}}](https://tex.z-dn.net/?f=v%20%3D%5Csqrt%7B%5Cdfrac%7B650%7D%7B0.563%5Ctimes%2010%5E%7B-3%7D%7D%7D)
v = 1074.49 m/s
The wire is fixed at both ends. Nodes occur at fixed ends.
For First harmonic when there is a node at each end and the longest possible wavelength will have condition
λ=2 L
λ=2 x 0.577 = 1.154 m
we now,
v = f λ
![f = \dfrac{v}{\lambda}](https://tex.z-dn.net/?f=f%20%3D%20%5Cdfrac%7Bv%7D%7B%5Clambda%7D)
![f = \dfrac{1074.49}{1.154}](https://tex.z-dn.net/?f=f%20%3D%20%5Cdfrac%7B1074.49%7D%7B1.154%7D)
f = 931.1 Hz
The frequency for first harmonic is equal to f = 931.1 Hz
1. A. 6.00 sec
The graph shows the velocity of an object (y-axis) versus the time (x-axis). In order to find when the magnitude of the velocity reaches 36.00 km/h, we should find the time t (x-coordinate) at which the velocity (y-coordinate) is 36.
By looking at the graph, we see that this occurs when t=6.00 s.
2. A. positive acceleration
In a velocity-time graph like this one, the slope of the curve corresponds to the acceleration of the object. In fact, acceleration is defined as:
![a=\frac{\Delta v}{\Delta t}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5CDelta%20v%7D%7B%5CDelta%20t%7D)
where
is the variation of velocity and
is the variation of time. We see that this quantity corresponds to the slope of the curve in the graph (in fact,
represents the increment of the y coordinate, while
represents the increment of the x coordinate). So, a positive slope means a positive acceleration: in this case, the slope is positive, so the acceleration is also positive.