Answer:
The expression for the initial speed of the fired projectile is:
![\displaystyle v_0=\frac{M+m}{m}(2gL[1-cos(\theta)]^{\frac{1}{2}})](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_0%3D%5Cfrac%7BM%2Bm%7D%7Bm%7D%282gL%5B1-cos%28%5Ctheta%29%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%29)
And the initial speed ratio for the 9.0mm/44-caliber bullet is 1.773.
Explanation:
For the expression for the initial speed of the projectile, we can separate the problem in two phases. The first one is the moment before and after the impact. The second phase is the rising of the ballistic pendulum.
First Phase: Impact
In the process of the impact, the net external forces acting in the system bullet-pendulum are null. Therefore the linear momentum remains even (Conservation of linear momentum). This means:
(1)
Second Phase: pendular movement
After the impact, there isn't any non-conservative force doing work in al the process. Therefore the mechanical energy remains constant (Conservation Of Mechanical Energy). Therefore:
![Em_i=Em_f\\\frac{1}{2}mv^2_i=mgH\\v_i=[2gH]^\frac{1}{2}](https://tex.z-dn.net/?f=Em_i%3DEm_f%5C%5C%5Cfrac%7B1%7D%7B2%7Dmv%5E2_i%3DmgH%5C%5Cv_i%3D%5B2gH%5D%5E%5Cfrac%7B1%7D%7B2%7D)
(2)
The height of the pendulum respect L and θ is:
(3)
Using equations (1),(2) and (3):
(4)
The initial speed ratio for the 9.0mm/44-caliber bullet is obtained using equation (4):
In a spectrograph, black lines can be seen going through the array of colors. The pattern of these lines indicate the composition of the star.
Different elements block different parts of the spectrum, resulting in black lines.
If the coefficient of static friction is 0.3, then the minimum force required to get it moving is equal in magnitude to the maximum static friction that can hold the body in place.
By Newton's second law,
• the net vertical force is 0, since the body doesn't move up or down, and in particular
∑ <em>F</em> = <em>n</em> - <em>mg</em> = <em>n</em> - 50 N = 0 ==> <em>n</em> = 50 N
where <em>n</em> is the magnitude of the normal force; and
• the net horizontal force is also 0, since static friction keeps the body from moving, with
∑ <em>F</em> = <em>F'</em> - <em>f</em> = <em>F'</em> - <em>µn</em> = <em>F'</em> - 0.3 (50 N) = 0 ==> <em>F'</em> = 15 N
where <em>F'</em> is the magnitude of the applied force, <em>f</em> is the magnitude of static friction, and <em>µ</em> is the friction coefficient.
Answer:
Say you are holding a thread to the end of which is tied a stone. Now when you start whirling it around you will notice that two forces have to be applied simultaneously. One which pulls the thread inwards and the other which throws it sideways or tangentially.
Both these forces will generate their respective accelerations.
The one pointed inwards will generate centripetal or radial acceleration.
The one pointing sideways will generate tangential acceleratio
Explanation:
A major difference between tangential acceleration and centripetal acceleration is their direction
Centripetal means “center seeking”. Centripetal acceleration is always directed inward.
Tangential acceleration is always directed tangent to the circle.
Tangential acceleration results from the change in magnitude of the tangential velocity of an
object. An object can move in a circle and not have any tangential acceleration. No tangential
acceleration simply means the angular acceleration of the object is zero and the object is moving
with a constant angular velocity
Answer:
The amplitude of the eardrum's oscillation is 6.65×10^-13 m.
Explanation:
Given data:
The sound has a frequency of 262 Hz
The sound level is 84 dB
The air density is 1.21 kg/m^3
The speed of sound is 346 m/s
Solution:
As, Intensity of sound is given by,
I = Io×10^(s/10 db)
I = 2×π^2×ρ×v×f^2×Sm^2
Thus,
Sm = √(Io×10^(s/10 db)) / √( 2×π^2×ρ×v×f^2)
Now, put the values,
Sm = √( 10^-12 × 10^(84/10) ) / √( 2×(3.14)^2×1.21×346×(262)^2 )
Sm = √(2.51×10^-4 / 5.66×10^8)
Sm = √0.443×10^-12
Sm = 6.65×10^-13 m.
