Answer:
a) t = 0.75 s, b) t = 2.25 s
Explanation:
The speed of sound is constant in a material medium
v = 340 m / s
we can use the relations of uniform motion to find the time
v = x / t
t = x / v
In the exercise, the observer's distance to the wall is indicated d = 510 m, it also indicates that the shot is fired at the midpoint
x = d / 2
a) direct sound the distance from the observer to the screen is
x = 510/2 = 255 m
t = 255/340
t = 0.75 s
b) echo sound.
In this case the sound reaches the wall bounces, the distance is
x = d / 2 + d
x = 3/2 d
x = 3/2 510
x = 765 m
the time is
t = x / v
t = 765/340
t = 2.25 s
Answer:
E = 1.19 N/C
Explanation:
Let's first determine the length of the arc which can be given as:
L= Rθ
where:
L = length of the arc
R = radius of curvature
θ = angle in radius
L = (9.09×10⁻²m)(2.59)
L = (0.0909)(2.59)
L = 0.235431 m
Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:
![E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B%5Clambda%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D-sin%28-%5Cfrac%7B%5Ctheta%7D%7B2%7D%29%5D)
![E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B%5Clambda%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%2Bsin%28%5Cfrac%7B%5Ctheta%7D%7B2%7D%29%5D)
![E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2%5Clambda%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%5D)
Since 
where;
L = length
Q = charge
λ = density of the charge;
then substituting 
for λ, we have :
![E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2%28%5Cfrac%7BQ%7D%7BL%7D%29%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%5D)
![E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2Q%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%5D%7D%7B4%20%5Cpi%20E_oLR%7D)
substituting our given parameter; we have:
![E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2%286.26%2A10%5E%7B-12%7DC%29%5Bsin%5Cfrac%7B2.59rad%7D%7B2%7D%5D%7D%7B4%20%5Cpi%20%288.85%2A10%5E%7B-12%7DC%5E2%2FN.m%5E2%29%280.235431%29%280.0909%29%7D)
E = 1.1889 N/C
E = 1.19 N/C
∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C
Answer:
0 (Zero)
Explanation:
Definition of work done states that ,
Work done on an object is the force applied on that object in the direction of Displacement of the object .
means that ,
W = F.dS (Dot product of Force and displacement)
When the wall is acted upon by the force of 20 N , Wall does not move ,
Thus , displacement of the wall is zero.
So, W = 20×0 = 0.
Thus, Work done on the wall is zero.
Answer:
Explanation:
From the question we are told that:
Force 
Time 
Length of pedal 
Radius of wheel 
Moment of inertia, 
Generally the equation for Torque on pedal 
is mathematically given by
Generally the equation for angular acceleration 
is mathematically given by
Therefore Angular speed is \omega
Generally the equation for Tangential velocity V is mathematically given by
Answer:
Conduction heat transfer is the transfer of <em>heat by means of molecular excitement within a material without bulk motion</em> of the matter.
Explanation:
Conduction heat transfer in gases and liquids is due to the collisions and diffusion of the molecules during heir random motion.
