Answer:
5.571 sec
Explanation:
angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s
Period To = 2π / angular frequency
Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got
T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec
t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been ( To / (√ (1 - (v²/c²))). where To = 2.00 sec
Answer:
No one is right
Explanation:
John Case:
The function 
is defined between -1 and 1, So it is not possible obtain a value 
greater.
In addition, if you move the function cosine a T Value, and T is the Period, the function take the same value due to the cosine is a periodic function.
Larry case:
Is you have 
, the domain of this is [0,2].
it is equivalent to adding 1 to the domain of the , and its mean that the function 
, in general, is not greater than .
A) See ray diagram in attachment (-6.0 cm)
By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:
where
q is the distance of the image from the lens
f = -10 cm is the focal length (negative for a diverging lens)
p = 15 cm is the distance of the object from the lens
Solving for q,
B) The image is upright
As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:
where 
are the size of the image and of the object, respectively.
Since q < 0 and p > o, we have that 
, which means that the image is upright.
C) The image is virtual
As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.
This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual
Answer:
A. 92.88 °C
B. 401.535 °C
Explanation:
So, the overheating system is going to be based on the principle of thermal linear expansion. The behavior of this principle is ruled by the equation:
∆L= L_i * ∆T * α
Where α is a coefficient of linear expansion. For a cylinder made from polycarbonate α = 70,2*10^(-6) °C^(-1) and for a cylinder made from cast iron α = 12*10^(-6) °C^(-1). If we isolate the term of the temperature’s difference, we have:
∆L/(L_i * α) = ∆T → T_f = T_i + ∆L/(L_i * α)
Replacing the values, for the case of the Polycarbonate we have:
T_f = T_i + ∆L/(L_i * α) = 23°C+0,0273cm/(6,01cm *70,2 * 10^(-6)°C^(-1) ) = 92,88 °C
Replacing the values, for the case of the Cast Iron we have:
T_f = T_i +∆L/(L_i * α) = 23°C + 0,0273cm/(6,01cm * 12 * 10^(-6) °C^(-1) ) = 401,535 °C
As we see, is way better to use the polycarbonate in this application.
Have a nice day. Let me know if I can help you with anything else. :D
Answer:
b
Explanation:
friction needs two objects
the two objects need to touch each other
the two objects need to either have relative movement or, tend to have relative movement urge. (kinetic friction or static friction)
