Answer:
the answer is 173 m/s
Explanation:
the the sound traveled from the dog to the building and back to the dog in 0.5 seconds, every second passes the sound travels 346 m/s but if half a second passes then we divide 346 by 2 and get 173 m/s.
Answer:
Option D. 0.061 J
Explanation:
We'll begin by converting 5 cm to m. This can be obtained as follow:
100 cm = 1 m
Therefore,
5 cm = 5 cm × 1 m / 100 cm
5 cm = 0.05 m
Next, we shall convert 250 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
250 g = 250 × 1 Kg / 1000
250 g = 0.25 Kg
Next, we shall determine the weight of the object. This can be obtained as follow:
Mass (m) = 0.25 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Weight (W) =?
W = m × g
W = 0.25 × 9.8
W = 2.45 N
Next, we shall determine the spring constant. This can be obtained as follow:
Compression (e) = 0.05 m
Weight = Force (F) = 2.45 N
Spring constant (K) =?
F = Ke
2.45 = K × 0.05
Divide both side by 0.05
K = 2.45 / 0.05
K = 49 N/m
Finally, we shall determine the potential energy contained in the compressed spring. This can be obtained as follow:
Compression (e) = 0.05 m
Spring constant (K) = 49 N/m
Energy (E) =?
E = ½Ke²
E = ½ × 49 × (0.05)²
E = 24.5 × 0.0025
E = 0.061 J
Therefore, the potential energy contained in the compressed spring is 0.061 J
Answer:
We can use equations of motion


To solve this problem we will apply the concepts related to the centripetal Force and the Force given by weight and formulated in Newton's second law. Through the two expressions we can find the radius of curve made in the hand. To calculate the normal force, we will include the concepts of sum of forces to obtain the net force on the body at the top and bottom of the maneuver. The expression for centripetal force acting on the jet is

According to Newton's second law, the net force acting on the jet is
F = ma
Here,
m = mass
a = acceleration
v = Velocity
r = Radius
PART A ) Equating the above two expression the equation for radius is


Replacing with our values we have that
![r = \frac{(1140km/hr[\frac{1000m}{1km}\frac{1hour}{3600s}])^2}{7(9.8m/s^2)}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B%281140km%2Fhr%5B%5Cfrac%7B1000m%7D%7B1km%7D%5Cfrac%7B1hour%7D%7B3600s%7D%5D%29%5E2%7D%7B7%289.8m%2Fs%5E2%29%7D)

PART B )
<u>- The expression for effective weight of the pilot at the bottom of the circle is</u>

![N = (69kg)(9.8m/s^2)+\frac{(69)(1140km/hr[\frac{1000m}{1km}\frac{1hour}{3600s}])^2}{1.462*10^3m}](https://tex.z-dn.net/?f=N%20%3D%20%2869kg%29%289.8m%2Fs%5E2%29%2B%5Cfrac%7B%2869%29%281140km%2Fhr%5B%5Cfrac%7B1000m%7D%7B1km%7D%5Cfrac%7B1hour%7D%7B3600s%7D%5D%29%5E2%7D%7B1.462%2A10%5E3m%7D)

<em>Note that the normal reaction N is directed upwards and gravitational force mg is directed downwards. At the bottom of the circle, the centripetal force is directed upwards. So the centripetal force is obtained from the gravitational force and the normal reaction. </em>
<u>- The expression for effective weight of the pilot at the top of the circle is</u>

![N = (69kg)(9.8m/s^2)-\frac{(69)(1140km/hr[\frac{1000m}{1km}\frac{1hour}{3600s}])^2}{1.462*10^3m}](https://tex.z-dn.net/?f=N%20%3D%20%2869kg%29%289.8m%2Fs%5E2%29-%5Cfrac%7B%2869%29%281140km%2Fhr%5B%5Cfrac%7B1000m%7D%7B1km%7D%5Cfrac%7B1hour%7D%7B3600s%7D%5D%29%5E2%7D%7B1.462%2A10%5E3m%7D)

<em>Note that at the top of the circle the centripetal force is directed downwards. So the centripetal force is obtained from normal reaction and the gravitational force. </em>
Answer:
1/√2
Explanation:
Kinetic energy of a body is expressed as
KE = 1/2mv²
If it's KE is doubled
2KE = 1/2mv1² where v1 is the new speed when the kinetic energy is doubled
To know the value of the amount the velocity has changed, we will divide both equations
2KE/KE = (1/2mv²)/(1/2mv1²)
2 = v²/v1²
(v/v1)² = 2
v/v1 = √2
v1 = v/√2
v1 = 1/√2 × v
The new velocity has changed by
1/√2vinitial