You will use the height of the bridge from the ground.
Solution:
Formula to be used is y=Viy(t)+g(t^2)/2
Where:
Vi=initial velocity which is 0 m/s
y=10 m
Gravitational acceleration or g =9.8m/s^2
T= time you need
Substitute all the given to the formula
10m=(0m/s)(t)+(9.8m/s^2)(t^2)/2
10mx2=9.8m/s^2(t^2)
Now isolate the variable you want to find which is T or time
10mx2/9.8m/s^2=t^2
20m/9.8m/s^2=t^2
Square root of 2.04= square root of t^2
T=1.43 secs
The answer is 1.43 seconds
Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
Answer:
A larger impulse. A 1-kg ball has twice as much speed as a 10-kg ball.
Explanation:
Answer:
(i) The angular speed of the small metal object is 25.133 rad/s
(ii) The linear speed of the small metal object is 7.54 m/s.
Explanation:
Given;
radius of the circular path, r = 30 cm = 0.3 m
number of revolutions, n = 20
time of motion, t = 5 s
(i) The angular speed of the small metal object is calculated as;
(ii) The linear speed of the small metal object is calculated as;
To solve this problem it is necessary to apply the concepts related to wavelength depending on the frequency and speed. Mathematically, the wavelength can be expressed as
Where,
v = Velocity
f = Frequency,
Our values are given as
L = 3.6m
v= 192m/s
f= 320Hz
Replacing we have that
The total number of 'wavelengths' that will be in the string will be subject to the total length over the size of each of these undulations, that is,
Therefore the number of wavelengths of the wave fit on the string is 6.
