2 Points<br> When must scientific theories be changed?

Superman is flying 54.5 m/s when he sees

Nady [450]

348.34 m/s. When Superman reaches the train, his final velocity will be 348.34 m/s.

To solve this problem, we are going to use the kinematics equations for constant aceleration. The key for this problem are the equations $d=v_{0} t+\frac{at^{2} }{2}$ and $v_{f} =v_{0} +at$ where $d$ is distance, $v_{0}$ is the initial velocity, $v_{f}$ is the final velocity, $t$ is time, and $a$ is aceleration.

Superman's initial velocity is $v_{0}=54.5\frac{m}{s}$ , and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know $d$ , $v_{0}$ and $t$ , we have to find the aceleration $a$ in order to find $v_{f}$ .

From the equation $d=v_{0} t+\frac{at^{2} }{2}$ we have to clear $a$ , getting the equation as follows: $a=\frac{2(d-v_{0}t) }{t^{2} }$ .

Substituting the values:

$a=\frac{2(850m-54.5\frac{m}{s}.4.22s) }{(4.22s)^{2}}=69.63\frac{m}{s^{2}}$

To find $v_{f}$ we use the equation $v_{f} =v_{0} +at$ .

Substituting the values:

$v_{f} =54.5\frac{m}{s} +(69.63\frac{m}{s^{2}}.4.22s)=348.34\frac{m}{s}$

5 0

2 years ago

Just need help with 1 and 2 please :D i’m having a bit of trouble :/

dexar [7]

1. Traveling by car means you have specific roads to follow. You won’t be able to go straight to Banning high from POLAHS. The 8.4km will be defined as distance. Traveling by helicopter you don’t have roads to follow that means you can fly directly to banning high. 6.8km will be defined as displacement.

2. A) 400m
B)0m
C)d=1/2(vi+vf)t
400=1/2(0+vf)92
8.7m/s
D) 0m/s
E) Not sure but instantaneous velocity refer to velocity at a given point. Average velocity is just the average. Usually instantaneous velocity won’t be same as the average velocity.
Plz like if it helped.

7 0

3 years ago

You are asked to design a spring that will give a 1020 kg satellite a speed of 2.25 m/s relative to an orbiting space shuttle. Y

julsineya [31]

Answer:

(a) 2.45×10⁵ N/m

(b) 0.204 m

Explanation:

Here we have that to have a velocity of 2.25 m/s then the relationship between the elastic potential energy of the spring and the kinetic energy of the rocket must be

Elastic potential energy of the spring = Kinetic energy of the rocket

$\frac{1}{2} kx^2 = \frac{1}{2} mv^2$