Answer:
45 s .
Explanation:
The accelerator will first accelerate , then move with uniform velocity and at last it will decelerate to rest .
displacement s = ?
acceleration a = 1 m /s²
Final speed v = 5 m/s
initial speed u = 0
v² = u² + 2as
5² = 0 + 2 x 1 x s
s = 12.5 m
B) Let time of acceleration or deceleration be t
v = u + a t
5 = 0 + 1 t
t = 5 s
Similarly displacement during deceleration = 12.5 m
Total distance during uniform motion = 200 - ( 12.5 + 12.5 ) = 175 m .
velocity of uniform motion = 5 m /s
time during which there was uniform velocity = 175 / 5 = 35 s
Total time = 5 + 35 + 5 = 45 s .
Answer:
91.017N
Explanation:
Parameters
L=4.67m, m=0.192kg, t = 0.794s, The pulse makes four trips down and back along the cord, we have 4 +4 =8 trips( to and fro)
so N= no of trips = 8, From Wave speed(V) = N *L/t , we have :
V= 8*4.67/0.794 = 47.0529 m/s.
We compute the cords mass per length, Let it be P
P = M/L = 0.192/4.67 = 0.04111 kg/m
From T = P * V^2 where T = Tension, we have
T = 0.04111 * (47.0529)^2
T = 91.017N.
The tension in the cord is 91.017N
When someone is holding something that has been struck or splashed by lightning, contact damage occurs.
We need additional information concerning lightning and injuries in order to identify the solution.
<h3>What types of injuries are brought on by lightning?</h3>
- Lightning is the name for a natural electrical discharge that occurs quickly and with a dazzling flash.
- It has a tremendous amount of energy.
- Lightning-related injuries can be divided into three categories: direct strikes, side splashes, and contact injuries.
- When someone is struck by lightning directly, they can get direct injury.
- When a current splashes from a neighboring object, it is called a side splash.
- When someone touches a lightning-hit object, contact harm results.
In light of this, we can say that contact injuries happen when a person is holding an object that has been struck by lightning or splashed by it.
Learn more about the lightning and harm here:
brainly.com/question/28055828
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I see the light moving exactly at speed equal to c.
In fact, the second postulate of special relativity states that:
"The speed of light in free space has the same value c<span> in all inertial frames of reference."
</span>
The problem says that I am moving at speed 2/3 c, so my motion is a uniform motion (constant speed). This means I am in an inertial frame of reference, so the speed of light in this frame must be equal to c.
