Oh jk haha LOL bye hahaha
ANSWER
C. 5 seconds to 7 seconds
EXPLANATION
This is a velocity time graph.
We can read the final velocity, v,the initial velocity u and the time easily from the graph.
See attachment for the initial and final velocities on each time interval.
So we can use either of the following equations to calculate the distance traveled.
Or
.
During the time interval 0 to 3, the distance traveled is,
.
During the time interval 3 to 5 seconds, the distance traveled is,
.
During time interval, 5 to 7 seconds the distance traveled is,
.
During time interval 7 to 8 seconds the distance traveled is
.
During time interval 8 to 10 seconds the distance traveled is
.
The correct answer is C.
Answer:
Let's see what to do buddy...
Explanation:
_________________________________
_________________________________
The body start from rest which means :
_________________________________
In third second means :
t = 2 -----¢ t = 3 -----¢ ∆t = 1
_________________________________
We have this equation to find the distance.
_________________________________
_________________________________
And we're done.
Thanks for watching buddy good luck.
♥️♥️♥️♥️♥️
The current drawn by the series circuit is(R₁ + R₂)/R₁R₂ times the current drawn by the parallel circuit.
Let the resistance of the two lamps are R₁ and R₂.
Then the equivalent resistance in series combination is: R = R₁ + R₂.
And, the equivalent resistance in parallel combination is:
r = R₁R₂/(R₁ + R₂).
So, if the supply voltage is V,
Then, current drown in series combination; = V/R = V/(R₁ + R₂)
And, current drown in parallel combination; = V/r = V(R₁ + R₂)/R₁R₂
So , = [ V/(R₁ + R₂)] /[V(R₁ + R₂)/R₁R₂]
= (R₁ + R₂)/R₁R₂
Hence, the ratio of current drawn in series and current drown in parallel is (R₁ + R₂)/R₁R₂. So, he current drawn by the series circuit is(R₁ + R₂)/R₁R₂ times the current drawn by the parallel circuit.
Learn more about electric current here:
brainly.com/question/2264542
#SPJ1
frequency fapproach is heard by a passenger = 302.05 Hz
frequency frecede is heard by a passenger = 228.37433 Hz
Given :
speed of train to the ground = 30.0 m/s
frequency emitted by the train whistle = 262 Hz
speed of sound in air = 344 m/s
To Find :
(A) frequency fapproach (B) frequency frecede
Solution :
The frequency of approach is given by
f' = f <u>v + v(relative)</u>
v - v(air)
= 262x <u>344 + 18</u>
344 - 30
f' = 302.05 Hz
The frequency of approach is 302.05 Hz
The frequency of recede is given by
f' = f <u>v - v(relative)</u>
v + v(air)
= 262 x <u>344 - 18</u>
344 + 30
The frequency of recede is 228.37433 Hz
Learn more about Frequency here:
brainly.com/question/254161
#SPJ4