So the area under a velocity time graph is distance or displacement, if you have done calculus yet you will understand that if you take the integral of a velocity function then you end up with displacement. Thats for later understanding however.
So this appears to be a right triangle so we can find the area of a triangle as:
0.5bh = A
Since our area is 10 meters lets alter our formula a bit to fit the situation:
Our base here is time and our height is velocity so:
0.5tv = Δx
So we can read off the graph that our velocity at the end, or our final velocity appears to be near 2.0 m/s
So we have v, and Δx so lets isolate for time by dividing by v and 0.5
t = Δx / 0.5v
Now lets plug all that in:
t = 10 / 0.5(2)
t = 10 seconds
Hope this helped!
Question: The force between a pair of 0.005 C is 750 N. What is the distance between them?
Answer:
17.32 m
Explanation:
From coulomb's Law,
F = kqq'/r²........................... Equation 1
Where F = Force between the force, q' and q = both charges respectively, k = coulomb's constant, r = distance between both charges.
make r the subject of the equation above
r = √(kqq'/F)..................... Equation 2
From the question,
Given: q = q' = 0.005 C, F = 750 N
Constant: k = 9.0×10⁹ Nm²/C².
Substitute these values into equation 2
r = √(9.0×10⁹×0.005×0.005/750)
r = √(300)
r = 17.32 m.
Hence the distance between the pair of charges = 17.32 m
Answer:
d = 380 feet
Explanation:
Height of man = perpendicular= 130 feet
Angle of depression = ∅ = 70 °
distance to bus stop from man = hypotenuse = d = 130 sec∅
As sec ∅ = 1 / cos∅
so d = 130 sec∅ or d = 130 / cos∅
d = 130 / cos(70°)
d = 380 feet
Asterism, a pattern of stars that is not a constellation. An asterism can be part of a constellation, such as the Big Dipper, which is in the constellation Ursa Major, and can even span across constellations, such as the Summer Triangle, which is formed by the three bright stars Deneb, Altair, and Vega.
a)You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.
from above statement we got
height = 78.4 m
since the ball is thrown, so its vertical velocity would be zero
u = 0
taking g = 9.8m/s^2
now, using the equation of motion
h = ut + gt^2/2
now putting all the values in it
we got ,
78.4 = 9.8 * t^2/ 2
by solving we got,
t = 4 sec
b) now, since along the horizontal , no force acting and accelaration is zero so
R = ut , R is RANGE
R = 5 * 4
range = 20 m
c) vertical components of the stone’s velocity just before it hits the ground = v sin θ =
horizontal components of the stone’s velocity just before it hits the ground = v cos θ
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