Answer:
H = Vy t - 1/2 g t^2 height of an object with an initial "vertical" velocity
at t sec after firing
Vy = 78 m/s * sin 40 = .643 * 78 m/s = 50.1 m/s
H = 50.1 * 6 - 1/2 * 9.8 * 6^2 = 300 m - 176 m = 124 m
Answer:
False because igneous rocks are formed from a volcano and sedimentary never move they stay in one spot
Answer:
A.) 8 m/s
B.) 7.0 m
Explanation:
Given that a block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane.
(a) What is its velocity when it reaches the top of the plane?
Since the plane is frictionless, the final velocity V will be the same as 8 m/s
The velocity will be 8 m/s as it reaches the top of the plane.
(b) How far horizontally does it land after it leaves the plane?
For frictionless plane,
a = gsinø
Acceleration a = 9.8sin28
Acceleration a = 4.6 m/s^2
Using the third equation of motion
V^2 = U^2 - 2as
Substitute the a and the U into the equation. Where V = 0
0 = 8^2 - 2 × 4.6 × S
9.2S = 64
S = 64/9.2
S = 6.956 m
S = 7.0 m
Answer:
5.571 sec
Explanation:
angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s
Period To = 2π / angular frequency
Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got
T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec
t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been ( To / (√ (1 - (v²/c²))). where To = 2.00 sec
