Frequency=1/T, where T is the time period.
Reciprocal of frequency = T or it can be written as T = 1/f, where f is frequency.
For this problem, we use the equations derived for rectilinear motion at constant acceleration. The equations are:
a = (v - v₀)/t
x = v₀t + 0.5at²
where
a is acceleration
v and v₀ are the final and initial velocities, respectively
x is the distance
t is the time
First, let's determine the a to be used in the second equation:
a = (7.5 m/s - 0 m/s)/1.7 s = 4.411 m/s²
x = (0)(1.7s) + 0.5(4.411 m/s²)(1.7 s)²
x = 6.375 m
Answer:
ice
Explanation:
If the toy car is on a smooth surface, there is less friction. Therefore, the car will most likely go faster. Ice has the least friction so the toy car would travel fast.
Answer:
C
Explanation:
Because...
A= Incideant ray
B= Angle of inciderance
<u>C= angle of reflection</u>
D= reflection ray
Answer:
6.69 m/s
4.483 m
1.42s
Explanation:
Given that:
Initial Velocity, u = 0
Final velocity, v =?
Acceleration, a = 35m/s²
1.) using the relation :
v² = u² + 2as
v² = 0 + 2(35) * 64*10^-2m
v² = 70 * 0.64
v = sqrt(44.8)
v = 6.693
v = 6.69 m/s
B.) height from the ground, h0 = 2.2
How high ball went , h:
Using :
v² = u² + 2as
Upward motion, g = - ve
0 = 6.69² + 2(-9.8)*(h - 2.2)
0= 6.69² - 19.6(h - 2.2)
44.7561 + 43.12 - 19.6h = 0
19.6h = 44.7561 - 43.12
h = 87.8761 / 19.6
h = 4.483 m
C.)
vt - 0.5gt² = h - h0
6.69t - 0.5(9.8)t²
6.69t - 4.9t² = 1.83 - 2.2
-4.9t² + 6.69t + 0.37 = 0
Using the quadratic equation solver :
Taking the positive root:
1.4185 = 1.42s
