Answer:
r = 58.44 [m]
Explanation:
To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.
a = v²/r
where:
a = centripetal acceleration = 15.4 [m/s²]
v = tangential speed = 30 [m/s]
r = radius or distance [m]
r = v²/a
r = 30²/15.4
r = 58.44 [m]
Explanation:
Given that,
Initial velocity, u = 11.3 m/s
Angle above the horizontal, 
Time of flight :

Horizontal distance traveled is given by :
x = ut
x = 11.3 m/s × 1.32 s
x = 14.916 m
Maximum height is given by :

Hence, time of flight is 1.32 s, horizontal distance is 14.916 m and maximum height is 2.14 m.
Answer:
The acceleration is 1 cm/s^2.
Explanation:
The acceleration is defined as the rate of change of velocity.
Here, initial velocity, u = 3/1 = 3 cm/s
final velocity, v = 4/1 = 4 cm/s
time, t = 1 s
Let the acceleration is a.
Use first equation of motion
v = u + at
4 = 3 + 1 x a
a = 1 cm/s^2
Answer:
scalar quantity
Explanation:
Vector quantities have two characteristics, a magnitude and a direction. Scalar quantities have only a magnitude.
Answer:
k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
w = √ (k / m)
angular velocity and rel period are related
w = 2π / T
substitution
T = 2π √(m / K)
in Experimental measurements give us the following data
m (g) A (cm) t (s) T (s)
100 6.5 7.8 0.78
150 5.5 9.8 0.98
200 6.0 10.9 1.09
250 3.5 12.4 1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
T = t / 10
To find the spring constant we linearize the equation
T² = (4π²/K) m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
m ’= 4π² / k
where m’ is the slope
k = 4π² / m'
the equation of the line of the attached graph is
T² = 0.00603 m + 0.0183
therefore the slope
m ’= 0.00603 s²/g
we calculate
k = 4 π² / 0.00603
k = 6547 g / s²
we reduce the mass to the SI system
k = 6547 g / s² (1kg / 1000 g)
k = 6,547 kg / s² =
k = 6,547 N / m
let's reduce the uniqueness
[N / m] = [(kg m / s²) m] = [kg / s²]