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zhenek [66]
3 years ago
12

On an amusement park ride passengers accelerate straight downward from rest to 22.9 m/s in 2.2 s.

Physics
2 answers:
Kamila [148]3 years ago
8 0

Given,  

Initial Velocity (u) = 0 m/s  

Final Velocity (v) = 25.9 m/s  

Time (t) = 2.5 sec  

avg. acceleration = ?  

avg. acceleration = v-u/t  

avg = 25.9/2.5 = 10.36 m/s²  

avg = 10.36 m/s²

Sindrei [870]3 years ago
5 0

Answer:

Acceleration, a=10.4\ m/s^2

Explanation:

Given that,

Initial velocity of the ride, u = 0 (at rest)

Final speed of the ride, v = 22.9 m/s

Time taken, t = 2.2 s

Let a is the acceleration of the passengers on this ride. It can be calculated using the formula of the acceleration as :

a=\dfrac{v-u}{t}

a=\dfrac{22.9-0}{2.2}

a=10.4\ m/s^2

So, the time required to bring the car to a stop is 10.4\ m/s^2. Hence, this is the required solution.

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4 years ago
•• CP Two blocks connected by a light horizontal rope sit at rest on a horizontal, frictionless surface. Block AA has mass 15.0
Firdavs [7]

Answer:

(a) T= 38.4 N

(b) m= 26.67 kg

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Kinematics

d= v₀t+ (1/2)*a*t² (Formula 2)

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

v₀=0, d=18 m , t=5 s

We apply the formula 2 to calculate the accelerations of the blocks:

d= v₀t+ (1/2)*a*t²

18= 0+  (1/2)*a*(5)²

a= (2*18) / ( 25) = 1.44 m/s² to the right

We apply Newton's second law to the block A

∑Fx = m*ax

60-T = 15*1.44

60 - 15*1.44 = T

T = 38.4 N

We apply Newton's second law to the block B

∑Fx = m*ax

T = m*ax

38.4 = m*1.44

m= (38.4) / (1.44)

m = 26.67 kg

7 0
4 years ago
What is the pendulum length whose period is 2.0s ?
Mashutka [201]
Formula\ for\ period:\\\ T=2 \pi \sqrt{\frac{L}{g}}\\\ g-gravity=9,8 \frac{m}{s^2} ,\ L-pendulum \ length \\\\ \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\\\ \frac{T^2}{2 \pi } = \frac{L}{g} \\\\\ \frac{T^2}{2 \pi }*g=L\\\\ L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24mT=2 \pi   \sqrt{\frac{L}{g}} \\
 \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\
 \frac{T^2}{2 \pi }  = \frac{L}{g} \\
 \frac{T^2}{2 \pi }*g=L\\
L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24m

7 0
3 years ago
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