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RSB [31]
2 years ago
12

Plz answer the question!

Physics
1 answer:
Sindrei [870]2 years ago
6 0
The light would reflect and go back up in a vortexian loop
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A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its
m_a_m_a [10]

Answer: 0.0146m

Explanation: The formula that defines the velocity of a simple harmonic motion is given as

v = ω√A² - x²

Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.

The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA

One half of maximum speed = speed of motion

3ωA/2 = ω√A² - x²

ω cancels out on both sides of the equation, hence we have that

A/2 = √A² - x²

(0.0169)/2 = √(0.0169² - x²)

0.00845 = √(0.0169² - x²)

By squaring both sides, we have that

0.00845² = 0.0169² - x²

x² = 0.0169² - 0.00845²

x² = 0.0002142

x = √0.0002142

x = 0.0146m

5 0
3 years ago
What is the difference between weight and mass? mass and weight are both intrinsic properties of an object. if you multiply the
MatroZZZ [7]

if you multiply the mass of an object by the acceleration due to gravity, you will obtain the object's weight. mass is an intrinsic property of matter


looks like a good answer ...

4 0
2 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
3 years ago
A kid on a trampoline has 1,000 J of potential energy when they are at the top of a jump. How much kinetic energy will the kid h
emmasim [6.3K]
C: 500 J
Hope this helped!
7 0
2 years ago
At what speed will a box be falling at a time t = 0.75 s after being dropped?
IRINA_888 [86]

Explanation:

Initial speed(u)= 0 m/s (Ball is dropped)

time(t)= 0.75 s

acceleration(a)= 10 m/s² (gravity)

Final speed(v)= u+at

v=0+(10)× 0.75

v=7.5 m/s

Speed is 7.5 m/s

8 0
3 years ago
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