Just do what u would do if u were at a stop sign
Class 1 lever
Explanation:
In a class 1 lever, the fulcrum is placed between the effort and the load. This lever systems is the most common.
- The effort is the force input and the load is the force output
- The fulcrum is a hinge between the load and effort.
- Movement of the effort and load are in opposite directions.
- There are other classes of lever like the class 2 and 3.
- They all have different load, fulcrum and effort configurations
learn more:
Load related problems brainly.com/question/9202964
Torque brainly.com/question/5352966
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Acceleration = (change in speed) / (time for the change)
Change in speed = (speed at the end) minus (speed at the beginning.
The cart's acceleration is
(0 - 2 m/s) / (0.3 sec)
= ( -2 / 0.3 ) (m/s²) = -(6 and 2/3) m/s² .
Newton's second law of motion says
Force = (mass) x (acceleration) .
For this cart: Force = (1.5 kg) x ( - 6-2/3 m/s²)
= ( - 1.5 x 20/3 ) (kg-m/s²)
<span> = </span>- 10 newtons .
<span>The force is negative because it acts opposite to the direction </span>
<span>in which the cart is moving, it causes a negative acceleration, </span>
<span>and it eventually stops the cart.</span>
Explanation :
It is given that :
The initial velocity of the car is 10 m/s.
Juliette sets the acceleration to zero.
We know that, acceleration 
\
where,
u is the initial velocity and v is the final velocity.
So, final velocity will become 10 m/s.
Hence, the car will move with the constant velocity i.e. 10 m/s.
So, Shakina described that the car will move with the constant velocity of 10 m/s as acceleration is zero.
Answer:
1) λ = 0.413 m
, 2)v = 25,213 m / s
, 3) T = 0.216 N
, 4) m = 22.04 10-3 kg
Explanation:
1) The resonance occurs when the traveling wave bounces at the ends and the two waves are added, the ends as they are fixed have a node, the wavelength and the length of the string are related
λ = 2L / n n = 1, 2, 3 ...
In this case L = 0.62 m and n = 3
Let's calculate
λ = 2 0.62 / 3
λ = 0.413 m
2) the velocity related to wavelength and frequency
v = λ f
v = 0.413 61
v = 25,213 m / s
3) let's use the equation
v = √T /μ
T = v² μ
T = 25,213² 3.4 10⁻⁴
T = 0.216 N
4) the rope tension is proportional to the hanging weight
T-W = 0
T = W
W = m g
m = W / g
m = 0.216 / 9.8
m = 22.04 10-3 kg
5) n = 2
λ = 2 0.62 / 2
λ = 0.62 m
6) v = λ f
v = 0.62 61
v = 37.82 m / s
7) T = v² μ
T = 37.82² 3.4 10⁻⁴
T = 0.486 N
8) m = W / g
m = 0.486 / 9.8
m = 49.62 10⁻³ kg
9) n = 1
λ = 2 0.62
λ = 1.24 m
v = 1.24 61
v = 75.64 m / s
T = v² miu
T = 75.64² 3.4 10⁻⁴
T = 2.572 10⁻² N
m = 2.572 10⁻² / 9.8
m = 262.4 10⁻³ kg
