The mass to be hung at a point 40cm to the right of the pivot point such that the stick hangs level is 15 g.
How to calculate mass?To solve this problem, use the principle of moments which states that the sum of the clockwise moments about a pivot point is equal to the sum of the counterclockwise moments about the same pivot point. In this case, take the pivot point to be the suspension point of the meter stick.
Let x be the mass that needs to hang at a point 40cm to the right of the pivot point. Then, set up the following equation:
(clockwise moment) = (counterclockwise moment)
(0.2 kg) × (0.3 m) × (9.81 m/s²) = (x kg) × (0.4 m) × (9.81 m/s²) + (0.1 kg) × (0.1 m) × (9.81 m/s²)
Simplifying this equation:
0.05886 = 3.924x
x = 0.015 kg or 15 g
Therefore, we need to hang a 15 g mass at a point 40 cm to the right of the pivot point such that the stick hangs level.
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what is the unit of time not based on a heavenly body
A block is attached to a spring and executes
simple harmonic motion according to x = 2.0
cos(50t), where x is in meters and t is in
seconds. The spring constant is k = 100 N/m.
What is the mass of the block?
Answer:
.04 kg
Explanation:
The equation for simple harmonic motion is x = A*cos(ωt), where A is the amplitude, ω is the angular frequency, and t is the time.
Comparing the given equation x = 2.0cos(50t) with the equation for simple harmonic motion, we see that A = 2.0 meters and ω = 50 radians/second.
The angular frequency is related to the spring constant and mass by the equation ω = sqrt(k/m), where sqrt denotes the square root.
Substituting the values given, we get:
50 = sqrt(100/m)
Squaring both sides and solving for m, we get:
m = 100/2500 = 0.04 kg
. When a large truck
hits a small car, the
forces are equal.
• However, the small
.
car experiences a
much greater
change in velocity
much more rapidly
than the big truck.
Which vehicle ends up
with more damage?
Answer:
Car
Explanation:
Based on Newton 2nd law, energy are conserves. Meaning that if the Force is equal, the car with lower mass must be travelling in a much greater acceleration.
F = m.a
where,
a = Δv/Δt
When talking about energy, there are 2 factor: mass and velocity.
The change of Kinetic energy experience by the car is
ΔEk = 1/2.m.Δv²
Eventhough the car has smaller mass, notice that the velocity will be squared. In this case the velocity is the a more dominant factor. It means that energy absorbed by the car is much larger.
This graph is a combination of atmospheric carbon dioxide measurements taken from ice cores in Antarctica and air samples atop Mauna Loa. The graph BEST shows how the atmosphere has been affected by an increase in
Responses
Industrial activity
Industrial activity
Photosynthesis
Photosynthesis
Sea Levels
Sea Levels
Ozone Layer
A ball is initially at rest and travels 7.8 m. The ball travels at an acceleration of 6.4 m/s². What is the final velocity of the ball? Give your answer to 1 decimal place.
The final velocity of the ball to one decimal place is approximately 10.0 m/s.
What is the final velocity of the ball?From the third equation of motion:
v² = u² + 2as
Where v is final velocity, u is initial velocity, a is acceleration and s is the distance covered.
Given that:
Ball was initially at rest, initial velocity u = 0acceleretaion a = 6.4 m/s²distance traveled s = 7.8 mFinal velocity v = ?Plug the given values into the abovr formula and solve for the final velocity v.
v² = u² + 2as
v² = 0² + ( 2 × 6.4 m/s² × 7.8 m )
v² = 2 × 6.4 m/s² × 7.8 m
v² = 99.84 m²/s²
v = √( 99.84 m²/s² )
v = 10.0 m/s
Therefore, the final velocity is 10.0 m/s.
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The velocity of three particles of masses 20g, 30g and 50g are 2i, 10j and 10k respectively. The velocity of the centre of mass the three particle) is
Answer:
The velocity of the center of mass (Vcm) of a system of particles can be calculated using the formula:
Vcm = (m1v1 + m2v2 + m3v3 + ... + mnvn) / (m1 + m2 + m3 + ... + mn)
where m1, m2, m3, ... mn are the masses of the particles and v1, v2, v3, ... vn are their velocities.
In this problem, we have three particles with masses of 20g, 30g, and 50g and velocities of 2i, 10j, and 10k respectively. We can convert the masses to kg to make the calculations easier:
m1 = 20g = 0.02kg
m2 = 30g = 0.03kg
m3 = 50g = 0.05kg
Using the formula above, we can calculate the velocity of the center of mass:
Vcm = (m1v1 + m2v2 + m3v3) / (m1 + m2 + m3)
Vcm = (0.02kg * 2i + 0.03kg * 10j + 0.05kg * 10k) / (0.02kg + 0.03kg + 0.05kg)
Vcm = (0.04i + 0.3j + 0.5k) / 0.1kg
Vcm = 0.4i + 3j + 5k m/s
Therefore, the velocity of the center of mass of the three particles is 0.4i + 3j + 5k m/s.
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a student is swinging a ball attatched to a string in a vertical circle
The magnitude of the acceleration of the ball applied at the bottom of the circle can be expressed in the form of FTension-FGravity/M.
Option D is correct.
When making a vertical circle with a ball on a string?Along the string's circular and vertical paths, the tension changes. As long as the total quantity of kinetic and potential energy is constant throughout, the ball's speed can change.
Centripetal force varies as a result of motion variations.
We can determine how tight a string that is traveling in a vertical circle is using the expression below:
FC = mv2 /r.
A moving item attached to a string experiences centripetal force, which is determined by the product of the object's mass (mg) and the string tension. (T).
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Complete question:
A student swings ball of mass M on the end of a string in vertical circle of radius R,as shown in the figure below. Also shown is diagram representing all the forces exerted on the ball at the bottom of the circle where its speed is What is the magnitude of the acceleration of the ball at the bottom of the circle? FTension FGravity
A)Fi /M
B)Fc /M
C)Fr+Fg/M
D) Ft- Fg/M
