Answer:
Natalie says that all things with mass have a gravitational field, but the force is very weak and cannot be perceived around small objects.
Explanation:
The force due to gravity is proportional to the mass of the object and inversely proportional to the square of the distance between objects. The Earth is so massive that the force due to its gravity is much greater than the force between objects on the counter.
If there were no friction, the objects might move toward each other, depending on what other masses were near them tending to cause them to move in other directions.
Natalie's explanation is about the best.
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Additional comment
The universal gravitational constant was determined by Henry Cavendish in the late 18th century using lead balls weighing 1.6 pounds and 348 pounds. His experiment was enclosed in a large wooden box to minimize outside effects. While these masses are somewhat greater than those of a glue bottle and stapler, the experiment shows the force of gravity between "small" objects can be measured.
To understand and apply the formula τ=Iα to rigid objects rotating about a fixed axis. To find the acceleration a of a particle of mass m, we use Newton's second law: F net=ma , where F net is the net force acting on the particle.To find the angular acceleration α of a rigid object rotating about a fixed axis, we can use a similar formula: τnet=Iα, where τnet=∑τ is the net torque acting on the object and I is its moment of inertia.
Part A:
Assume that the mass of the swing bar, is negligible. Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, l, as well as the acceleration due to gravity g.
Part B:
Now consider a similar situation, except that now the swing bar itself has mass mbar.Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, mbar, l, as well as the acceleration due to gravity g.
Answer:
Hello your question is incomplete attached below is the missing part of the question
In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Assume that the pivot is attached tot he center of the bar.
You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2.I
answer : part A = 2*[(M1 - M2)/(M1 + M2)]*g/L
part A = attached below
Explanation:
Part A :
Assuming that mass of swing is negligible
α = T/I
where ; T = torque, I = inertia,
hence T = L/2*9*(M1 - M2)
also; I = [tex]M1*(L/2)^2 + M2*(L/2)^2[/tex]= ( M1 + M2) * (L/2)^2
Finally the magnitude of the angular acceleration α
α = 2*[(M1 - M2)/(M1 + M2)]*g/L
Part B attached below
A bat at rest sends out ultrasonic sound waves at 46.2 kHz and receives them returned from an object moving directly away from it at 21.8 m/s, what is the received sound frequency?
f= ? Hz
Answer:
f" = 40779.61 Hz
Explanation:
From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;
from the Doppler effect equation, we can calculate the initial observed frequency as:
f' = f(1 - (v_o/v))
We are given;
f = 46.2 kHz = 46200 Hz
v_o = 21.8 m/s
v is speed of sound = 343 m/s
Thus;
f' = 46200(1 - (21/343))
f' = 43371.4285 Hz
In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;
Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;
f" = f'/(1 + (v_o/v))
f" = 43371.4285/(1 + (21.8/343))
f" = 40779.61 Hz
A rectangular reflecting pool is 85.0 ft wide and 120 ft long. What is the area of the pool in square meters?
A car is traveling at a constant speed of 20 m/s for 3 seconds. Then the driver puts on the brakes. The total distance the car travels is 100 m. What is the total time the car was moving?
Answer:
15 seconds
Explanation:
If car was moving at 20m/s for 3 sec.
if car traveled 100m = 15 sec total
Two 0.60-kilogram objects are connected by a thread that passes over a light, frictionless pulley. The objects are initially held at rest. If a third object with a mass of 0.30 kilogram is added on top of one of the 0.60-kilogram objects and the objects are released, the magnitude of the acceleration of the 0.30-kilogram object is most nearly:______
Answer:
2 m/s²
Explanation:
From the given information:
The first mass m_1 = 0.6 kg
The second mass m_2 = 0.3 kg
The magnitude for the acceleration of 0.3 kg is:
a = net force/ effective mass
Mathematically, it can be computed as follows:
[tex]a = \dfrac{F}{m}[/tex]
[tex]a = \dfrac{(m_2 +m_1 -m_1) }{(m_2+m_1+m_1)}(g)[/tex]
[tex]a = \dfrac{0.3 +0.6 -0.6}{(0.3 +0.6+0.6)}(9.8)[/tex]
a ≅ 2 m/s²
Our Sun’s mass is 1.0 and our Earth’s mass is 2.0. The distance is standard as given on the simulation. Describe the path of the Earth.
Answer:
Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).
Explanation:
The speed limit on some segments of interstate 4 is 70 mph. What is this in km/h?
Answer:
112.63km/hr
Explanation:
The given dimension is :
70mph
We are to convert this to km/hr
1 mile = 1.609km
so;
70mph x 1.609 = 112.63km/hr
So,
The solution is 112.63km/hr
true or false A person's speed around the Earth is faster at the poles than it is at the equator.
Answer:False
Explanation:The Earth rotates faster at the equator than at the poles.
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.69 times a second. A tack is stuck in the tire at a distance of 0.331 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed.
Answer:
the tack's tangential speed is 5.59 m/s
Explanation:
Given that;
R = 0.331 m
wheel rotates 2.69 times a second which means, the wheel complete 2.69 revolutions in a second, so
ω = 2.69 rev/s × 2π/1s = 16.9 rad/s
using the relation of angular speed with tangential speed
tangential speed v of the tack is expressed as;
v = R × ω
so we substitute
v = 0.331 m × 16.9 rad/s
v = 5.59 m/s
Therefore, the tack's tangential speed is 5.59 m/s