This is a poorly written question.
<span>Out of the choices listed, the first one is the only one that includes
a true statement ... the greater the mass of two objects, the
greater
the gravitational attraction is between them.</span>
-- Newton's law of universal gravitation doesn't "suggest" that. It states it ...
boldly and unequivocally.
-- The law doesn't refer to the "greatness" of the mass of the two objects.
It refers to the product of their masses.
-- It's true that the law of universal gravitation can be massaged and
manipulated to reveal the existence of gravitational planetary orbits.
But there's a lot more to it than simply the masses.
For example ... the gravitational force between two objects is inversely
proportional to
(the distance between the objects)² .
It turns out that IF that exponent were not precisely, exactly 2.000000... ,
then gravitational orbits could not exist.
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
2) Newton's third law<span>: For every action, there is an equal and opposite reaction.
3) Please provide the diagram </span>
Answer:
The two value of the wavelength for the out of tune guitar is
Explanation:
From the question we are told that
The wavelength of the note is 
The difference in beat frequency is 
Generally the frequency of the note played by the guitar that is in tune is
Where 
is the speed of sound with a constant value 
The difference in beat is mathematically represented as
Where 
is the frequency of the sound from the out of tune guitar
substituting values
The wavelength for this frequency is
For the second value of the second frequency
The wavelength for this frequency is
Answer:
D
Explanation:
From the information given:
The angular speed for the block 
Disk radius (r) = 0.2 m
The block Initial velocity is:
Change in the block's angular speed is:
However, on the disk, moment of inertIa is:
The time t = 10s
∴
Frictional torques by the wall on the disk is:
Finally, the frictional force is calculated as:
