Answer:
light rays reflect off an object ,strike the mirror ,and are reflected into your eyes
Answer:
The equation of equilibrium at the top of the vertical circle is:
\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}
The speed experimented by the car is:
\frac{N}{m}+g=\frac{v^{2}}{R}
v = \sqrt{R\cdot (\frac{N}{m}+g) }
v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}
v\approx 9.302\,\frac{m}{s}
The equation of equilibrium at the bottom of the vertical circle is:
\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}
The normal force on the car when it is at the bottom of the track is:
N=m\cdot (\frac{v^{2}}{R}+g )
N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)
N=21.690\,N
Answer:
I believe the answer is B.
Explanation:
Newton's First Law of Gravity states, "The greater the weight (or mass) of an object, the more inertia it has. Heavy objects are harder to move than light ones because they have more inertia.
"
If the gymnast mass were doubled, her height (h) from the top of the board would be as follows,
с Stay the same
Explanation:
- The Mass of an object or body does not affect the acceleration due to gravity in any kind of way.
- Light weight objects accelerate more slowly than the heavy objects because when the forces other than the gravity also plays a major role.
- Mass increases of a body when an object has higher velocity or the speed.
- The greater the force of gravity, it would give a direct impact on the object's acceleration; thus considering only a force, the heavier the object is, it would accelerate faster. But an acceleration depends upon the two factors which are force and mass.
- Newton's second law of motion states that the acceleration of an object is dependent upon the two factors which are, the net force of an object and the mass of the object.
Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
