Answer:
r = 0.0173 m = 1.73 cm
Explanation:
Here, the centripetal force of the block will be providing the required breaking tension in the string:
[tex]Tension = Centripetal Force\\T = F_c\\\\T = \frac{mv^2}{r} \\\\r = \frac{mv^2}{T}\\[/tex]
where,
r = radius = ?
m = mass of block = 0.13 kg
v = tangential spee of block = 4 m/s
T = Breaking Strength = 30 N
Therefore,
[tex]r = \frac{(0.13\ kg)(4\ m/s)^2}{30\ N}[/tex]
r = 0.0173 m = 1.73 cm
When finding the radius of the string at the point it breaks, the tangential
velocity is assumed to be constant.
The radius when the string breaks is [tex]\underline{6.9 . \overline 3 \times 10^{-3}} \ m[/tex]Reasons:
The mass of the small block, m = 0.130 kg
Initial radius of the circle of rotation = 0.800 m
Tangential velocity, v = 4.00 m/s
The radius of the path of rotation is reduced as the string is pulled
Breaking strength of the string = 30.0 N
Required:
The radius of the circle when the string brakes
Solution:
[tex]Centripetal \ force = \dfrac{m \cdot v^2}{r}[/tex]
Where;
r = The radius of the circle of rotation
When the string brakes, w have;
Centripetal force = Breaking strength of the string = 30.0 N
Which gives;
[tex]\displaystyle r = \mathbf{\dfrac{m \cdot v^2}{Centrifugal \ force}} = \frac{0.130 \times 4^2}{30} =6.9\overline 3 \times 10^{-2}[/tex]
The radius of the circle when, the string breaks r = [tex]\underline{6.9\overline 3 \times 10^{-2}} \ m[/tex]
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A student placed four objects on a plastic tray: a rock, an eraser, a wood block, and an ice cube. The student slowly lifts the tray and measures the height at which each object begins to slide in order to compare the friction of each object. The table above shows the results. In a separate experiment the student used the same objects and tray, but glued a piece of gritty sandpaper to the tray. What results would be expected? A) The items would slide faster at the same heights. B) The items would start to slide when the tray is not lifted as high. C) The items would not slide at all, no matter how high the tray is raised. D) The tray would need to be raised higher before the items start to slide.
Answer:
D) The tray would need to be raised higher before the items start to slide.
Explanation:
The tray would need to be raised higher due to the gritty sandpaper causing it to have more grip or fraction between the two surfaces. In the first experiment it would have a smooth surface to slide across but in the second experiment it would have a rougher surface to slide across.
Example: It would be like taking two different kinds of shoes like tennis shoes and football cleets. Then seeing which one has more traction when going up a hill.
Hope this helps!!
An ideal horizontal spring-mass system is set into motion. At an instant when the mass passes through its equilibrium position: The potential energy in the spring is at its _____. The kinetic energy of the mass is at its ______. The magnitude of net force acting on the mass is at its ______.
Answer:
the potential energy is zero, and the kinetic energy must be maximum
F = 0
Explanation:
In this exercise you are asked to complete the sentences of a simple harmonic movement of a mass-spring system.
In this system mechanical energy is conserved
at the most extreme point the carousel potential energy is
K_e = ½ k x²
the kinetic energy is zero for that stopped.
At the equilibrium point
the spring elongation is x = 0 so the potential energy is zero
and the kinetic energy must be maximum since total energy of the system is conserved
the spring force is
F =- k x
as in the equilibrium position x = 0 this implies that the force is also zero
F = 0
In this exercise we have to use the knowledge of force to calculate the energy of a spring, in this way we find that:
The potential energy in the spring is at its [tex]K_e = 1/2 k x^2[/tex]. The kinetic energy of the mass is at its zero . The magnitude of net force acting on the mass is at its Zero.
In this system mechanical energy is conserved, at the most extreme point the carousel potential energy is:
[tex]K_e = 1/2 k x^2[/tex]
The kinetic energy is zero for that stopped or when at the equilibrium point, so:
the spring elongation is x = 0 so the potential energy is zero the kinetic energy must be maximum since total energy of the system is conserved
the spring force is:
[tex]F =- k x\\F=0[/tex]
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the speed of light in a certain medium is 0.6c. find critical angle , if the index of refraction is 1.67
Answer:
[tex]\theta_c = 36.78^o[/tex]
Explanation:
The relationship between the refractive index and the critical angle is given as follows:
[tex]\eta = \frac{1}{Sin\ \theta_c} \\\\Sin\ \theta_c = \frac{1}{\eta}\\\\\theta_c = Sin^{-1}(\frac{1}{\eta} )[/tex]
where,
η = refractive index = 1.67
θc = critical angle =?
Therefore,
[tex]\theta_c = Sin^{-1}(\frac{1}{1.67} )[/tex]
[tex]\theta_c = 36.78^o[/tex]