We have that for the Question "What is the tension in the string connecting block 100 to block 99? What is the tension in the string connecting block 50 to block 51?"
it can be said that
From the question we are told
Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first block is pulled with a force of 100 N.
Assuming mass of each block is 1 kg
The equation for the force is given as
Now, between block 100 and 99,
Now between block 50 and 51. There are 50 blocks behind 51 st block, so,
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Answer:
The tension in the string connecting block 50 to block 51 is 50 N.
Explanation:
Given that,
Number of block = 100
Force = 100 N
let m be the mass of each block.
We need to calculate the net force acting on the 100th block
Using second law of newton
We need to calculate the tension in the string between blocks 99 and 100
Using formula of force
We need to calculate the total number of masses attached to the string
Using formula for mass
We need to calculate the tension in the string connecting block 50 to block 51
Using formula of tension
Put the value into the formula
Hence, The tension in the string connecting block 50 to block 51 is 50 N.
B) Good insulators
C) Highly malleable
D) Good conductors
Answer:
3.06 seconds
Explanation:
To find the maximum height above the ground level, we can use the kinematic equation for vertical motion. The equation is:
h = (v^2 - u^2) / (2g)
Where:
h is the maximum height,
v is the final velocity (which is 0 when the ball reaches its highest point),
u is the initial velocity (30 m/s),
and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values, we get:
h = (0^2 - 30^2) / (2 * 9.8)
Simplifying the equation gives us:
h = -900 / 19.6
The maximum height above the ground level is approximately -45.92 meters. Since the height cannot be negative, the maximum height is 45.92 meters.
To find the time it takes for the ball to reach the ground, we can use another kinematic equation:
t = (v - u) / g
Where:
t is the time,
v is the final velocity (which is 0 when the ball reaches the ground),
u is the initial velocity (30 m/s),
and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values, we get:
t = (0 - 30) / 9.8
Simplifying the equation gives us:
t = -30 / 9.8
The time it takes for the ball to reach the ground is approximately -3.06 seconds. Since time cannot be negative, the ball takes approximately 3.06 seconds to reach the ground.
The metricsystem was first put into practice in 1799, during the French Revolution, when the existing system of measures became impractical for trade and was supplanted by a decimal system based on the kilogram and the meter.
During the FrenchRevolution, the existing system of measures became impractical for trade and was replaced by a decimalsystem based on the kilogram and the meter, and the metricsystem was born.
In 1793, the meter was defined as one ten-millionth of the distance from the equator to the NorthPole along a great circle, implying that the Earth's circumference is approximately 40000 km.
The meter was redefined in 1799 in terms of a prototype meter bar.
The meter was introduced as a new unit of length, defined as one ten-millionth of the shortestdistance between the NorthPole and the Equator passing through Paris, assuming an Earth flattening of 1/334.
Thus, this is the history of the metric system as it applies to the meter.
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Answer and explanation;
In 1670 Gabriel Mouton, Vicar of St. Paul’s Church and an astronomer proposed the swing length of a pendulum with a frequency of one beat per second as the unit of length.
In 1791 the Commission of the French Academy of Sciences proposed the name meter to the unit of length. It would equal one tens-millionth of the distance from the North Pole to the equator along the meridian through Paris.It is realistically represented by the distance between two marks on an iron bar kept in Paris.
In 1889 the 1st General Conference on Weights and Measures define the meter as the distance between two lines on a standard bar that made of an alloy of 90%platinum with 10%iridium.
In 1960 the meter was redefined as 1650763.73 wavelengths of orange-red light, in a vacuum, produced by burning the element krypton (Kr-86).
In 1984 the Geneva Conference on Weights and Measures has defined the meter as the distance light travels, in a vacuum, in 1299792458⁄ seconds with time measured by a cesium-133 atomic clock which emits pulses of radiation at very rapid, regular intervals.
to block ultraviolet rays.
b.
to block infrared rays.
c.
to block glare from reflections (for example off lakes or roads).
d.
to look cool.
The primary purpose of polarized sunglasses is to block glare from reflections (for example off lakes or roads). The correct option is C.
Polarized sunglasses are a type of eyewear that has a special filter that blocks intense, reflected light and reduces glare. The filter is oriented vertically, allowing only the vertical component of light waves to pass through, while blocking the horizontal component. This reduces the glare from flat surfaces, such as water or snow, which can be distracting and even dangerous. Polarized sunglasses also enhance contrast and provide clearer vision, making them popular among athletes and outdoor enthusiasts. They are especially useful for activities such as fishing, boating, skiing, and driving. Polarized sunglasses can come in a variety of styles and lens colors, and can be worn in both prescription and non-prescription forms.
eEre in the Question,
Option a, to block ultraviolet rays, is true to some extent. Polarized sunglasses may have UV-blocking properties, but their primary purpose is not to block UV rays.
Option b, to block infrared rays, is not true. Polarized sunglasses are not designed to block infrared rays.
Option d, to look cool, is not true. While some people may wear polarized sunglasses for fashion purposes, their primary purpose is to reduce glare and improve visibility.
Therefore, The correct option is C. to block glare from reflections (for example off lakes or roads).
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