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s344n2d4d5 [400]

3 years ago

14

A stone thrown off a bridge 10m above a river has an initial velocity of 8ms at an angle of 25 degrees above the horizontal. Wha

t is the range of the stone?

1 answer:

Darina [25.2K]3 years ago

7 0

Explanation:

It is given that,

Height above which the stone was thrown, h = 10 m

Initial velocity of the stone, u = 8 m/s

Angle above the horizontal, $\theta=25^{\circ}$

The horizontal component of velocity is, $u_x=v\ cos\theta=8\ cos(25)=7.25\ m/s$

The vertical component of velocity is, $u_y=v\ sin\theta=8\ cos(25)=3.38\ m/s$

Let t is the time of flight in vertical motion. The second equation of motion is :

$h=u_yt-\dfrac{1}{2}gt^2$

$10=3.38t-\dfrac{1}{2}\times 9.8t^2$

t = 0.34 seconds

Let s is the range of the stone. It can be calculated as :

$s=u_x\ cos\theta \times t$

$s=7.25\times 0.34$

s = 2.46 meters

So, the range of the stone is 2.46 meters. Hence, this is the required solution.

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Classify each statement below as an observation or as an inference based on the current Big Bang model. (note that the helium ab

Aleks04 [339]

Answer:

Observations:

Large scale structures look about the same in all directions.
The temperatures of the cosmic microwave background varies slightly with direction.
The cosmic microwave background temperature is approximately 2.73 Kelvin.
The Helium abundance is at least 25% in every galaxy studied so far.

While Inference includes:

Photons of the cosmic microwave background is radiation left over from the Big bang.
The cosmic microwave background is radiation left over from the Big bang.
Fusion during the universe's first five minutes produced 75% hydrogen and 25% helium.
Large-scale structure grew around density variations present in the early universe.

Explanation:

Inference here refers to the process of drawing a conclusion based on what is already known while Observation involves the perception and recording of data via the use of scientific instruments and procedures.

The insights below highlights the explanations for some of the insights used in determining its Observation or Inference status.

The Cosmic Microwave Background, which contains the afterglow of light and radiation is left over from the Big bang.

The Cosmic Microwave Background is black body radiation at a temperature of 2.725 Kelvin.

In the year 1965, a pair of radio astronomers named Arno Penzias and Bob Wilson respectively were working at Bell labs, in New Jersey. Using an antenna that is sensitive to the radiation of microwaves, they discovered an unknown source of static noise. After they tried doing everything possible to reduce the noise, they realized that the microwaves had a black body spectrum, with a characteristic temperature of only T = 2.725 Kelvin ( and maximum wavelength of 1 millimeter). These ever present microwaves seen in every direction are called "Cosmic Microwave Background". This explains the reason behind statement five(5) being an Observation.

Finally, The Cosmos Background Explorer (COBE) which launched a satellite in the year 1989 found that there were slight fluctuations in the radiation spectrum. The temperature varies very slightly from one place to another by a few parts in 100,000 i.e (10^-5). The excerpts above clarify the reason for the answer to statement four(4) being listed as an Observation.

6 0

4 years ago

What is the change in length of a 1400. m steel, (12x10^-6)/(C0) , pipe for a temperature change of 250.0 degrees Celsius? Remem

8090 [49]

Answer:

$\Delta L = 4.2 m$

Explanation:

As per the formula of thermal expansion we know that

$L = L_o(1 + \alpha\Delta T)$

so here we will have

$L_o = 1400 m$

$\alpha = 12 \times 10^{-6} per ^oC$

$\Delta T = 250 degree C$

so here change in the length of the rod is given as

$\Delta L = L - L_o$

$\Delta L = L_o \alpha \Delta T$

$\Delta L = 1400 (12 \times 10^{-6})(250)$

$\Delta L = 4.2 m$

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Which type of map would you most likely use to locate mineral deposits? A. topographic B. geologic C. satellite D. hazard

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I think you would be using a topographic Map, So the answer should be A

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3 years ago

What is the terminal velocity of a 6.00-kg mass object in falling with a drag force with a magnitude that depends on speed, v, a

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When object reached the terminal speed then its acceleration is zero

So as per Newton's II law we can say

$F_{net} = 0$

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so we can write

$mg - F_d = 0$

so here we are given that

$F_d = 30$

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$58.8 - 30 *v = 0$

$v = \frac{58.8}{30}$

$v = 1.96 m/s$

so terminal speed will be nearly 2 m/s

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