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3 years ago

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If questions asked in a research study to do not accurately relate to the subject or construct being studied, the study’s ______

____ is compromised. A. population B. evaluation C. administration D. validity Please select the best answer from the choices provided A B C D

1 answer:

Elina [12.6K]3 years ago

4 0

Answer: d

Explanation: Did it on Ed2020 got it right

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When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 2.5 x 10-15 m before

Tasya [4]

Explanation:

It is given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, $d=2.5\times 10^{-15}\ m$

Speed of light, $c=3\times 10^{8}\ m\s$

Let t is the time interval required for the strong interaction to occur. The speed is given by :

$c=\dfrac{d}{t}$

$t=\dfrac{d}{c}$

$t=\dfrac{2.5\times 10^{-15}\ m}{3\times 10^{8}\ m/s}$

$t=8.33\times 10^{-24}\ s$

So, the time interval required for the strong interaction to occur is $8.33\times 10^{-24}\ s$ . Hence, this is the required solution.

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3 years ago

A car travels a distance of 100 km. For the first 30 minutes it is driven at a constant speed of 80 km/hr. The motor begins to v

gregori [183]

Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, $d_{\frac{1}{2}}$ .

Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below

$d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}$

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, $d_{\text{remain}}$ , in which the driver reduces the speed to 40km/hr is

$d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}$ .

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by $t_{\text{remian}}$ .

$t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}$ .

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

$\text{speed} \ = \ \displaystyle\frac{\Delta d}{\Delta t} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{(0.5 \ \text{hr} \ + \ 1.5 \ \text{hr})} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{2 \ \text{hr}} \\ \\ \\ \text{speed} \ = \ 50 \ \text{km hr}^{-1}$

Therefore, the average speed of the car is 50 km/hr.

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3 years ago

Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at

Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

$v = \dfrac{2\pi R}{T}$

$v = \dfrac{2\pi\times 8}{4.30}$

v = 11.69 m/s

now, Force does the ring push on her at the top

$- N - m g = \dfrac{-mv^2}{R}$

$N + m g = \dfrac{mv^2}{R}$

$N = \dfrac{mv^2}{R}- m g$

$N = m(\dfrac{v^2}{R}- g)$

$N = 58\times (\dfrac{11.69^2}{8}- 9.8)$

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

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The answer is False give thanks for the answer m8 and happy Halloween

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Direction!

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