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faust18 [17]
3 years ago
9

A dog walking to the right at 1.5\,\dfrac{\text m}{\text s}1.5sm​1, point, 5, space, start fraction, m, divided by, s, end fract

ion spies a cat ahead, and begins chasing the cat with a constant acceleration of 12\,\dfrac{\text m}{\text s^2}12s2m​12, space, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction.
What is the velocity of the dog after running for 3.0\,\text m3.0m3, point, 0, space, m?

​
Physics
1 answer:
Tatiana [17]3 years ago
6 0

Answer:

8.6 m/s

Explanation:

We can find the final velocity of the dog by using the following SUVAT equation:

v^2-u^2=2ad

where

u is the initial velocity

a is the acceleration

d is the distance covered

For the dog in the problem, we have

u = 1.5 m/s

a = 12 m/s^2

And the distance covered is

d = 3.0 m

Therefore, we can re-arrange the equation to find the final velocity, v:

v=\sqrt{u^2+2ad}=\sqrt{1.5^2+2(12)(3.0)}=8.6 m/s

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And on the moon, the with is 595/6 N
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A proton and an electron are held in place on the x axis. The proton is at x = -d, while the electron is at x = +d. They are rel
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6 0
3 years ago
A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the f
hram777 [196]

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

h=ut+\frac{1}{2}gt^2

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

h=ut+\frac{1}{2}gt^2

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2    .... (2)

By equation the equation (1) and (2), we get

41.7=1.12 u +4.9 \times 1.12^{2}

u = 31.75 m/s

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Determine the density of a rectangular piece of concrete that measures 3.7 cm by 2.1 cm by 5.8 cm and has a mass of 43.8 grams.
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It is customary to work in SI units.

Calculate the volume of the concrete.
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Answer: 971.9 kg/m³
5 0
3 years ago
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