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Bad White [126]
2 years ago
9

3. A fighter jet covers a distance of 895m while accelerating from 22m/s to 35m/s. How long (in

Physics
1 answer:
larisa86 [58]2 years ago
6 0

Answer:

31.404 seconds

Explanation:

To answer this equation, SUVAT is your best option utilizing and rearranging the known values to solve for the unknown.

here we have the values for

s=895

u=22

v=35

t= the unknown value

in this instant the equation s=0.5 x (u+v)t is the best equation to use

so we sub in the known values

895=0.5 x (22+35)t

rearrange to solve for t

895=28.5t

895/28.5=t

t=31.404 seconds (rounded to 3 decimal places)

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A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h
hram777 [196]

Answer:

v_{2}=3.5 m/s

Explanation:

Using the conservation of energy we have:

\frac{1}{2}mv^{2}=mgh

Let's solve it for v:

v=\sqrt{2gh}

So the speed at the lowest point is v=7 m/s

Now, using the conservation of momentum we have:

m_{1}v_{1}=m_{2}v_{2}

v_{2}=\frac{1*7}{2}

Therefore the speed of the block after the collision is v_{2}=3.5 m/s

I hope it helps you!

       

8 0
3 years ago
What is the oxidation state of a hydrogen atom bound to an iron atom.?
tresset_1 [31]
the answer is rust so the answer is rust
8 0
3 years ago
Which of the following materials is attracted to a magnet?
Butoxors [25]

Answer:

A. aluminum

Explanation:

because it's metal

4 0
2 years ago
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Surface currents and waves are powered by what
Allisa [31]
It is powered by the Earth's rotation and the moon gives a little boost.
4 0
3 years ago
Positive Charge is distributed along the entire x axis with a uniform density 12 nC/m. A proton is placed at a position of 1.00
lions [1.4K]

Answer:

b.  \Delta KE = 390 eV

Explanation:

As we know that the electric field due to infinite line charge is given as

E =\frac{\lambda}{2\pi \epsilon_0 r}

here we can find potential difference between two points using the relation

\Delta V = \int E.dr

now we have

\Delta V = \int(\frac{\lambda}{2\pi \epsilon_0 r}).dr

now we have

\Delta V = \frac{\lambda}{2\pi \epsilon_0}ln(\frac{r_2}{r_1})

now plug in all values in it

\Delta V = \frac{12\times 10^{-9}}{2\pi \epsilon_0}ln(\frac{1+5}{1})

\Delta V = 216ln6 = 387 V

now we know by energy conservation

\Delta KE = q\Delta V

\Delta KE = (e)(387V) = 387 eV

3 0
3 years ago
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