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Licemer1 [7]
3 years ago
6

A 75.5 kg diver drops from a diving board 10.0 m above the waters surface. Find the divers speed just before he strikes the wate

r.
Physics
1 answer:
jenyasd209 [6]3 years ago
8 0

Answer:

<em>v=14 m/s</em>

Explanation:

<u>Mechanical Energy </u>

The kinetic energy of a body (K) is the capacity of doing work due to its speed. It can be expressed as

\displaystyle K=\frac{mv^2}{2}

The potential energy (U) is the capacity of doing work due to its height respect to a certain reference level.

U=mgh

The  mechanical energy is the sum of both

\displaystyle E_m=\frac{mv^2}{2}+mgh

The principle of conservation of mechanical energy states it must remain the same if no external force is acting on it. The diver drops from the diving board, which means its initial speed is zero (and so its initial kinetic energy). Thus, the mechanical energy at the jumping time is

\displaystyle E_m=mgh=(75)(10)(9.8)=7350\ J

When the diver is about to get into the water, his height reaches zero and the speed is at maximum. All the potential energy became kinetic energy, so

\displaystyle \frac{mv^2}{2}=7350\ J

Rearranging

\displaystyle v^2=\frac{2(7350)}{75}=196

v=\sqrt{196}=14\ m/s

The final speed of the diver is

\boxed{v=14\ m/s}

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In the diagram, q1= -2.60*10^-9 C and
Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

E_{net}=90.37\: N/c

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}

Where:

  • k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
  • q1 is the first charge
  • d1 is the distance from q1 to P

|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}

|E_{1}|=80.84\: N/C

And E2 will be:

|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}

|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}

|E_{2}|=40.39\: N/C

Finally, we need to use the  Pythagoras theorem to find the magnitude of the net electric field.

E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}

E_{net}=\sqrt{80.84^{2}+40.39^{2}}

E_{net}=90.37\: N/c

I hope it helps you!

7 0
3 years ago
WHO WANts TO HAVE SOME GLIZZY ACTION
Lilit [14]
Bruh huh.............
6 0
2 years ago
A cat climbs 10 m directly up a tree.
Harman [31]
 <span>there is no horizontal displacement if he went straight up 

straight up means vertical, so his vertical displacment is 20 m</span>
6 0
3 years ago
Please help I will mark you brainliest
Radda [10]

I believe the answer is a

7 0
3 years ago
Read 2 more answers
Find the speed of light in each of the following materials. (a) gallium phosphide m/s (b) carbon disulfide m/s (c) benzene
Oksanka [162]

Explanation:

We need to calculate the speed of light in each materials

(I). Gallium phosphide,

The index of refraction of Gallium phosphide is 3.50

Using formula of speed of light

v=\dfrac{c}{\mu}....(I)

Where, \mu = index of refraction

c = speed of light

Put the value into the formula

v=\dfrac{3\times10^{8}}{3.50}

v=8.6\times10^{7}\ m/s

(II) Carbon disulfide,

The index of refraction of Gallium phosphide is 1.63

Put the value in the equation (I)

v=\dfrac{3\times10^{8}}{1.63}

v=1.8\times10^{8}\ m/s

(III). Benzene,

The index of refraction of Gallium phosphide is 1.50

Put the value in the equation (I)

v=\dfrac{3\times10^{8}}{1.50}

v=2\times10^{8}\ m/s

Hence, This is the required solution.

7 0
3 years ago
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