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3 years ago

A 50Kg girl jumps off a 5-meter-high diving board. What is her kinetic energy right before she hits the water?

Physics

1 answer:

Kryger [21]3 years ago

6 0

Answer:

2450 J

Explanation:

Given that,

A 50Kg girl jumps off a 5-meter-high diving board.

We need to find the kinetic energy of the girl before she hits the water. At this point the kinetic energy becomes equal to the potential energy such that,

$K=mgh\\\\K=50\times 9.8\times 5\\\\K=2450\ J$

So, her kinetic energy right before she hits the water is equal to 2450 J.

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The elements that do not ordinarily form compounds are

ololo11 [35]

The noble gasses, they are the elements on group 18 of the periodic table and do not usually form compounds. The noble gases are as follows: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og).

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8 0

3 years ago

Read 2 more answers

a block has a volume of 0.9m3 and a density of 4,000kg/m3. whats the force of gravity acting on the block in water?(A)3,528N(B)8

Rina8888 [55]

There is a correct answer to this question, and then there is the one they want you to choose. Which one should I give you first ?

Let's start with the correct answer:

The force of gravity on any object is

   (mass of the object)
times
       (acceleration of gravity on the planet where the object is) .

The acceleration of gravity on Earth is about 9.8 m/s²,
so the force of gravity on this object is

(4000 kg/m³) x (0.9 m³) x (9.8 m/s²)

   = (3,600 kg) x (9.8 m/s²)

   = 35,280 kg-m/s² = 35,280 Newtons.

That's the force of gravity attracting this block and the Earth
toward each other. It makes no difference whether the block
is in your bedroom closet, in the back yard under a pile of mulch,
inside a steel safe resting on a bed of styrofoam peanuts and
slivered almonds, or underwater in the neighbor's pool.
That's the force of gravity on this block, and it's the correct answer
to the question.

It's not one of the choices, though. That's because the question
is poorly written. The person who wrote the question is unclear on
the concepts, and the more you work with the question, the more
unclear and confused YOU'LL become.

When the block is in water, the force of gravity on it doesn't change.
BUT ... there's ANOTHER force on it ... the buoyant force ... acting
upward on it, and canceling part of the force of gravity.

The buoyant force is the weight of the displaced water.
The displaced water is the water that has to get out of the way
when you drop the block in, so the volume of displaced water
is the volume of the block.

-- The volume of the block is (0.9 m³).

-- The density of water is 1000 kg/m³, so the mass of 0.9 m³ of water
is 900 kg.

-- The weight of 900 kg of water is (mass) x (gravity)

   = (900 kg) x (9.8 m/s²)

   = 8,820 Newtons.

When the block is in water, it feels like it's that much LIGHTER,
because that's the force of the water pushing UP on the block.
It's the same reason why your big brother seems so light in the
pool that you can pick him up and carry him.

So how heavy does this block FEEL in water ?

The force of gravity pulling down on it:    35,280 newtons
The force of water pushing up on it: 8,820 newtons
How heavy the block feels (the difference) 26,460 newtons

The question is written so poorly that even THIS number
is not one of the choices.

Again, the thing to realize is that being in the water does NOT
change the force of gravity on anything. It only creates another
force, that acts against gravity.

Just like . . . When you walk up some stairs, how does it happen
that you suddenly move upward, opposite to gravity. Does the force
of gravity acting on you change ? No ! But you use your leg muscles
to create another force in the opposite direction, that works against
gravity, and makes you seem so light that you can actually move up,
opposite to gravity.

7 0

2 years ago

The graph shows the acceleration, velocity, and displacement of a marble attached to a metal spring that has been secured to a t

SpyIntel [72]

<span>D. When velocity at a maximum, acceleration is also at a maximum.

Acceleration is the rate of change of velocity so, it is directly proportional to it, hence, when it will increase velocity will also increase

Hope this helps!</span>

7 0

3 years ago

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ

lesantik [10]

Answer:

The magnitude of the force on positive charges will be $\bf{227.06~N}$ and the magnitude of the force on the negative charge is $\bf{302.7~N}$ .

Explanation:

Given:

The value of the charges, $q = 3.5~\mu C$ .

The length of each side of the triangle, $l = 2.9~cm$ .

Consider a equilateral triangle $\bigtriangleup ABC$ , as shown in the figure. Let two point charges of magnitude $q$ are situated at points $A$ and $B$ and another point charge $-q$ is situated at point $C$ .

The value of the force on the charge at point $A$ due to charge at point $C$ is given by

$F_{CA} = \dfrac{kq^{2}}{l^{2}},~~along~CA$

The value of the force on the charge at point $A$ due to charge at point $B$ is given by

$F_{BA} = \dfrac{kq^{2}}{l^{2}},~~along~BA$

The net resultant force on the charge at point $A$ is given by

$~~~~F_{A} = \sqrt{F_{BA}^{2} + F_{CA}^{2} + 2F_{BA}F_{CA}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{A}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(1)$

The value of the force on the charge at point $B$ due to charge at point $C$ is given by

$F_{CB} = \dfrac{kq^{2}}{l^{2}},~~along~CB$

The value of the force on the charge at point $B$ due to charge at point $A$ is given by

$F_{AB} = \dfrac{kq^{2}}{l^{2}},~~along~AB$

The net resultant force on the charge at point $B$ is given by

$~~~~F_{B} = \sqrt{F_{AB}^{2} + F_{CB}^{2} + 2F_{AB}F_{CB}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{B}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(2)$

The value of the force on the charge at point $C$ due to charge at point $A$ is given by

$F_{AC} = \dfrac{kq^{2}}{l^{2}},~~along~AC$

The value of the force on the charge at point $C$ due to charge at point $B$ is given by

$F_{BC} = \dfrac{kq^{2}}{l^{2}},~~along~BC$

The net resultant force on the charge at point $C$ is given by

$F_{C} = 2F_{BC} \sin 60^{0}~~along~the~line~perpendicular~to~AB\\~~~~~= \dfrac{2kq^{2}}{l^{2}}\sin 60^{0}~~~~~~~~~~~~~~~~~~~~(3)$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (1), we have

$F_{A} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (2), we have

$F_{B} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (3), we have

$F_{C} = \dfrac{2(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}} \sin 60^{0}\\~~~~~= 302.7~N$

8 0

2 years ago

List the planets of the solar system, from closest to farthest from the sun.

kogti [31]

Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune

7 0

3 years ago