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3 years ago

A swimming pool measures 5.0 m long x 4.0 m wide x 3.0 mdeep.

Physics

1 answer:

QveST [7]3 years ago

3 0

Answer:

a) 588,000 N

b) 294000 N

Explanation:

Given that

Density of water = 1000kg/m3

(g) = 9.8m/s2

volume is given as (V)= 5m*4m*3m

a) force will be equal to weight of water

$W = mg = \rho g V = 1000\times 9.8 \times (5*4*3) = 588,000 kg m/s^2$

b) at either end

$P = \frac{F}{A} = P_o + \rho gh$

$dF = PdA$

$dF = \rho g w \int h dh$

$F = \rho g w \frac{h^2}{2}$

$F = \rho g A \frac{h}{2}$ [A = wh]

$F = 1000\times 9.8 \times 5\times 4 \times\frac{3}{2}$

F = 294000 N

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Aisha is sitting on frictionless ice and holding two heavy ski boots. Aisha weighs 637 N, and each boot has a mass of 4.50 kg. A

Studentka2010 [4]

Answer:

-0.73 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum.

In fact, in absence of external forces (the ice is frictionless, so no friction), the total momentum of Aisha + two boots is conserved.

At the beginning, their total momentum is zero, since they are at rest:

$p_i = 0$ (1)

After, their total momentum is:

$p_f = Mv + 2mv'$ (2)

where:

M is Aisha's mass

v is Aisha's velocity relative to the ground

m = 4.50 kg is the mass of each boot

v' is the boot's velocity relative to the ground

We can find:

$M=\frac{W}{g}=\frac{637 N}{9.8 N/kg}=65 kg$ is Aisha's mass (where W = 637 N was her weight)

v' can be rewritten as:

$v'=v+6$

because 6 m/s is the velocity of the boots relative to her, while v' is their velocity relative to the ground.

Substituting and combining (1) and (2) we find:

$0=Mv+2m(v+6)\\0=Mv+2mv+12m\\v=\frac{-12m}{M+2m}=\frac{-12(4.50)}{65+2(4.50)}=-0.73 m/s$

and the negative sign indicates that the direction is opposite to that of the boots.

8 0

3 years ago

A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav

gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

$K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2$ (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

$v^2=v_o^2+2gy$

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}$

With this value you can find the spring constant k from the equation (1):

$mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}$

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

$W=F_e\\\\mg=kx$

you solve the last expression for x:

$x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m$

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}$

Next, you calculate x from the equation (1):

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m$

The net fire is stretched 2.72 m

5 0

3 years ago

Particle A of charge 2.79 10-4 C is at the origin, particle B of charge -5.64 10-4 C is at (4.00 m, 0), and particle C of charge

kirill [66]

Answer:

a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N g) 16.9 N

h) 24.3 N θ = 44.2º

Explanation:

a) As the electric force is exerted along the line that joins the charges, due to any of the charges A or C has non-zero x-coordinates, the force has no x components either.

So, Fcax = 0

b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.

Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²

Fyca = 29. 9 N

c) In order to get the magnitude of the force exerted by B on C, we need to know first the distance between both charges:

rbc² = (3.00 m)² + (4.00m)² = 25.0 m²

⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²

⇒ Fbc = 21.7 N

d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:

Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:

cos θ = x/r = 4.00 / 5.00 m =

Fcbx = 21.7*(-0.8) = -17.4 N

e) The y component can be calculated in the same way, projecting the force over the y-axis, as follows:

Fcby = Fcb* sin (-θ) = 21.7* (-3.00/5.00) = -13.0 N

f) The sum of both x components gives :

Fcx = 0 + (-17.4 N) = -17.4 N

g) The sum of both y components gives :

Fcy = 29.9 N + (-13.0 N) = 16.9 N

h) The magnitude of the resultant electric force acting on C, can be found just applying Pythagorean Theorem, as follows:

Fc = √(Fcx)²+(Fcy)² = (17.4)² + (16.9)² $\sqrt{((17.4)^{2} +(16.9)^{2}}$ = 24.3 N

The angle from the horizontal can be found as follows:

Ф = arc tg (16.9 / 17.4) = 44.2º

4 0

3 years ago

Who is the most genius scientist in the world

aliya0001 [1]

Answer:

Marie Curie

Explanation:

I hope to see you helped :D?

6 0

2 years ago

Darwin is sitting on top of a 10 m tree and decided to drop a 3 kg baseball to the ground. What is the velocity when the PE turn

Nimfa-mama [501]

The AMOUNT of energy the ball has doesn't change. It's 294 joules in Darwin's hand, and it's still 294 joules when the ball hits the ground. It's all PE before he let's it go, and it steadily changes from PE to KE all the way down.

It BEGINS to turn into KE immediately, when Darwin lets go of the ball, and it starts to fall.

More and more PE turns into KE as the ball falls, all the way down.

When the ball hits the ground, it has no more PE left. All of its mechanical energy is then KE.

8 0

3 years ago