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antiseptic1488 [7]
2 years ago
7

If you swim from one point at the edge of the pool to another, along a straight line, what is the longest distance d you can swi

m?
Physics
1 answer:
densk [106]2 years ago
4 0

The longest distance that a person can swim is 5.64 m.

<h3>What is the longest distance?</h3>

We know that the diameter of a circle is a line that is drawn from one point to another in the circle. Now we are told that the pool is circular in nature. That implies that the longest that a person can swim could be obtained form the diameter of the circle.

Given that;

A =  πr^2

A = area of teh circular pool

r  = radius of the pool

r = √A/ π

r = √25/3.142

r = 2.82m

Diameter of the circular pool = 2 r = 2 (2.82 cm) = 5.64 m

Learn more about circle:brainly.com/question/11833983

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Missing parts;

An ad for an above-ground pool states that it is 25 m2. From the ad, you can tell that the pool is a circle. If you swim from one point at the edge of the pool to another, along a straight line, what is the longest distance d you can swim? Express your answer in three significant figures.

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You apply a horizontal force of 25N to push a shopping cart across the parking lot at a constant velocity. a) what is the net fo
AlekseyPX

(a) The net force on the shopping cart is zero.

(b) The the force of friction on the shopping cart is 25 N.

(c) When same force is applied to the shopping cart on a wet surface, it will move faster.

<h3>Net force on the shopping cart</h3>

The net force on the shopping cart is calculated as follows;

F(net) = F - Ff

where;

  • F is the applied force
  • Ff is the frictional force

ma = F - Ff

where;

  • a is acceleration of the cart
  • m is mass of the cart

at a constant velocity, a = 0

0 = F - Ff

F(net)  = 0

F = Ff = 25 N

Net force is zero, and frictional force is equal to applied force.

<h3>On wet surface</h3>

Coefficient of kinetic friction of solid surface is greater than that of wet surface.

Since frictional force limit motion, when the frictional force is smaller, the object tends to move faster.

Thus, the cart will move faster on a wet surface due to decrease in friction.

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4 0
2 years ago
A jet plane lands with a speed of 100 m/s and can
kiruha [24]

Answer:

a) t = 20 [s]

b) Can't land

Explanation:

To solve this problem we must use kinematics equations, it is of great importance to note that when the plane lands it slows down until it reaches rest, ie the final speed will be zero.

a)

v_{f}=v_{i}-(a*t)

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = desacceleration = 5 [m/s^2]

t = time [s]

Note: the negative sign of the equation means that the aircraft slows down as it stops.

0 = 100 - 5*t

5*t = 100

t = 20 [s]

b)

Now we can find the distance using the following kinematics equation.

x -x_{o}=(v_{o}*t)+\frac{1}{2}*a*t^{2}

x - xo = distance [m]

x -xo = (0*20) + (0.5*5*20^2)

x - xo =  1000 [m]

1000 [m] = 1 [km]

And the runaway is 0.8 [km], therefore the jetplane needs 1 [km] to land. So the jetpalne can't land

4 0
3 years ago
The strength of the force of gravity depends on
givi [52]

The masses of the objects and how much distance there is between them

3 0
3 years ago
Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer sec
GaryK [48]

Mary and her younger brother Alex decide to ride the carousel at the State Fair, Mary's and Alex's  angular speed M and tangential speed vM is mathematically given as

Mary's and Alex's  angular speed=1.43

Tangential speed mary=3.22 m/s

Tangential speed alex =2.260m/s

<h3>What is Mary's and Alex's angular speed M and tangential speed vM?</h3>

Generally, the equation for angular speed is mathematically given as

w=2\pi /T\\\\Therefore\\\\w=2\pi/3.9

w = 1.61 rev/see 3.9

Centripetal acc mary = v^2/r

Centripetal acc mary  = w^2r

Centripetal acc mary = w^2x 2m

Centripetal acc. of Alex = w²x L.u

Therefore

\frac{(ac) mary }{(ac) plex}= 1.43

Hence

tang. speed V=Wr

tang. speed of mary = 1.61x2 = 3.22 m/s

tang. speed of Alex: 1.61X1·4 =2.260m/s

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8 0
2 years ago
What is the kinetic energy of a 9.0 kg steelhead if its speed is 16 m/s?
Vesnalui [34]

<u>We are given:</u>

Mass of the Steelhead(m) = 9 kg

Velocity of the Steelhead(v) = 16 m/s

<u>Calculating the Kinetic Energy:</u>

KE = 1/2mv²

replacing the variables

KE = 1/2 * 9 * (16)²

KE = 1152 Joules

8 0
3 years ago
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