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4 years ago

If a 70kg swimmer pushes off a pool wall with a force of 250N. What is her acceleration?

Physics

2 answers:

alexgriva [62]4 years ago

7 0

Since , F = ma 250 = 70 × a a = 25/7 m/s²

Alenkasestr [34]4 years ago

6 0

Newton taught us: Force = (mass) x (acceleration)

Divide each side by (mass) : Acceleration = (force) / (mass) .

The only problem here is: This formula applies when the "Force" is the
only force on the object. When the objects in these school problems are
falling out of airplanes, shot from guns, or being hit by baseball bats, we
routinely ignore the force of air resistance against the object. We're
comfortable with that, maybe because it's become a habit. But now,
we're not so comfortable about ignoring the force of water resistance.

All I can tell you is that if you DO ignore the water resistance, that is,
if the water were not there, her acceleration would be

(250 newtons) / (70 kg) = 3.57 m/s² = about 0.36 g .

But what is it really, in the water ?

If you've spent any substantial amount of time anywhere near competitive
swimmers, then you know that it depends on their position coming off the
wall, what they do with their knees and knuckles, how straight they hold
their body, how deep the texture of their swim-cap is, and how well they've
shaved their legs.

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fgiga [73]

Answer:

Explanation:

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8 0

3 years ago

d. e. Study the given diagram and calculate the following: i. work done by load ii. work done by effort iii. M.A iv. V.R v. effi

il63 [147K]

i. The work done by the load is load x distance moved by load.

ii. The work done by effort is effort applied x distance moved by effort.

iii. The mechanical advantage of the simple machine is Load/effort.

iv. The velocity ratio of the simple machine is 2.

v. The efficiency of the machine is M.A/V.R x 100%.

The work done by the load is the product of the load and the distance through which the load is moved. The magnitude is calculated as follows;

Work done by the load = load x distance moved by load

<h3>Work done by effort</h3>

The work done by the effort is the product of the effort and the distance through which the effort is applied. The magnitude is calculated as follows;

Work done by effort = effort applied x distance moved by effort

<h3>Mechanical advantage of the simple machine</h3>

M.A = Load/Effort

<h3>Velocity ratio of the simple machine</h3>

V.R = distance moved by effort / distance moved by load

V.R = 30 cm/15 cm

V.R = 2

<h3>Efficiency of the machine</h3>

E = (M.A/V.R) x 100%

Thus, the work done by the load is load x distance moved by load.

The work done by effort is effort applied x distance moved by effort.

The mechanical advantage of the simple machine is Load/effort.

The velocity ratio of the simple machine is 2.

The efficiency of the machine is M.A/V.R x 100%.

Learn more about mechanical advantage here: brainly.com/question/25984831

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4 0

2 years ago

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n200080 [17]

I see a pillow

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8 0

3 years ago

PLZ HELP ME ITS MY LAST QUESTION I GOT LIKE 6 MIN LEFT

spin [16.1K]

Answer:

12.7m/s

Explanation:

Given parameters:

Mass of the diver = 77kg

Height = 8.18m

Unknown:

Speed of the diver before hitting the water = ?

Solution:

The speed of the diver before hitting the water is the final velocity.

To solve this problem, we use the expression below;

v² = u² + 2gH

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

H is the height

Now insert the parameters and solve;

v² = 0² + 2 x 9.8 x 8.18

v² = 160.328

v = 12.7m/s

3 0

3 years ago

A laser beam is incident on two slits with a separation of 0.230 mm, and a screen is placed 4.75 m from the slits. If the bright

Mariulka [41]

Answer:

$7.55\times 10^{-7} m$

Explanation:

We are given that

d=0.23 mm= $0.23\times 10^{-3} m$

$1mm=10^{-3} m$

Screen is placed from the slits at distance ,L=4.75 m

The bright interference fringes on the screen are separated by 1.56 cm.

$\Delta y=1.56 cm=1.56\times 10^{-2} m$

1 m=100 cm

We have to find the wavelength of laser light.

We know that

$\Delta y=\frac{\lambda L}{d}$

Substitute the values

$1.56\times 10^{-2}=\frac{\lambda\times 4.75}{0.23\times 10^{-3}}$

$\lambda=\frac{1.56\times 10^{-2}\times 0.23\times 10^{-3}}{4.75}$

$\lambda=7.55\times 10^{-7} m$

4 0

3 years ago