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3 years ago

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if a swimmer is traveling at a constant speed of 0.85 m/s how long would it take to swim the length of 50 meter olymic sized poo

l

1 answer:

VARVARA [1.3K]3 years ago

7 0

Answer:

58.82 Seconds

Explanation:

50m/0.85m/s=58.82s

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How is the levitation of the SCMagLev train achieved? Just write a sentence.

Harman [31]

Answer:

<h2>The SCMaglev (superconducting maglev, formerly called the MLU) is a magnetic levitation (maglev) railway system developed by Central Japan Railway Company (JR Central) and the Railway Technical Research Institute</h2>

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2 years ago

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A projectile is launched from ground level at angle u and speed v0 into a headwind that causes a constant horizontal acceleratio

ale4655 [162]

Answer:

Explanation:

Given

Launch angle =u

Initial Speed is $v_0$

Horizontal acceleration is $a_x=a$

At maximum height velocity is zero therefore

$v_f=v_i-gt$

$0=v_0\sin u-gt$

$t=\frac{v_0\sin u}{g}$

Total time of flight $T=2t=\frac{2v_0\sin u}{g}$

During this time horizontal range is

$R=v_o\cos u\cdot 2t-\frac{a(2t)^2}{2}$

$R=\frac{2v_0^2\sin u\cos u}{g}-\frac{2av_0^2\sin ^u}{g^2}$

For maximum range $\frac{\mathrm{d} R}{\mathrm{d} u}=0$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2\cos 2u}{g}-\frac{4av_0^2\sin u\cos u}{g^2}$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0$

$\tan 2u=\frac{g}{a}$

$u=\frac{1}{2}tan ^{-1}\frac{g}{a}$

(b)If a =10% g

$a=0.1g$

thus $u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}$

$u=42.14^{\circ}$

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3 years ago

English or metric <br> What is 16in to cm

il63 [147K]

40.64 cm is the answer to your question.

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3 years ago

A mass of 3.0 kg rests on a smooth surface inclined 34° above the horizontal. It is kept from sliding down the plane by a spring

azamat

Answer:

The spring stretched by x = 13.7 cm

Explanation:

Given data

Mass = 3 kg

k = 120 $\frac{N}{m}$

Angle $\theta$ = 34°

From the free body diagram

Force acting on the box = mg sin $\theta$

⇒ F = 3 × 9.81 × $\sin34$

⇒ F = 16.45 N ------- (1)

Since box is attached with the spring so a spring force also acts on the box.

$F_{sp}$ = k x

$F_{sp}$ = 120 $x$ -------- (2)

The net force acting on the body is given by

$F_{net} = ma$

Since acceleration of the box is zero so

$F_{net} = 0$

$F - F_{sp} = 0$

$F = F_{sp}$

Put the values from equation (1) & (2) we get

16.45 = 120 $x$

x = 0.137 m

x = 13.7 cm

Therefore the spring stretched by x = 13.7 cm

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3 years ago

A string is stretched between a fixed support and a pulley a distance 111 cm apart. The tension on the string is controlled by a

kvv77 [185]

Answer:

88.8 m/s= Speed of wave propagation in the required mode.(3 loops)

Explanation:

When there are 3 loops.

the total length = L = 3 λ /2

⇒ λ = 2 L / 3 = 2 ( 1.11 ) / 3 = 0.74 m

Velocity = v = f λ = (120)(0.74) = 88.8 m/s

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3 years ago