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3 years ago

7

A skipper on a boat notices wave crests passing the anchor chain every 6.0 seconds. The skipper estimates the distance between c

rests at 30.0
m. What is the speed of the water waves?

1 answer:

kow [346]3 years ago

7 0

Explanation:

given Time Period = 5 Sec

Amplitude = 1m, Wave length = 15 m

using time period = 1/ frequency , Frequency = 1/5 cycles per sec = 0.2 sec = 12 per min

wave speed= Frequency * Wave length

speed = 0.2*15 = 3m/s

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What is the mass of the object if it has a density of 657 g/mL and a volume of 32 mL?<br> Show work!

Luden [163]

Answer:

The answer is 21024g/mL

Explanation:

Multiply 657 by 32:

657

× 32

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

21024

↳ 21024 g/mL

3 0

3 years ago

A ball of radius r rolls on the inside of a track of radius R. If the ball starts from rest at the vertical edge of the track, f

meriva

Answer:

$v=\sqrt{\dfrac{10g(R-r)}{7}}$

Explanation:

Given that

Radius of track = R

Radius of ball = r

The ball can be treated as solid sphere, so

The moment of inertia of ball

$I=\dfrac{2}{5}mr^2$

When the ball reach at the lowest position then it will have both angular and linear speed.

Condition for rolling without slipping v= ωr

Form energy conservation

$mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

v= ωr

$I=\dfrac{2}{5}mr^2$

$mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2$

$mg(R-r)=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mv^2$

$2mg(R-r)=mv^2+\dfrac{2}{5}mv^2$

$2g(R-r)=\dfrac{7}{5}v^2$

$v=\sqrt{\dfrac{10g(R-r)}{7}}$

3 0

3 years ago

What is the initial velocity of a go-kart traveling at a uniform acceleration of 0.5 m/s^2 for 5s as it slows down to a stop?

Arlecino [84]

The go-kart's velocity $v$ after time $t$ is given by

$v=v_0+at$

where $v_0$ is its initial velocity and $a$ is its acceleration. After $t=5\,\mathrm s$ , the go-kart stops completely, so

$0\,\dfrac{\mathrm m}{\mathrm s}=v_0+\left(-0.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(5\,\mathrm s)\implies v_0=2.5\,\dfrac{\mathrm m}{\mathrm s}$

where $a$ because we know the go-kart is slowing down.

7 0

3 years ago

You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 6 m at your feet, then a

Zigmanuir [339]

Answer:

4.5 m/s

Explanation:

The rock must barely clear the shelf below, this means that the horizontal distance covered must be

$d_x = 5 m$

while the vertical distance covered must be

$d_y = 6 m$

The rock is thrown horizontally with velocity $v_x$ , so we can rewrite the horizontal distance as

$d_x = v_x t$

where t is the time of flight. Re-arranging the equation,

$t=\frac{d_x}{v_x}$ (1)

The vertical distance covered instead is

$d_y = \frac{1}{2}gt^2$

where we omit the term $ut$ since the initial vertical velocity is zero. From this equation,

$t=\sqrt{\frac{2d_y}{g}}$ (2)

Equating (1) and (2), we can solve the equation to find $v_x$ :

$\frac{d_x}{v_x}=\sqrt{\frac{2d_y}{g}}\\\frac{d_x^2}{v_x^2}=\frac{2d_y}{g}\\v_x = d_x \sqrt{\frac{g}{2d_y}}=5\sqrt{\frac{9.8}{2(6)}}=4.5 m/s$

6 0

3 years ago

s) A body of mass 2 kilograms moves on a circle of radius 3 meters, making one revolution every 5 seconds. Find the magnitude of

jeyben [28]

Answer:

Centripetal force acting on the body = 9.47 N

Explanation:

Mass of body, m = 2 kg

Radius, r = 3 m

It makes one revolution in 5 seconds.

Period, T = 5 s

$\texttt{Angular velocity, }\omega =\frac{2\pi }{T}=\frac{2\pi }{5}=1.256rad/s$

Centripetal force, F = mrω²

F = 2 x 3 x 1.256² = 9.47 N

Centripetal force acting on the body = 9.47 N

8 0

3 years ago