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3 years ago

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A rowboat heads directly across a river at a speed of 3 m/s. Select the correct equations that show that if the river flows at 4

m/s, the speed of the boat relative to the riverbank is 5 m/s.

1 answer:

olasank [31]3 years ago

6 0

Answer:

Explanation:

Check attachment for solution

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A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra

blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

$y=\sqrt{1+x^3}$

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Now at (2,3)

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s$

7 0

3 years ago

Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an

8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 $m/s^2$ where as radial the component of acceleration is 8.77 $m/s^2$

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

$r=e^u$

So the transverse component of acceleration are given as

$a_{\theta}=(ru''+2r'u')\\$

Here

$r=e^u\\r=e^{\pi/4}\\r=2.1932 m$

$r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m$

So

$a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\$

The transverse component of acceleration is 26.32 $m/s^2$

The radial component is given as

$a_r=r''-r\theta'^2$

Here

$r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m$

So

$a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2$

The radial component of acceleration is 8.77 $m/s^2$

6 0

3 years ago

A 1150 kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 7.69 m before contacting the beam,

Natasha_Volkova [10]

Answer:

the average force 11226 N

Explanation:

Let's analyze the problem we are asked for the average force, during the crash, we can find this from the impulse-momentum equation, but this equation needs the speeds and times of the crash that we could look for by kinematics.

Let's start looking for the stack speeds, it has a free fall, from rest (Vo=0)

Vf² = Vo² - 2gY

Vf² = 0 - 2 9.8 7.69 = 150.7

Vf = 12.3 m / s

This is the speed that the battery likes when it touches the beam. They also give us the distance it travels before stopping, let's calculate the time

Vf = Vo - g t

0 = Vo - g t

t = Vo / g

t = 12.3 / 9.8

t = 1.26 s

This is the time to stop

Now let's use the equation that relates the impulse to the amount of movement

I = Δp

F t = pf-po

The amount of final movement is zero because the system stops

F = - po / t

F = - mv / t

F = - 1150 12.3 / 1.26

F = -11226 N

This is the average force exerted by the stack on the vean

7 0

3 years ago

Why society might initially reject a new scientific theory?

Irina18 [472]

I was about to say: because people generally get comfortable with
what they think they know, and don't like the discomfort of being told
that they have to change something they're comfortable with.

But then I thought about it a little bit more, and I have a different answer.

"Society" might initially reject a new scientific theory, because 'society'
is totally unequipped to render judgement of any kind regarding any
development in Science.

First of all, 'Society' is a thing that's made of a bunch of people, so it's
inherently unequipped to deal with scientific news. Anything that 'Society'
decides has a lot of the mob psychology in it, and a public opinion poll or
a popularity contest are terrible ways to evaluate a scientific discovery.

Second, let's face it. The main ingredient that comprises 'Society' ... people ...
are generally uneducated, unknowledgeable, unqualified, and clueless in the
substance, the history, and the methods of scientific inquiry and reporting.

There may be very good reasons that some particular a new scientific theory
should be rejected, or at least seriously questioned. But believe me, 'Society'
doesn't have them.

That's pretty much why.

6 0

3 years ago

10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when

SVEN [57.7K]

Answer:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

Explanation:

For this case we can use the second law of Newton given by:

$\sum F = ma$

The friction force on this case is defined as :

$F_f = \mu_k N = \mu_k mg$

Where N represent the normal force, $\mu_k$ the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

$F_f = ma$

Replacing the friction force we got:

$\mu_k mg = ma$

We can cancel the mass and we have:

$a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}$

And now we can use the following kinematic formula in order to find the distance travelled:

$v^2_f = v^2_i - 2ad$

Assuming the final velocity is 0 we can find the distance like this:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

5 0

3 years ago