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4 years ago

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Now assume that the stream flows east at 4.0 m/s. draw the vectors v⃗ w, representing the velocity of the stream, and v⃗ tot, re

presenting the velocity of your rowboat relative to the stream bank. be sure to draw both vectors. draw v⃗ tot and v⃗ w starting at the tail and tip of v⃗ still respectively. the location, orientation and length of the vectors will be graded. each vector's length is displayed in meters per second.

1 answer:

ELEN [110]4 years ago

3 0

Remember that the total velocity of the motion is the vector sum of the velocity you would have in still water and the stream. Always place the vectors carefully to be able to come up with an accurate sum vector.

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A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

jeka57 [31]

Answer:

a. A = 0.735 m

b. T = 0.73 s

c. ΔE = 120 J decrease

d. The missing energy has turned into interned energy in the completely inelastic collision

Explanation:

a.

4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax

vmax = 4.0 m/s

¹/₂ * m * v²max = ¹/₂ * k * A²

m * v² = k * A² ⇒ 10 kg * 4 m/s = 100 N/m * A²

A = √1.6 m ² = 1.26 m

At = 2.0 m - 1.26 m = 0.735 m

b.

T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s

T = 2π *√ 10 / 100 *s² = 1.99 s

T = 1.99 s -1.26 s = 0.73 s

c.

E = ¹/₂ * m * v²max =

E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J

E₂ = ¹/₂ * 10 * 4² = 80 J

200 J - 80 J = 120 J decrease

d.

The missing energy has turned into interned energy in the completely inelastic collision

3 0

3 years ago

An engineer has the task of producing an aluminum alloy with a density of 3.0 grams per cubic centimeter. She comes up with the

pochemuha

Answer:

The best option is for the following option m = 15 [g] and V = 5 [cm³]

Explanation:

We have that the density of a body is defined as the ratio of mass to volume.

$Ro =m/V$

where:

Ro = density = 3 [g/cm³]

Now we must determine the densities with each of the given values.

<u>For m = 7 [g] and V = 2.3 [cm³]</u>

$Ro=7/2.3\\Ro=3.04 [g/cm^{3} ]$

<u>For m = 10 [g] and V = 7 [cm³]</u>

<u /> $Ro=10/7\\Ro=1.42[g/cm^{3} ]\\$ <u />

<u>For m = 15 [g] and V = 5 [cm³]</u>

<u /> $Ro=15/5\\Ro=3[g/cm^{3} ]\\$ <u />

<u>For m = 21 [g] and V = 8 [cm³]</u>

<u /> $Ro=21/8\\Ro=2.625[g/cm^{3} ]\\$ <u />

5 0

3 years ago

Heat transfer in liquids or gases that happens due to currents of hot and cold is called

yKpoI14uk [10]

Answer:

Convection

Explanation:

three types of heat transfer

Heat is transfered via solid material (conduction), liquids and gases (convection), and electromagnetical waves (radiation).

3 0

3 years ago

Read 2 more answers

A 62.00-cm guitar string under a tension of 70.000 N has a mass per unit length of 0.10000 g/cm. What is the highest resonant fr

quester [9]

Answer:

17,947.02 Hz

Explanation:

length (L) = 62 cm = 0.62 m

tension (T) = 70 N

mass per unit length (μ) = 0.10000 g/cm = 0.010000 kg/m

maximum frequency = 18,000 Hz

f = $\frac{n}{2L} x \sqrt{\frac{T}{μ}}$

f = $\frac{n}{2 x 0.62} x \sqrt{\frac{70}{0.01} }$

f = n x 67.47

18,000 = n x 67.47

n = 266.8≈ 266

the 267th overtone is the highest overtone that can be heard by this person, and its frequency would be 26 x 67.47 = 17,947.02 Hz

8 0

3 years ago

A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0

3 years ago