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2 years ago

7

A transverse wave on a string has an amplitude A. A tiny spot on the string is colored red. As one cycle of the wave passes by,

what is the total distance traveled by the red spot?a.) 4Ab.) 1/2 Ac.) 1/4 Ad.) Ae.) 2A

1 answer:

DiKsa [7]2 years ago

6 0

Since it's a transverse wave, a particle on the string moves left and right as the wave passes by, but the particle doesn't travel forward or backward at all.

So the little red dot moves 'A' to the left, then 'A' back to the center, then 'A' to the right, then 'A' back to the center again.

All together, the red dot moves a total distance of <em>4A . (choice 'a')</em>

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A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea

Mice21 [21]

Answer:

$v = 3369.2 m/s$

Explanation:

As we know that Beacon is rotating with angular speed

$f = 10 rev/min$

so we have

$\omega = 2\pi f$

$\omega = 2\pi(\frac{10}{60})$

$\omega = 1.047 rad/s$

now we know that

$v = r \omega$

here we will have

$r = 2 miles = 2(1609 m)$

$r = 3218 m$

so we have

$v = 3218(1.047)$

$v = 3369.2 m/s$

6 0

1 year ago

A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car p

DedPeter [7]

Answer:

a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

Explanation:

<em><u>The knowable variables are </u></em>

$d_{t}=25m$

$t_{1}=1.3 s$

$t_{2}=3.9 s$

$t_{3}=5.5 s$

Since the three traffic signs are <u>equally spaced</u>, the <u>distance between each sign is $\frac{25}{3} m$ </u>

a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

Since we know the velocity in two points and the time the car takes to pass the traffic signs

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

6 0

1 year ago

two people, each with a mass of 70 kg, are wearing inline skates and are holding opposite ends of a 15m rope. One person pulls f

Zina [86]

Answer:

7.75 s

Explanation:

Newton's second law:

∑F = ma

35 N = (70 kg) a

a = 0.5 m/s²

Given v₀ = 0 m/s and Δx = 15 m:

Δx = v₀ t + ½ at²

(15 m) = (0 m/s) t + ½ (0.5 m/s²) t²

t = 7.75 s

5 0

2 years ago

A distance of 2.00 mm separates two objects of equal mass. If the gravitational force between them is 0.0104 N, find the mass of

aleksklad [387]

Given the distance r = 2/1000 m, the force between them F = 0.0104 N, the mass of the two object can be calculated using formula:

F = G(m1m2)/r^2 since the mass are equal F = G (m^2)/r^2

And where G = is the gravitational constant (6.67E-11 m3 s-2 kg-1)

The mass of the two objects are 24.96 kg

6 0

1 year ago

A vehicle has an initial velocity of v0 when a tree falls on the roadway a distance xf in front of the vehicle. The driver has a

Korvikt [17]

Answer:

$v^2=v_o^2-2\times a\times (v_o.t)$

Explanation:

Given:

Initial velocity of the vehicle, $v_o$

distance between the car and the tree, $x_f$

time taken to respond to the situation, $t$

acceleration of the car after braking, $a$

Using equation of motion:

$v^2=u^2+2a.s$ ..............(1)

where:

$v=$ final velocity of the car when it hits the tree

$u=$ initial velocity of the car when the tree falls

$a=$ acceleration after the brakes are applied

$s=$ distance between the tree and the car after the brakes are applied.

$s=v_o\times t$

Now for this situation the eq. (1) becomes:

$v^2=v_o^2-2\times a\times (v_o.t)$ (negative sign is for the deceleration after the brake is applied to the car.)

5 0

1 year ago