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lesantik [10]
3 years ago
10

A transverse wave is propagated in a string stretched along the x-axis. The equation of the wave, in SI units, is given by:

Physics
1 answer:
irina [24]3 years ago
6 0

Answer:

The wave speed in SI units, is closest to - 2.7 m/s

Explanation:

given information:

y = 0.005 cos [π(38t - 14x)]

we know that the wave function is

y =  A cos (kx ± ωt)

where

A = amplitude

w = angular velocity

k = wave number

thus,

y = 0.005 cos [(14πx - 38πt)]

ω = - 38π

k = 14π

to find the speed v, we use the equation below

v = ω/k

so,

v = - 38π/14π

  = - 2.7 m/s

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A monument has a height of 348 ft, 8 in. Express this height in meters. Answer in units of m.
tensa zangetsu [6.8K]

Answer:

The height of mountain in meter will be 106.2732 m

Explanation:

We have given height of mountain = 348 ft,8 in

We know that 1 feet = 0.3048 meter

So 348 feet =348\times 0.3048=106.07meter

And we know that 1 inch = 0.0254 meter

So 8 inch 8\times 0.0254=0.2032m

So the total height of mountain in meter = 106.07+0.2032 = 106.2732 m

The height of mountain in meter will be 106.2732 m

4 0
3 years ago
Felix expends 100 W of power to clomb 10 meters in 20 seconds how much force does he exert
yulyashka [42]

Answer:

200 N

Explanation:

Power = work / time

Work = force × distance

Therefore:

Power = force × distance / time

100 W = F × 10 m / 20 s

F = 200 N

He exerted 200 N.

3 0
3 years ago
Read 2 more answers
Interactive Solution 6.39 presents a model for solving this problem. A slingshot fires a pebble from the top of a building at a
mariarad [96]

(a) 29.8 m/s

To solve this problem, we start by analyze the vertical motion first. This is a free fall motion, so we can use the following suvat equation:

v_y^2 - u_y^2 = 2as

where, taking upward as positive direction:

v_y is the final vertical velocity

u_y = 0 is the initial vertical velocity (zero because the pebble is launched horizontally)

a=g=-9.8 m/s^2 is the acceleration of gravity

s = -25.0 m is the displacement

Solving for vy,

v_y = \sqrt{u^2+2as}=\sqrt{0+2(-9.8)(-25)}=-22.1 m/s (downward, so we take the negative solution)

The pebble also have a horizontal component of the velocity, which remains constant during the whole motion, so it is

v_x = 20.0 m/s

So, the final speed of the pebble as it strikes the ground is

v=\sqrt{v_x^2+v_y^2}=\sqrt{20.0^2+(-22.1)^2}=29.8 m/s

(b) 29.8 m/s

In this case, the pebble is launched straight up, so its initial vertical velocity is

u_y = 20.0 m/s

So we can find the final vertical velocity using the same suvat equation as before:

v_y^2 - u_y^2 = 2as

v_y = \sqrt{u^2+2as}=\sqrt{(20.0)^2+2(-9.8)(-25)}=-29.8 m/s (downward, so we take the negative solution)

The horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

(c) 29.8 m/s

This case is similarly to the previous one: the only difference here is that the pebble is launched straight down instead than up, therefore

u_y = -20.0 m/s

Using again the same suvat equation:

v_y^2 - u_y^2 = 2as

v_y = \sqrt{u^2+2as}=\sqrt{(-20.0)^2+2(-9.8)(-25)}=-29.8 m/s (downward, so we take the negative solution)

As before, the horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

We notice that the final value of the speed is always the same in all the three parts, so it does not depend on the direction of launching. This is due to the law of conservation of energy: in fact, the initial mechanical energy of the pebble (kinetic+potential) is the same in all three cases (because the height h does not change, and the speed v does not change either), and the kinetic energy gained during the fall is also the same (since the pebble falls the same distance in all 3 cases), therefore the final speed must also be the same.

7 0
3 years ago
Which of the following can be the first step to take in organizing data
Natalka [10]

the first step would be gathering information , and making sure its correct

5 0
3 years ago
-. A 2kg cart moving to the right at 5m/s collides with an 8kg cart at rest. As a
bulgar [2K]

Answer:

<em>The velocity of the carts after the event is 1 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

Since the total momentum is conserved, then:

P = P'

In a system of two masses, the equation simplifies to:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

The m1=2 kg cart is moving to the right at v1=5 m/s. It collides with an m2= 8 kg cart at rest (v2=0). Knowing they stick together after the collision, the common speed is:

\displaystyle v'=\frac{2*5+8*0}{2+8}=\frac{10}{10}=1

The velocity of the carts after the event is 1 m/s

3 0
2 years ago
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