In which type of wave do the particles in a medium move perpendicular to the direction that the wave moves?

What forms of technology are scientists using to study El Niño? need help fast!?!?!?!?

Natali5045456 [20]

Noaa. The buoys transmit some of the data on a daily basis to NOAA through a satellite in space.

6 0

3 years ago

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Work done= ________ transferred

mamaluj [8]

Answer:

Work done= Energy transferred

Explanation:

Work is the transfer of energy. In physics we say that work is done on an object when you transfer energy to that object. If you put energy into an object, then you do work on that object (mass).

4 0

3 years ago

A 15g bullet is fired horizontally into a 3kg block of wood suspended by a long cord. Assume that the bullet remains in the bloc

Varvara68 [4.7K]

Answer:

261.3 m/s

Explanation:

Mass of bullet=m=15 g= $\frac{15}{1000}=0.015 kg$

1 kg=1000g

Mass of block=M=3 kg

d=0.086 m

Total mass =M+m=3+0.015=3.015 kg

K.E at the time strike=Gravitational potential energy at the end of swing

$\frac{1}{2}(m+M)^2V^2=(m+M)gh$

Using g= $9.8m/s^2$

Substitute the values

$\frac{1}{2}(3.015)V^2=3.015\times 9.8\times 0.086$

$V^2=\frac{2\times 3.015\times 9.8\times 0.086}{3.015}$

$V=\sqrt{2\times 3.015\times 9.8\times 0.086}}$

$V=1.3m/s$

Velocity after collision=V=1.3 m/s

Velocity of block=v'=0

Using conservation law of momentum

$mv+Mv'=(m+M)V$

Using the formula

$0.015v+3(0)=3.015(1.3)$

$0.015v=3.015(1.3)$

$v^2=\frac{3.015(1.3)}{0.015}=261.3$

$v=261.3 m/s$

5 0

3 years ago

A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri

exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

a)

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

$U_i +K_i = U_f + K_f$

where

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at halfway

$K_f$ is the final kinetic energy, at halfway

The equation can be rewritten as

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m = 2.7 kg is the mass of the block

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

$h_i = 0.54$ is the initial height

u = 0 is the initial speed

$h_f = 0.25 m$ is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

$v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s$

b)

The total mechanical energy of the block can be calculated from the initial conditions, and it is

$E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J$

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

$E=E_e = \frac{1}{2}kx^2$

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

$x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m$

c)

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

$E=E_e = 14.3 J$

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

$E=U_f = mg h_f$

where $h_f$ is the maximum height reached. Solving for this quantity, we find

$h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m$

which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.

Learn more about kinetic and potential energy:

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#LearnwithBrainly

5 0

3 years ago

A spy camera is said to be able to read the numbers on a car's license plate. If the numbers on the plate are 4.30 cm apart, and

Maslowich

Answer:

D = 2.38 m

Explanation:

This exercise is a diffraction problem where we must be able to separate the license plate numbers, so we must use a criterion to know when two light sources are separated, let's use the Rayleigh criterion, according to this criterion two light sources are separated if The maximum diffraction of a point coincides with the first minimum of the second point, so we can use the diffraction equation for a slit

a sin θ = m λ

Where the first minimum occurs for m = 1, as in these experiments the angle is very small, we can approximate the sine to the angle

θ = λ / a

Also when we use a circular aperture instead of slits, we must use polar coordinates, which introduce a numerical constant

θ = 1.22 λ / D

Where D is the circular tightness

Let's apply this equation to our case

D = 1.22 λ / θ

To calculate the angles let's use trigonometry

tan θ = y / x

θ = tan⁻¹ y / x

θ = tan⁻¹ (4.30 10⁻² / 140 10³)

θ = tan⁻¹ (3.07 10⁻⁷)

θ = 3.07 10⁻⁷ rad

Let's calculate

D = 1.22 600 10⁻⁹ / 3.07 10⁻⁷

D = 2.38 m

4 0

3 years ago

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