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3 years ago

What is the formula for finding the frequency in longitudinal waves and also transverse waves? Thank you!

Physics

2 answers:

sweet [91]3 years ago

5 0

Frequency of any wave =

(the SPEED of the wave)/(wavelength)

insens350 [35]3 years ago

3 0

The frequency of a wave is the number of wavelengths per second. The period of a longitudinal wave is the time taken by the wave to move one wavelength. As for transverse waves, the symbol T is used to represent period and period is measured in seconds

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Work output of a large machine in a factory is 89,000 joules, and it’s input is 102,000 joules. Work output of a similar machine

mojhsa [17]

(89000/102000)×100
=87.25%

(92000/104000)×100
=88.46%

efficiency is (output/input)×100
if u get confused which way input and output should go, remember the smaller value is always output and it's above in the fraction, then only it's possible to get a efficiency lower than 100.

7 0

3 years ago

Read 2 more answers

A 1.0-kg block moving to the right at speed 3.0 m/s collides with an identical block also moving to the right at a speed 1.0 m/s

____ [38]

Answer:

Speed of both blocks after collision is 2 m/s

Explanation:

It is given that,

Mass of both blocks, m₁ = m₂ = 1 kg

Velocity of first block, u₁ = 3 m/s

Velocity of other block, u₂ = 1 m/s

Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$v=\dfrac{1\ kg\times 3\ m/s+1\ kg\times 1\ m/s}{2\ kg}$

v = 2 m/s

Hence, their speed after collision is 2 m/s.

7 0

3 years ago

The vector product of vectors A⃗ and B⃗ has magnitude 12.0 m2 and is in the +z-direction.Vector A⃗ has magnitude 4.0 m and is in

g100num [7]

Answer:

θ=180°

Explanation:

The problem says that the vector product of A and B is in the +z-direction, and that the vector A is in the -x-direction. Since vector B has no x-component, and is perpendicular to the z-axis (as A and B are both perpendicular to their vector product), vector B has to be in the y-axis.

Using the right hand rule for vector product, we can test the two possible cases:

If vector B is in the +y-axis, the product AxB should be in the -z-axis. Since it is in the +z-axis, this is not correct.

If vector B is in the -y-axis, the product AxB should be in the +z-axis. This is the correct option.

Now, the problem says that the angle θ is measured from the +y-direction to the +z-direction. This means that the -y-direction has an angle of 180° (half turn).

8 0

3 years ago

If a ball is tossed straight up into the air, at what position is its potential energy the greatest? Question 2 options: a) when

yarga [219]

Q1: At the highest point

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3 years ago

Two racing boats set out from the same dock and speed away at the same constant speed of 104 km/h for half an hour (0.500 h), th

Zinaida [17]

Answer:

The blue boat traveled 6.1 km farther west than the green boat

The green boat traveled 10.7 km farther south than the blue boat

Explanation:

The equation for linear uniform speed movement is

X(t) = X0 + v * t

Since we have two coordinates (X, Y) we use

X(t) = X0 + vx * t

Y(t) = Y0 + vy * t

The dock will be the origin of coordinates so X0 and Y0 will be zero. The X axis will be pointing west and the Y axis south.

The blue boat moves with a direction 24° south of west, so it will have speeds:

vxb = 104 * cos(24) = 95 km/h

vyb = 104 * sin(24) = 42.3 km/h

And the green boat:

vxg = 104 * cos(37.7) = 82.3 km/h

vyg = 104 * sin(37.7) = 63.6 km/h

After half hour the boats will have arrived at positions

Xb = 95 * 0.5 = 47.5 km

Yb = 42.3 * 0.5 = 21.1 km

And

Xg = 82.3 * 0.5 = 41.4 km

Yg = 63.6 * 0.5 = 31.8 km

The difference in positions of the boats

47.5 - 41.4 = 6.1 km

31.8 - 21.1 = 10.7 km

5 0

3 years ago