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3 years ago

15

If the frequency of a given wave increases,what happens to the wavelength?

2 answers:

Greeley [361]3 years ago

8 0

It shortens so that the tips reach faster

Blababa [14]3 years ago

8 0

The product of frequency and wavelength is always the
same number . . . the wave's speed.

So if the frequency increases, the wavelength must decrease,
in order for their product not to change.

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A material that can easily flow is called a...

A. Fluid.

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If an isotope has 8 protons and 10 neutrons in it's nucleus, what is it's atomic mass?

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3 years ago

If two objects of unequal mass collide, both objects will continue to move in the direction of

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4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two

Dvinal [7]

Answer:-2.61 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=mV_{o}+MU_{o}$ (2)

$p_{f}=(m+M)V_{f}$ (3)

$m=1200 kg$ is the mass of the first car

$V_{o}=20 m/s$ is the velocity of the first car, to the North

$M=1400 kg$ is the mass of the second car

$U_{o}=-22 m/s$ is the mass of the second car, to the South

$V_{f}$ is the final velocity of both cars after the collision

$mV_{o}+MU_{o}=(m+M)V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

$V_{f}=\frac{(1200 kg)(20 m/s)+(1400 kg)(-22 m/s)}{1200 kg+1400 kg}$ (6)

Finally:

$V_{f}=-2.61 m/s$ (7) This is the resulting velocity of the wreckage, to the south

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3 years ago

A 35.4 kg girl is riding on the back of a 15.23 kg cart. the cart and the kid are both moving eastward at 4.25 m/s when she step

Grace [21]

Answer:

The final velocity of the cart is $v_c = 7.02 \ m/s$

Explanation:

From the question we are told that

The mass of the girl is $m_g = 35.4 \ kg$

The mass of the cart is $m_c = 15.23 \ kg$

The speed of the cart and kid(girl) is $v = 4.25 \ m/s$

The final velocity of the girl is $v_g = 3.06 \ m/s$

Let assume that velocity eastward is positive and velocity westward is negative (Note that if we assume vise versa it wouldn't affect the answer )

The total momentum of the system before she steps off the back of the cart

is mathematically evaluated as

$p__{T1}} = (m_g + m_c) * v$

substituting values

$p__{T1}} = (35.4 + 15.23) * 4.25$

$p__{T1}} =215.17 \ kg m /s$

The total momentum after she steps off the back of the cart is mathematically evaluated as

$p__{T2}} = (m_g * v_g ) +( m_c * v_c )$

Where $v_c$ is the final velocity of the cart

substituting values

$p__{T2}} = (35.4 * 3.06 ) +( 15.23 * v_c )$

$p__{T2}} = 108. 324 + 15.23 v_c$

Now according to the law of conservation of momentum

$p__{T1}} =p__{T2}}$

So

$215.17 \ kg m /s = 108. 324 + 15.23 v_c$

=> $v_c = 7.02 \ m/s$

Since the value is positive it implies that the cart moved eastward

7 0

3 years ago