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3 years ago

Does anyone know how to answer this

Physics

2 answers:

Anuta_ua [19.1K]3 years ago

4 0

You're well on your way to solving this, and closer to the solution than you think !

-- To work with the y-components of the velocities, use the sines of the
angles, not the cosines. The cosines are the horizontal components.

-- Make that change in your last line.
   Make the 3v₂ term negative ... the vertical part of v₂ is downward.

-- The m₁ had no vertical momentum before the collision,
    so the sum of all vertical components after the collision is still zero.
    Set your last line equal to zero, and it'll practically solve itself for v₂ .

Alenkinab [10]3 years ago

3 0

I got 8.57m per s. just use conservation of momentum of y direction

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The asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 5.0 earth years. assume a

My name is Ann [436]

Using Kepler's 3rd law which is: T² = 4π²r³ / GM
Solved for r :
r = [GMT² / 4π²]⅓
Where G is the universal gravitational constant,M is the mass of the sun,T is the asteroid's period in seconds, andr is the radius of the orbit.
Change 5.00 years to seconds :
5.00years = 5.00years(365days/year)(24.0hours/day)(6... = 1.58 x 10^8s
The radius of the orbit then is computed:
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3 years ago

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Nets used to catch falling boulders on the side of rocky hillside roadways

Natalija [7]

Answer:

more time to change the momentum of falling rocks

Explanation:

Momentum is related to "mass in motion." So, if an object is moving, then it has momentum as it has its mass in motion. The amount of momentum is dependent upon how much and how fast the object is moving.

If an object is moving slowly, it means that the object is losing momentum.

Nets used to catch falling boulders on the side of rocky hillside roadways are more effective than rigid fences because their breakage is reduced by more time to change the momentum of falling rocks.

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4 years ago

If the maximum energy given to an electron during compton scattering is 30 kev, what is the wavelength of the incident photon?

hammer [34]

The wavelength of the incident photon is $1.19x10^{-2} nm$ .

What is wavelength?

The wavelength, or the distance over which the shape of a periodic wave repeats, is the spatial period in physics. It is a property of both traveling waves and standing waves, as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two neighboring crests, troughs, or zero crossings. The Greek letter lambda is frequently used to denote wavelength. The term wavelength is also sometimes used to describe modulated waves, their sinusoidal envelopes, or waves created by the interference of several sinusoids.

The relationship between wavelength and frequency is inverse, assuming a sinusoidal wave flowing at a constant speed.

Calculations:

The energy loss Δλ=h/ $m_{e} c$ (1-cos∅)

Conservation of momentum gives,

$h_{f} =P_{e} c-hf'\\hf+hf'=P_{e} c\\hf+hf'=\sqrt{(m_{e} c^{2} } +k^{2} )-me^{2} c^{4} \\\\=\sqrt {(511+30)^{2} -(511)^{2} } \\=178keV$

$hf=hf'+k=hf-hf'=30keV\\$

$2hf=178+30\\2hf=208\\hf=104keV\\$

Wavelength(λ)= $\frac{hc}{104keV}$ = $\frac{1.24keV.nm}{104keV}$

Wavelength(λ)= $1.19x10^{-2} nm$

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brainly.com/question/12924624

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2 years ago

An electric motor is used to run an elevator. The total mass of the elevator car and passenger is 1600 kg. The elevator moves up

Masja [62]

Answer:

Approximately $2.7 \times 10^{4}\; {\rm W}$ , assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$

Explanation:

The weight of the elevator is:

$\begin{aligned}& \text{weight} \\ =\; & m\, g \\ =\; & 1600\; {\rm kg} \times 9.81\; {\rm m \cdot s^{-2}} \\ \approx\; & 15700\; {\rm N}\end{aligned}$ .

Since the speed of the elevator is constant, the acceleration of this elevator would be $0$ .

By Newton's Second Law of Motion, the net force on the elevator (proportional to acceleration) would also be $0\!$ . All external forces on the elevator need to be balanced in every direction.

The only two vertical forces on the elevator are:

the weight of the elevator (downward gravitational pull from the earth,) and
the upward pull from the motor.

These two forces need to balance one another. Since the weight of the elevator is approximately $15700\; {\rm N}$ , the upward pull of the motor would be $15700\; {\rm N}\!$ . in magnitude.

The direction of this upward pull is the same as the direction of the motion of this elevator. Thus, the work that the motor did on the elevator would be positive:

$\begin{aligned}& \text{work} \\ =\; & F\, s \\ \approx\; & 15700\; {\rm N} \times 15.0\; {\rm m} \\ \approx\; & 2.35 \times 10^{5}\; {\rm J}\end{aligned}$ .

Since the velocity of the elevator is constant, instantaneous power output of the motor would be equal to the average power of the motor:

$\begin{aligned}& \text{power} \\ =\; & \frac{\text{work}}{t} \\ \approx\; & \frac{2.35 \times 10^{5}\; {\rm J}}{8.82\; {\rm s}} \\ \approx\; & 2.7 \times 10^{4}\end{aligned}$ .

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Hope I helped! ( Smiles )

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4 years ago

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