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2 years ago

A student, Daniela, is arguing that since a meteor is weightiess in space, if it crashes

Physics

1 answer:

zmey [24]2 years ago

5 0

Answer:

she's wrong because she is and there it doesn't say she's right

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how do i write this to where it is true? in the blue is a statement and the attached piece is false so i have to make it true an

OlgaM077 [116]

Answer:

cross out the false piece in blue and write the true piece in red

4 0

3 years ago

Particles that attract each other are

photoshop1234 [79]

B. I belive :)
Hopes this helps

3 0

2 years ago

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A student drops a rubber ball onto a surface. Assume that this is a closed system. The ball bounces, but each successive bounce

spayn [35]

I would say D.) The ball bounces many times suggesting the energy is used up efficiently

5 0

2 years ago

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A photon of red light (wavelength = 710 nm) and a ping-pong ball (mass = 2. 40 × 10^-3 kg) have the same momentum. At what speed

Paha777 [63]

The speed of the ball moving is

$v = 3.94 \times 10 {}^{ - 25}m/s$

what is momentum?

The momentum p of a classical object of mass m and velocity v is given by pclassical =mv.

For photons with wavelength λ,this equation does not hold.Instead, the momentum of the Photon is given by p Photon = h/λ

where,h is the planck's constant.

The momentum of the red Photon is

given:

$h = 6.626 \times 10 {}^{ - 34}kgm {}^{2}/s(plancks \: constant)$

$λ = 700 \times 10 {}^{ - 9} m(photons \: wavelength)$

$p \: photons = \frac{6.626 \times 10 {}^{ - 34}kgm {} ^{2}/s }{700 \times 10 {}^{ - 9}m }$

$= 9.47 \times 10 {}^{ - 28}kgm /s$

since,the Photon and the ping-pong ball have the same momentum,we have

$pball = pphotons = \frac{6.626 \times 10 {}^{ - 34}kgm/s }{700 \times 10 {}^{ - 9} }$

$pball = mv,m = 2.40 \times 10 {}^{ - 3}kg$

$v = 3.94 \times 10 {}^{ - 25} m/s$

Therefore, if the red photon and the ping-pong ball have the same momentum, the ping-pong ball must have a speed of approximately

$v = 3.94 \times 10 {}^{ - 25}m/s$

learn more about momentum of photon from here: brainly.com/question/28197406

#SPJ4

3 0

1 year ago

What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?

Nastasia [14]

Answer:

$F=1.26*10^{-3}N$

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

$F=\frac{kq_1q_2}{d^2}$

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges( $q_1,q_2$ ) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

$F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N$

8 0

2 years ago