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3 years ago

What happens to the piece of steel?

Physics

2 answers:

anyanavicka [17]3 years ago

6 0

The answer should be c....

Viefleur [7K]3 years ago

5 0

The answer is D because it would move toward piece of iron

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A projectile is launched vertically from the surface of the Moon with an initial speed of 1360 m/s. At what altitude is the proj

LenaWriter [7]

Answer:

485520 m

Explanation:

$v_{o}$ = initial velocity of the projectile = 1360 m/s

$v_{f}$ = final velocity of the projectile = $\left ( \frac{2}{5} \right )v_{_{o}}$ = $\left ( \frac{2}{5} \right )(1360)$ = 544 m/s

a = acceleraton due to gravity on moon = - 1.6 m/s²

h = Altitude of the projectile

Using the kinematics equation

$v_{f}^{2} = v_{o}^{2} + 2 a h$

Inserting the values

$544^{2} = 1360^{2} + 2 (-1.6) h$

h = 485520 m

3 0

3 years ago

What happens to energy and matter when a wave moves through space?

vivado [14]

Answer:

energy can move from one location to another, the particles of matter in the medium return to their fixed position. A wave transports its energy without transporting matter.

4 0

2 years ago

gayle cooks a roast in her microwave oven. the klystron tube in the oven emits photons whose energy is 1.20 x 10^-3 ev. what are

AveGali [126]

Answer:

$\lambda=1.03\times 10^{-3}\ m$

Explanation:

Given that,

The energy of the microwave oven is $1.2\times 10^{-3}\ eV$ .

We need to find the wavelength of these photons.

$1.2\times 10^{-3}\ eV=1.2\times 10^{-3}\times 1.6\times 10^{-19}\\\\=1.92\times 10^{-22}\ J$

The energy of a wave is given by :

$E=\dfrac{hc}{\lambda}\\\\\lambda=\dfrac{hc}{E}$

Put all the values,

$\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{1.92\times 10^{-22}}\\\\\lambda=1.03\times 10^{-3}\ m$

So, the wavelength of these photon is $1.03\times 10^{-3}\ m$ .

7 0

2 years ago

A car moving with an initial speed v collides with a second stationary car that is one-half as massive. After the collision the

Mashutka [201]

Answer:

4v/3

Explanation:

Assume elastic collision by the law of momentum conservation:

$m_1v = m_1v_1 + m_2v_2$

where v is the original speed of car 1, v1 is the final speed of car 1 and v2 is final speed of car 2. m1 and m2 are masses of car 1 and car 2, respectively

Substitute $m_2 = m_1/2 \& v_1 = v/3$