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2 years ago

If you break a piece of glass, the shape of the glass changes, but the properties in the

Physics

1 answer:

lutik1710 [3]2 years ago

6 0

Answer:

It would be a physical change because you aren't changing to chemical properties of the glass, just the shape.

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A moving particle encounters an external electric field that decreases its kinetic energy from 9520 eV to 7060 eV as the particl

Sati [7]

Given Information:

KEa = 9520 eV

KEb = 7060 eV

Electric potential = Va = -55 V

Electric potential = Vb = +27 V

Required Information:

Charge of the particle = q = ?

Answer:

Charge of the particle = +4.8x10⁻¹⁸ C

Explanation:

From the law of conservation of energy, we have

ΔKE = -qΔV

KEb - KEa = -q(Vb - Va)

-q = KEb - KEa/Vb - Va

-q = 7060 - 9520/27 - (-55)

-q = 7060 - 9520/27 + 55

-q = -2460/82

minus sign cancels out

q = 2460/82

Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹

q = 2460(1.60x10⁻¹⁹)/82

q = +4.8x10⁻¹⁸ C

6 0

3 years ago

During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0

oksian1 [2.3K]

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

$d=\frac{1,280,000,000km}{37,000}=34,594.59km$

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

$v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}$

hence, the speed of the Hubble is approximately 384km/min

5 0

2 years ago

What is the direction of acceleration due to gravity

Paul [167]

Answer:

Negative, because the speed is decreasing.

4 0

2 years ago

A merry-go-round with a a radius of R = 1.9 m and moment of inertia I = 209 kg-m2 is spinning with an initial angular speed of ω

RideAnS [48]

Answer:

340.67 kgm²/s

Explanation:

R = Radius of merry-go-round = 1.9 m

I = Moment of inertia = 209 kgm²

$\omega_i$ = Initial angular velocity = 1.63 rad/s

m = Mass of person = 73 kg

v = Velocity = 4.8 m/s

Initial angular momentum is given by

$L=I\omega_i\\\Rightarrow L=209\times 1.63\\\Rightarrow L=340.67\ kgm^2/s$

The initial angular momentum of the merry-go-round is 340.67 kgm²/s

8 0

2 years ago

The rate of chage of displacement is called __________

faltersainse [42]

Answer:

Velocity is the rate of change of displacement.

8 0

2 years ago

Read 2 more answers