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ryzh [129]
3 years ago
13

A 200.0 kg piano is elevated by a crane to a height of 10.0 meters above a sidewalk. If the rope holding the piano breaks, what

will the potential energy of the piano be when it has fallen 5.00 meters?
a) 0 J
b) 9,810 J
c) 10,000 J
d) 19,620 J
Physics
1 answer:
Nitella [24]3 years ago
5 0
All that business about the crane and the rope and the falling
is only there to confuse us.

The piano ended up 5 meters above the ground.

           Potential energy = (mass) (gravity) (height)

                                   = (200 kg) (9.81 m/s²) (5 m)

                                   = (200 · 9.81 · 5)  (kg-m²/s²)

                                   =   9,810 joules   . 
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A 12-kg piece of metal displaces 1.6 L of water when submerged. Part A Find its density. Express your answer to two significant
Tatiana [17]

Answer:

ρ = 7500 kg/m³

Explanation:

Given that

mass ,m = 12 kg

Displace volume ,V= 1.6 L

We know that

1000 m ³ = 1 L

Therefore V= 0.0016 m ³

When metal piece is fully submerged

We know that

mass = Density x volume

m=\rho \times V

Now by putting the values in the above equation

\rho=\dfrac{12}{0.0016}\ kg/m^3

ρ = 7500 kg/m³

Therefore the density of the metal piece will be  7500 kg/m³.

6 0
4 years ago
What must happen to an atom of magnesium in order to become a magnesium ion Mg+2?
igomit [66]

Answer:

Answer is: c. It must lose two electrons and become an ion.

Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).

Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.

Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.

Explanation:

4 0
3 years ago
Read 2 more answers
When the mass of the bottle is 0.125 kg, the average maximum height of the beanbag is m. When the mass of the bottle is 0.250 kg
Jet001 [13]

Answer:

when the mass of the bottle is 0.125 kg, the average height of the beanbag is 0.35 m.

when the mass of the bottle is 0.250 kg, the average maximum height of the beanbag is 0.91m.

when the mass of the bottle is 0.375 kg, the average maximum height of the beanbag is 1.26m.

when the mass of the bottle is 0.500 kg, the average maximum height of the beanbag is 1.57m.

Explanation:

5 0
3 years ago
The crew of an enemy spacecraft attempts to escape from your spacecraft by moving away from you at 0.259 of the speed of light.
Vladimir79 [104]
If you do not have to use relative physics but classic physics, this is how you solve it:

Speed of light = c = 3 * 10^5 km/s

Speed of your foe respect to you: 0.259c

Speed of the torpedo respect to you: 0.349c

Speed of the torpedo respect your foe: 0.349c - 0.259c = 0.09c

Conversion to km/s = 0.09 * 3.0 * 10^5 km/s = 27000 km/s

Note that this solution, using classic physics do not take into account time and space dilation.

Answer: 27000 km/s
4 0
4 years ago
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

\int E dS=0

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

The area of the surface is

A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

The charge included within the sphere in this case is the charge on the paint layer, therefore

q=-17.0\mu C=-17.0\cdot 10^{-6}C

So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}

And the charge included within the sphere is again the charge on the paint layer,

q=-17.0\mu C=-17.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
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