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marin [14]
3 years ago
5

Calculate the kinetic energy of a 0.145-kg baseball with a velocity of 80 mil/h. Express the answer in Joules.

Physics
1 answer:
Alexus [3.1K]3 years ago
5 0

Answer: 92.7Joules//

Explanation:

Kinetic energy of moving body is 1/2mv²

K=1/2mv²

Mass = 0.145kg

V=80mil/h = 80×1609.34/3600

=35.76m/s, since 1 mile = 1609.34m and 1 hour = 3600seconds.

From the equation K=1/2mv² you can input.

K=1/2×0.145×35.76²

K=1/2×185.422

K= 185.422/2

K=92.7J//

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Describe seven physical properties that help distin-<br> guish one mineral from another.
Lilit [14]

Answer:

Color, Streak, luster, cleavage and fracture, hardness, crystal shape, and density.

Explanation:

6 0
3 years ago
Two engineering students, John with a weight of 96 kg and Mary with a weight of 48 kg, are 30 m apart. Suppose each has a 0.04%
djyliett [7]

Answer:

6.8370869499\times 10^{20}\ N

Explanation:

N_A = Avogadro's number = 6.022\times 10^{23}

e = Charge of electron = 1.6\times 10^{-19}\ C

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Z = Atomic number of water = 18

M = Molar mass of water = 0.018 kg/mol

m = Mass of person

The charge is given by

q=imbalance\times n\times e

Total number of protons and electrons in each sphere

n=\dfrac{mN_AZe}{M}

q=imbalance\times \dfrac{mN_AZe}{M}

q_1=0.0004\times \dfrac{96\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=3699916.8\ C

q_2=0.0004\times \dfrac{48\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=1849958.4\ C

Electrical force is given by

F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times 3699916.8\times 1849958.4}{30^2}\\\Rightarrow F=6.8370869499\times 10^{20}\ N

The electrostatic force of attraction between them is 6.8370869499\times 10^{20}\ N

4 0
3 years ago
A banana peel has lots of friction.<br> True or False
olga_2 [115]

Answer:

False

Explanation:

I learned it the hard way trust me T^T

3 0
2 years ago
At which of the following temperature and pressure levels would a gas be most likely to follow the ideal gas law? A. 0 K and 100
bulgar [2K]
The Ideal Gas Law makes a few assumptions from the Kinetic-Molecular Theory. These assumptions make our work much easier but aren't true under all conditions. The assumptions are,

1) Particles of a gas have virtually no volume and are like single points.
2) Particles exhibit no attractions or repulsions between them.
3) Particles are in continuous, random motion.
4) Collisions between particles are elastic, meaning basically that when they collide, they don't lose any energy.
5) The average kinetic energy is the same for all gasses at a given temperature, regardless of the identity of the gas.

It's generally true that gasses are mostly empty space and their particles occupy very little volume. Gasses are usually far enough apart that they exhibit very little attractive or repulsive forces. When energetic, the gas particles are also in fairly continuous motion, and without other forces, the motion is basically random. Collisions absorb very little energy, and the average KE is pretty close.

Most of these assumptions are dependent on having gas particles very spread apart. When is that true? Think about the other gas laws to remember what properties are related to volume.

A gas with a low pressure and a high temperature will be spread out and therefore exhibit ideal properties.

So, in analyzing the four choices given, we look for low P and high T.

A is at absolute zero, which is pretty much impossible, and definitely does not describe a gas. We rule this out immediately.

B and D are at the same temperature (273 K, or 0 °C), but C is at 100 K, or -173 K. This is very cold, so we rule that out.

We move on to comparing the pressures of B and D. Remember, a low pressure means the particles are more spread out. B has P = 1 Pa, but D has 100 kPa. We need the same units to confirm. Based on our metric prefixes, we know that kPa is kilopascals, and is thus 1000 pascals. So, the pressure of D is five orders of magnitude greater! Thus, the answer is B.
6 0
3 years ago
A glass window 0.33 cm thick measures 87 cm by 36 cm. How much heat flows through this window per minute if the inside and outsi
mojhsa [17]

Answer:

6.38 x 10^4 J

Explanation:

d = 0.33 cm = 0.33 x 10^-2 m, Area = 87 x 36 cm^2 = 0.87 x 0.36 m^2

ΔT = 14 degree C, t = 1 min = 60 second

K = 0.8 W / m K

Heat  = K A ΔT t / d

H = 0.8 x 0.87 x 0.36 x 14 x 60 / (0.33 x 10^-2)

H = 6.38 x 10^4 J

7 0
3 years ago
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