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SpyIntel [72]
3 years ago
13

Your boat departs from the bank of a river that has a swift current parallel to its banks. If you want to cross this river in th

e shortest amount of time, you should direct your boat If you want to cross this river in the shortest amount of time, you should direct your boat:_________
a. perpendicular to the current.
b. upstream.
c. downstream.
d. so that it drifts with the current.
Physics
1 answer:
REY [17]3 years ago
8 0

Answer:

D.

Explanation:

Given that your boat departs from the bank of a river that has a swift current parallel to its banks. If you want to cross this river in the shortest amount of time, you should direct your boat: so that it drifts with the current.

If the boat moves perpendicular to the current, the current flow will be the resistance to the movement of the boat. So, it's better for the boat to drifts perpendicularly with the current.

The best answer is therefore option D.

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qwelly [4]

Answer:Eating. Your muscles in your arms and mouth use energy to feed itself. Then your body digest the food which also takes energy.

Sleep. When your tired, you don’t have much energy. It is said that you use more energy while your sleeping. But how do you become energized if you were using even more energy than before?

6 0
2 years ago
Naturally occurring element X exists in three isotopic forms: X-28 (27.977 amu, 92.23% abundance), X-29 (28.976 amu, 4.67% abund
vova2212 [387]

Answer:

C. 28.09 amu

Explanation:

The natural occurring element exist in 3 isotopic forms: namely X-28 (27.977 amu, 92.23% abundance),  X-29 (28.976 amu, 4.67% abundance) and  X-30 (29.974 amu, 3.10% abundance).

The atomic weight of elements depends on the isotopic abundance. If you know the fractional abundance and the mass of the isotopes the atomic weight can be computed.

The atomic weight is computed as follows:

atomic weight = mass of X-28 × fractional abundance + mass of X-29 × fractional abundance + mass of  X-30 × fractional abundance

atomic weight = 27.977 × 0.9223 + 28.976 × 0.0467 + 29.974 ×  0.0310

atomic weight = 25.8031871 + 1.3531792 + 0.929194

atomic weight = 28.0855603 amu

To 2 decimal place atomic weight = 28.09 amu

6 0
3 years ago
A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

r \approx 18.897\times 10^{6}\,m

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

F = 6841.905\,N

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0
3 years ago
A worker drives a 0.562 kg spike into a rail tie with a 2.26 kg sledgehammer. The hammer hits the spike with a speed of 64.4 m/s
zubka84 [21]

Answer:

Explanation:

Given that,

Mass of sledge hammer;

Mh =2.26 kg

Hammer speed;

Vh = 64.4 m/s

The expression fot the kinetic energy of the hammer is,

K.E(hammer) = ½Mh•Vh²

K.E(hammer) = ½ × 2.26 × 64.4²

K.E ( hammer) = 4686.52 J

If one forth of the kinetic energy is converted into internal energy, then

ΔU = ¼ × K.E(hammer)

∆U = ¼ × 4686.52

∆U = 1171.63 J

Thus, the increase in total internal energy will be 1171.63 J.

4 0
3 years ago
I'm not sure what equation to use.
Lelechka [254]
I would think that you would multiply then divide
7 0
3 years ago
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