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maksim [4K]
3 years ago
6

Four football players are running down the field at the same speed. Player 1 weighs 180 lbs and is running toward the south goal

, player 2 weighs 200 lbs and is running toward the north goal, player 3 weighs 190 lbs and is running toward the north goal, and player 4 weighs 165 lbs and is running toward the south goal. Which player has the most momentum?
Physics
1 answer:
Serggg [28]3 years ago
7 0
Player 2 because moment is mass times acceleration and since they are all going the same speed. Speed doesn't matter so the only thing that is left is mass/ weight and he has the most
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A nerve signal travels 150 meters per second. Determine the number of kilometers that the nerve signal will travel in the same t
joja [24]

Explanation:

It is given that,

A nerve signal travels 150 meters per second. It is the speed of the nerve signal. We need to convert the number of kilometers that the nerve signal will travel in the same time.

We know that,

1 kilometer = 1000 meter

1 hour = 3600 seconds

150\ \dfrac{m}{s}=150\times \dfrac{1/1000\ km}{1/3600\ h}

150\ m/s=540\ km/h

So, the nerve signal will travel at the rate of 540 km/h. Hence, this is the required solution.

6 0
3 years ago
An aluminum wire having a cross-sectional area equal to 3.90 10-6 m2 carries a current of 6.00 A. The density of aluminum is 2.7
stich3 [128]

Answer:

Vd = 1.597 ×10⁻⁴ m/s

Explanation:

Given: A = 3.90×10⁻⁶ m², I = 6.00 A, ρ = 2.70 g/cm³

To find:

Drift Velocity Vd=?

Solution:

the formula is Vd = I/nqA         (n is the number of charge per unit volume)

n = No. of electron in a mole ( Avogadro's No.) / Volume

Volume = Molar mass / density   ( molar mass of Al =27 g)

V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³

n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)

n= 6.02 × 10 ²⁸

Now

Vd = (6A) / (  6.02 × 10 ²⁸ ×  1.6 × 10⁻¹⁹ C ×  3.9×10⁻⁶ m²)

Vd = 1.597 ×10⁻⁴ m/s

6 0
3 years ago
Which of the following wavelengths will produce standing waves on a string that is 3.5 m long?
denpristay [2]

In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:

\lambda=\frac{2}{n} L


The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:

\lambda=\frac{2}{1} \cdot 3.5 m=7.0 m


The wavelength of the 2nd harmonic is:

\lambda=\frac{2}{2} \cdot 3.5 m=3.5 m


The wavelength of the 4th harmonic is:

\lambda=\frac{2}{4} \cdot 3.5 m=1.75 m


It is not possible to find any integer n such that \lambda=5 m, therefore the correct options are A, B and D.

3 0
3 years ago
Read 2 more answers
Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 wi
Flauer [41]

Answer:

a. Fnet =37.67N

b. The direction = 133.4 from the x axis counter clockwise.

c. Option 2

Explanation:

Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.

Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..

Given that F1 is 71N at 20°, then it is in the first quadrant.

a. Fnet= F1+F2+F3

Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j

Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j

Resolving the vectors into x and y components.

Fnet= -2.04i+37.62j

Magnitude of the vector

Fnet= √((-2.04)^2+(37.62)^2)

Fnet= 37.67N

Fnet is approximately 38N.

b. Direction of the Fnet.

Angle=arctan(y/x)

Angle=arctan(-37.61/2.04)

Angle= -43.37°

The angle is in the negative x axis and positive y axis.

Then the direction becomes 180-43.37

Therefore, the direction of the net force is 133.37°.

c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.

Fnet=ma

Fnet= mv/t

So the velocity is in the direction of the Fnet.

3 0
3 years ago
Find the poing of center of gravity<br><br>plz show the steps...​
olasank [31]

Answer:

C

Explanation:

For a uniformly distributed mass, the center of gravity is also the geometric center.  For this shape, the center is at point C.

7 0
3 years ago
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