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natka813 [3]
3 years ago
5

A football player runs from his own goal line to the opposing team's goal line, returning to his ten-yard line, all in 26.6 s. C

alculate his average speed and the magnitude of his average velocity. (Enter your answers in yards/s.) HINT
Physics
1 answer:
Advocard [28]3 years ago
8 0

Answer:

a) Average Speed = 190yards/26.6s = 7.14 yards/s

b) Average Velocity = 10/26.6 = 0.38 yard/s

Complete Question;

A football player runs from his own goal line to the opposing team's goal line, returning to his ten-yard line, all in 26.6 s. Calculate his average speed and the magnitude of his average velocity. (Assume a 100 yard football field; note that the player's X-yard line is X yards from his own goal line. Enter your answers in yards/s.)

Explanation:

a) Speed = distance/time

time t = 26.6s

distance;

d1 = 100 yards (own goal line to the opposing team's goal line)

d2 = 90 yards (returning to his ten-yard line)

d = d1 + d2

d = 100 + 90 = 190 yards

Speed = 190yards/26.6s = 7.14 yards/s

b) velocity = displacement/time

time = 26.6s

displacement = 100-90 yards = 10yards

Velocity = 10/26.6 = 0.38 yard/s

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a truck drives up a hill with a 15 incline. if the truck has a constant speed of 22m/s, what are the horizontal and vertical com
Kamila [148]

Answer:

speed of truck (v) =  22 m/s ,

angle of hill (Θ) =15°

Find

Vertical component (Fv) = ?

Harizontal component (Fh) =?

               Vertical component (Fh) = V cosΘ

                                                        = 22. cos 15

                                                        = 21.25 m/s.

               Harizontal component (Fv) = V sinΘ

                                                            = 22. sin 15

                                                            = 5.69 m/s.

3 0
4 years ago
A uniform plank of mass 10kg and length 10m rests on two supports, A and B as shown. A boy of weight 500N stands at a distance o
kifflom [539]

Answer:

U² = 142.86 N

U¹ = 357.14 N

Explanation:

Taking summation of the moment about point A, we get the following equilibrium equation: (taking clockwise direction as positive)

W(2\ m) - U^2(7\ m) = 0

where,

W = weight of boy = 500 N

U² = reaction ay B = ?

Therefore,

(500\ N)(2\ m)-(U^2)(7\ m)=0\\U^2=\frac{1000\ Nm}{7\ m}\\

<u>U² = 142.86 N</u>

Now, taking summation of forces on the plank. Taking upward direction as positive, for equilibrium position:

W-U^1-U^2=0\\500\ N - 142.86\ N = U^1\\

<u>U¹ = 357.14 N</u>

3 0
3 years ago
Determine a valid way of finding the wire’s diameter if you know the resistivity of the material, \rho , and can measure the cur
olganol [36]

Answer:

To find the diameter of the wire, when the following are given:

Resistivity of the material (Rho), Current flowing in the conductor, I, Potential difference across the conductor ends, V, and length of the wire/conductor, L.

Using the ohm's law,

Resistance R = (rho*L)/A

R = V/I.

Crossectional area of the wire A = π*square of radius

Radius = sqrt(A/π)

Diameter = Radius/2 = [sqrt(A/π)]

Making A the subject of the formular

A = (rho* L* I)V.

From the result of A, Diameter can be determined using

Diameter = [sqrt(A/π)]/2. π is a constant with the value 22/7

Explanation:

Error and uncertainty can be measured varying the value of the parameters used and calculating different values of the diameters. Compare the values using standard deviation

7 0
3 years ago
In an ideal gas, collisions between molecules
chubhunter [2.5K]

Answer:

IDHHHHH

Explanation:

vfdvbggggggggggrhhhhhttttttfdsgt

5 0
3 years ago
A typical 100-watt light bulb consumes 100 j of energy per second. If a light bulb converts all of this energy to 400 nm light
VladimirAG [237]

The missing question is: how many photons are produced per second?

Answer:

2.0\cdot 10^{20}

Explanation:

The energy of a photon is given by

E_1 = \frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda is the wavelength of the photon

In this problem, the photons have wavelength

\lambda = 400 nm = 400 \cdot 10^{-9} m, so each photon has an energy of

E_1 = \frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{400\cdot 10^{-9}}=4.97\cdot 10^{-19} J

The total energy emitted by the bulb in 1 second is

E = 100 J

Therefore, the number of photons emitted per second is

n=\frac{E}{E_1}=\frac{100}{4.97\cdot 10^{-19}}=2.0\cdot 10^{20}

3 0
4 years ago
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