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Hitman42 [59]
3 years ago
12

While playing football, Chris runs 4.5 m at 20⁰ south of west. If an opponent was trying to tackle him how far west and south wo

uld he have to run to reach him?
Physics
1 answer:
rewona [7]3 years ago
5 0

Answer:

South = 1.5m

West =4.2m

Explanation:

Kindly see attached a rough draft of the situation

Step one

Given data

From the sketch the direction of the player is along the resultant of the triangle, corresponding to the Hypotenuse

Step two:

Hence for an opponent to tackle him towards the south, he must be at

sin θ= opp/hyp

sin 20=x/4.5

x=sin 20*4.5

x=0.342*4.5

x= 1.5m

Also, for an opponent to tackle him towards the south, he must be at

cos θ= adj/hyp

cos 20=y/4.5

y=cos 20*4.5

y=0.93*4.5

y= 4.2m

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Answer:

Centripetal force is the force that keeps the yoyo going in a circle, if the string breaks, the yoyo would would fly off in a direction that is different to the point on the circle.

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3 years ago
HELP PLEASE 20 POINTS SHOW WORK, ALL EQUATIONS
nataly862011 [7]

Answer:

s = 3 m

Explanation:

Let t be the time the accelerating car starts.

Let's assume the vehicles are point masses so that "passing" takes no time.

the position of the constant velocity and accelerating vehicles are

s = vt = 40(t + 2)  cm

s = ½at² = ½(20)(t)² cm

they pass when their distance is the same

½(20)(t)² = 40(t + 2)

10t² = 40t + 80

0 = 10t² - 40t - 80

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t = (4±√(4² - 4(1)(-8))) / 2(1)

t = (4± 6.928) / 2  ignore the negative time as it has not occurred yet.

t = 5.464 s

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300 cm when rounded to the single significant digit of the question numerals.

7 0
3 years ago
What is the frequency of a clock waveform whose period is 750 microseconds?
Allushta [10]
Use this formula to find your answer...

Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms

frequency (f)=1/( Time period).

Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.

Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.

6 0
3 years ago
Determine the change in electric potential energy of a system of two charged objects when a -2.1-C charged object and a -5.0-C c
elena55 [62]

Answer:

Change in electric potential energy ∆E = 365.72 kJ

Explanation:

Electric potential energy can be defined mathematically as:

E = kq1q2/r ....1

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q1 = charge 1 = -2.1C

q2 = charge 2 = -5.0C

∆r = change in distance between the charges

r1 = 420km = 420000m

r2 = 160km = 160000m

From equation 1

∆E = kq1q2 (1/r2 -1/r1) ......2

Substituting the given values

∆E = 9.0×10^9 × -2.1 ×-5.0(1/160000 - 1/420000)

∆E = 94.5 × 10^9 (3.87 × 10^-6) J

∆E = 365.72 × 10^3 J

∆E = 365.72 kJ

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Lightning strikes the ground

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