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galben [10]
3 years ago
15

HELP PLEASE 20 POINTS SHOW WORK, ALL EQUATIONS

Physics
1 answer:
nataly862011 [7]3 years ago
7 0

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

0 = t² - 4t - 8

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

s = 40(5.464 + 2) = 298.564 cm

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

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A girl runs once around a circular track with a radius of 100m at speed of 10m/s.
goblinko [34]

Answer:

0 m

Explanation:

Displacement is the shortest distance from one ppont to another. Here, the girl returns to the same spot after running. Hence, displacement is <u>0</u><u> </u><u>m</u>.

5 0
2 years ago
You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
Anarel [89]

Answer:

R = 24.3 m

Explanation:

As we know that the orbital speed is given as

v = \sqrt{\frac{GM}{R + h}}

here we know that

v = 5500 m/s

R = 4.48 \times 10^6 m

h = 630 km

now we have

5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}

M = 2.34 \times 10^24 kg

now acceleration due to gravity of planet is given as

a = \frac{GM}{R^2}

a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}

a = 7.7 m/s^2

now range of the projectile on the surface of planet is given as

R = \frac{v^2 sin2\theta}{g}

R = \frac{14.6^2 sin(2\times 30.8)}{7.7}

R = 24.3 m

3 0
3 years ago
What is the lowest possible temperature
HACTEHA [7]
The lowest possible temperature is absolute zero. However scientists have not reached this temperature, rather they have come very close to absolute zero.

6 0
2 years ago
Read 2 more answers
A car travels straight for 20 miles on a road that is 30° north of east. What is the east component of the car’s displacement
Over [174]

The east component of the cars displacement is 17.3 miles.

Trigonometric ratio is used to show the relationship between the sides of a right angled triangle and its angles.

Let x represent the east component of the cars displacement.

Using trigonometric ratio:

cos(30) = x / 20

x = 20 * cos(30)

x = 17.3 miles

The east component of the cars displacement is 17.3 miles.

Find out more on Trigonometric ratio at: brainly.com/question/1201366

4 0
2 years ago
Help? Please? Thanks!
Katena32 [7]
1,3 and 5 are the answers 
5 0
2 years ago
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