Answer:
5.72 s
Explanation:
From Newton's law, F = ma
The East is +ve direction, Hence,
F = +8930 N
m = 2290 kg
a = ?
8930 = 2290 × a
a = 8930/2290 = 3.90 m/s²
So, we will find the time it takes the car to stop using the equations of motion
a = 3.90 m/s²
u = initial velocity of the car = - 22.3 m/s (the velocity is to the west)
v = final velocity of the car = 0 m/s (since the car comes to rest)
t = time taken for the car to come to rest = ?
v = u + at
0 = - 22.3 + (3.90)(t)
3.9t = 22.3
t = 5.72 s
Answer:
E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.
Explanation:
We can answer this question by using Kepler's second law of planetary motion, which states that:
"A line connecting the center of the Sun with the center of each planet sweeps out equal areas in equal intervals of time"
This means that when a planet is further away from the Sun, it will move slower (because the line is longer, so it must move slower), while when the planet is closer to the Sun, it will move faster (because the line is shorter, so it must move faster).
In the text of this problem, it is written that the planet moves at 31 km/s when is close to the star and 35 km/s when it is farthest: this is in disagreement with what we said above, therefore the correct option is
E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.
Answer:
<em>10.90km</em>
Explanation:
Magnitude of the total displacement is expressed using the equation
d = √dx²+dy²
dx is the horizontal component of the displacement
dy is the vertical component of the displacement
dy = -6.7sin27°
dy = -6.7(0.4539)
dy = -3.042
For the horizontal component of the displacement
dx = -4.5 - 6.7cos27
dx = -4.5 -5.9697
dx = -10.4697
Get the magnitude of the bicyclist's total displacement
Recall that: d = √dx²+dy²
d = √(-3.042)²+(-10.4697)²
d = √9.2538+109.6146
d = √118.8684
<em>d = 10.90km</em>
<em>Hence the magnitude of the bicyclist's total displacement is 10.90km</em>
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