A car moves with constant velocity along a straight road. Its position is x1 = 0 m at t1 = 0 s and is x2 = 32 m at t2 = 2.0 s .
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The cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².
Answer:
Explanation:
Acceleration exerted by an object is the measure of change in speed or velocity of that object with respect to time. So the initial and final velocities play a major role in determining the acceleration of the cyclist. As here the initial velocity of the cyclist is the speed at rest and that is given as 0 m/s. Then after 3 seconds, the velocity of the cyclist changes to 9 m/s.
Then acceleration = change in velocity/Time.
Acceleration = (9-0)/3=9/3=3 m/s².
So the cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².
The box has 3 forces acting on it:
• its own weight (magnitude <em>w</em>, pointing downward)
• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)
• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)
Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have
• net parallel force:
∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>
• net perpendicular force:
∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0
Solve the net perpendicular force equation for the normal force:
<em>n</em> = <em>w</em> cos(35°)
<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)
<em>n</em> ≈ 120 N
Solve for the mag. of friction:
<em>f</em> = <em>µ</em> <em>n</em>
<em>f</em> = 0.25 (120 N)
<em>f</em> ≈ 30 N
Solve the net parallel force equation for the acceleration:
-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>
<em>a</em> ≈ (54.3157 N) / (15 kg)
<em>a</em> ≈ 3.6 m/s²
Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:
<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>
<em>v</em>² = 2 (3.6 m/s²) (3 m)
<em>v</em> = √(21.7263 m²/s²)
<em>v</em> ≈ 4.7 m/s
The astronaut will take 300 seconds
Explanation:
We can solve this problem by using the law of conservation of momentum.
In fact, the total momentum of the astronaut+object system must be conserved.
Initially, they are both at rest, so their total momentum is zero:
After the astronaut throws the object, their total momentum is:
where:
M = 80 kg is the mass of the astronaut
V is the final velocity of the astronaut
m = 500 g = 0.5 kg is the mass of the object
v = 8.0 m/s is the velocity of the object
Since momentum is conserved, we can write
And solving for V,
Which means that he starts moving at 0.05 m/s in the direction opposite to the object.
Now the astronaut needs to cover a distance of
d = 15.0 m
And his speed is
v = 0.05 m/s
Therefore, the time taken is
Learn more about momentum here:
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