Answer:
if the friction was removed from the ramp, the acceleration would drasticly increase, making it hard to stop.
Explanation:
Answer: an organic base, a phosphate group, and a sugar
Explanation:
Answer: 70712.44 ly
Using the following time dilation formula:
where is the time measured in the moving frame, is the speed of the moving frame, is the speed of light and is time measured in other frame.
Inserting the values:
Therefore, when a trip to a star 1000 ly years away is made with speed 0.99 c, 70712.44 ly would have passed on Earth.
Approximately 999 years will pass on Earth while you are gone.
The formula for time dilation is t′ = t1 - (vc)2 / c.
To calculate how many years will pass on Earth while you are on this trip, we need to determine the time experienced by the astronaut and then use the time dilation formula.
According to the information provided, the astronaut travels at an average speed of 0.99c, so v = 0.99c. The star is located 1,000 light-years away, so t1 = 1,000 years.
Plugging these values into the time dilation formula, we get:
t′ = 1,000 years - (0.99c)2 / c = 1,000 years - 0.9801 years = 999.0199 years.
Therefore, approximately 999 years will pass on Earth while you are gone.
#SPJ3
and Theodore at 17 cm/s.
When the Alvin crosses the finish line, how
far behind is Theodore?
Answer in units of cm.
The box has 3 forces acting on it:
• its own weight (magnitude w, pointing downward)
• the normal force of the incline on the box (mag. n, pointing upward perpendicular to the incline)
• friction (mag. f, 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:
∑ F = -f + w sin(35°) = m a
• net perpendicular force:
∑ F = n - w cos(35°) = 0
Solve the net perpendicular force equation for the normal force:
n = w cos(35°)
n = (15 kg) (9.8 m/s²) cos(35°)
n ≈ 120 N
Solve for the mag. of friction:
f = µn
f = 0.25 (120 N)
f ≈ 30 N
Solve the net parallel force equation for the acceleration:
-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) a
a ≈ (54.3157 N) / (15 kg)
a ≈ 3.6 m/s²
Now solve for the block's speed v given that it starts at rest, with v₀ = 0, and slides down the incline a distance of ∆x = 3 m:
v² - v₀² = 2 a ∆x
v² = 2 (3.6 m/s²) (3 m)
v = √(21.7263 m²/s²)
v ≈ 4.7 m/s
False