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shepuryov [24]
3 years ago
13

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

e toy rocket to the back of a sled, and take the modified sled to a large, flat, snowy field. You ignite the rocket, and observe that the sled accelerates from rest in the forward direction at a rate of 13.5 m/s2 for a time period of 3.30 s. After this time period, the rocket engine abruptly shuts off, and the sled subsequently undergoes a constant backward acceleration due to friction of 6.15 m/s2.After the rocket turns off, how much time does it take for the sled to come to a stop?By the time the sled finally comes to a rest, how far has it traveled from its starting point?
Physics
1 answer:
valentina_108 [34]3 years ago
4 0

Answer:

-time it takes for the sled to come to a stop after launch of rocket = 7.244 s

-distance sled has travelled from its starting point by the time it finally comes to rest is = 234.8655 m

Explanation:

From the question, looking at the motion while accelerating, we have;

Initial velocity; u = 0 m/s

Acceleration; a = 13.5 m/s²

Time; t = 3.3 s

Let's use first equation of motion to find final velocity (v).

v = u + at

v = 0 + (13.5 × 3.3)

v = 44.55 m/s

In this forward direction, let's calculate the displacement(d1) using newton's 3rd equation of motion.

d1 = ut + ½at²

d1 = 0(3.3) + ½(13.5 × 3.3²)

d1 = 73.5075 m

Now, let's consider the motion while slowing down and our final velocity will be 0 m/s while initial velocity will now be 44.55 m/s while acceleration is 6.15 m/s².

Thus, from v = u + at, we can find the time it take for the sled to come to a stop.

Now, since it's coming to rest acceleration will be negative. Thus;

0 = 44.55 + (-6.15t)

0 = 44.55 - 6.15t

t = 44.55/6.15

t = 7.244 s

Now we want to find out how far the sled has travelled from its starting point by the time it finally comes to rest.

Thus, we'll use the equation;

v² = u² + 2as

Where s will be the second displacement which we will call d2.

Thus;

0² = 44.55² + (-2 × 6.15 × s)

0 = 1984.7025 - 12.3s

12.3s = 1984.7025

s = 1984.7025/12.3

s = 161.358

Thus, d2 = s = 161.358 m

Thus, distance sled has travelled from its starting point by the time it finally comes to rest is ;

= d1 + d2 = 73.5075 + 161.358 = 234.8655 m

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3 years ago
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Answer:

F = 2,894 N

Explanation:

For this exercise let's use Newton's second law

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The radius r and the length of the rope is related

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Let's reduce the magnitudes to the SI system

     m = 101.7 g (1 kg / 1000g) = 0.1017 kg

      θ = 5 rev (2π rad / rev) = 31,416 rad

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     w = 31.416 / 5.1

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The maximum value of F is for  θ equal to zero

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7 0
3 years ago
Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 9.50 cm apart. You see an object jump from sid
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Answer: 12.67 cm, 8 cm

Explanation:

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Normal distance of separation of eyes, d(n) = 6 cm

Distance of separation is your eyes, d(y) = 9.5 cm

Angle created during the jump, θ = 0.75°

To solve this, we use the formula,

θ = d/r, where

θ = angle created during the jump

d = separation between the eyes

r = distance from the object

θ = d/r

0.75 = 9.5 / r

r = 9.5 / 0.75

r = 12.67 cm

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8 0
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Which type of force absorbs shock in vehicles
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hope this helps

5 0
3 years ago
a car with a mass of 2000 kilograms is moving around a circular curve at a uniform velocity of 25 meters per second. The curve h
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In the given problem, we say various information's that are going to help us reach the ultimate answer to the question. Let us first write the information's that have been presented in front of us.
Mass of the car = 2000 kg
Velocity of the car = 25 m/s^2
Radius of the circle = 80 m
Now we already know the equation for calculating the centripetal force and that is
Centripetal Force = [mass * (velocity)^2]/Radius
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So the centripetal force on the car is 15625 Newtons
  
4 0
3 years ago
Read 2 more answers
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