A 2500 kg mustang car accelerates from rest to 30 m/s in 6 sec.What is acceleration of the mustang car?

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

valentina_108 [34]

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

4 0

3 years ago

A 2.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 3.00-kg object B. Object A re

slamgirl [31]

Answer:

tension: 19.3 N
acceleration: 3.36 m/s^2

Explanation:

Given

mass A = 2.0 kg

mass B = 3.0 kg

θ = 40°

Find

The tension in the string

The acceleration of the masses

Solution

Mass A is being pulled down the inclined plane by a force due to gravity of ...

F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

F/m = a = F/m

(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

T = (2(29.4) +3(12.5986))/5 = 19.3192 N

__

Then the acceleration of B is ...

a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.

3 0

3 years ago

No matter how well we get along with our parents as adolescents, there comes a time when we need to break away and assert our ow

KonstantinChe [14]

Answer: I never said anything to them but if i did it would be "I need a break from you, i love you but sometimes I have to find the path on my own." i would say this if i wasn't scared to hurt them.

Explanation:

5 0

3 years ago

Describe a scientific method

laila [671]

The process of scientific method involves making hypotheses , driving predictions from them as logical consequences , and then carrying out experiments or empirical observations based on those predictions. A hypotheses is a conjecture , based on knowledge obtained while seeking answers to the questions.

7 0

3 years ago

I lost 1 point on this and I don’t know what I did wrong.

Viktor [21]

Answer:

i think its because u gave the almost every answer the same exact thing. all the questions have different ways of moving which means different forces for each one i hope this helps :)

Explanation:

4 0

3 years ago

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