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alex41 [277]
3 years ago
9

Suppose that the acceleration of a model rocket is proportional to the difference between 100 ft/sec and the rocket's velocity.

If it is initially at rest and its initial acceleration is 150 ft/sec2, how long will it take to accelerate to 80 ft/s
Physics
1 answer:
sp2606 [1]3 years ago
3 0

Answer:

Explanation:

initial velocity, u = 0

final velocity, v = 80 ft/s

acceleration, a = 150 ft/s²

Let the time taken is t.

v = u + at

80 = 0 + 150 x t

t = 0.53 second

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Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and
djverab [1.8K]

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

<h3>How can we calculate the magnitude of the average friction force exerted on the collar?</h3>

To calculate the magnitude of the average friction force exerted on the collar we are using the formula,

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Here we are given,

k = The spring has a spring constant.

= 25.5 N/m.

x_f = Final length of the spring .

= \sqrt{1.25^2+1.8^2}  -0.60

= 1.591 m

x_i= The initial length of the spring.

= 1.25−0.60

=0.65 m

y=The collar then travels downward a distance.

=  1.80 m.

m= The mass of the collar.

=3.55 kg

v_c = the velocity of the collar.

= 3.39 m/s.

g = The acceleration due to gravity.

= 9.81 m/s²

We have to calculate the magnitude of the average friction force exerted on the collar = F

Now we put the known values in the above equation, we get;

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Or, \frac{1}{2} \times 25.5 \times((1.591)^2 - (0.65)^2 ) + F\times 1.80 + \frac{1}{2}\times 3.55\times (3.39)^{2}  = 3.55\times 9.81\times 1.80

Or, F= 8.641 N

From the above calculation we can conclude that,

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

Learn more about friction:

brainly.com/question/24338873

#SPJ4

Disclaimer: This question is incomplete in the portal. Here is the complete question.

Question:

The 3.55 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.80 m. The spring has a spring constant of k = 25.5 N/m. The distance a is given as 1.25 m. The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and the spring has an unstretched length of 0.60 m .

6 0
2 years ago
Which of Newton's Three Laws Applies? Law 1, 2, or 3?
frozen [14]

Answer:

1. Newton's first law

2.Newton's second law

3.Newton's third law

Explanation:

1. Newton's first law stated, In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force... this is base of the concept of inertia.

2. Newton's second law stated, In an inertial frame of reference, the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration a of the object: F = ma, or in easier words, F is directly proportional to a.

3. Newton's third law stated, When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body., In this case, the Normal Are opposite with gravititional force.

7 0
3 years ago
a force of 25 newtons moves a box a distance of 4 meeters in 5 seconds.the work done on the box is ? NM and the power is. ? nm/
miskamm [114]

Answer:

The work done on the box is 100 Nm

The power is 20 Nm/s

Explanation:

There is a force 25 newtons moves a box a distance of 4 meters in

5 seconds

The work done on the box is the product of the force and the distance

that the box moves ⇒ <em>work = force × distance</em>

The force = 25 newtons

the distance = 4 meters

Work = 25 × 4 = 100 NM

<em>The work done on the box is 100 Nm</em>

<em></em>

The force moves the box 4 meters in 5 seconds

The power is the rate of work

<em>The power = work ÷ time</em>

The work = 100 Nm

The time = 5 seconds

The power = 100 ÷ 5 = 20 Nm/s

<em>The power is 20 Nm/s</em>

6 0
3 years ago
Help please I don't understand this
BARSIC [14]
It's either staying there or is going at the same pace
4 0
3 years ago
Object A is placed on top of object B. Object A is the same temperature as object B. How will heat flow between object A and obj
mash [69]
"No heat will flow between object A and object B" is the one among the following choices given in the question that describes how heat will <span>flow between object A and object B. The correct option among all the options that are given in the question is the second option or option "B". I hope it helps you.</span>
6 0
2 years ago
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