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3 years ago

Calculate the acceleration of an airplane that starts at rest and reaches a speed 45m/s in 9 seconds

Physics

2 answers:

PolarNik [594]3 years ago

6 0

Here's the explanation of HOW to find the answer. That's much more important than the answer.

The two BOLD lines are the parts you need to remember. They're the things you LEARN when you STUDY. If you remember them, then you can solve almost ANY acceleration problem.

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) minus (speed at the beginning)

Change in speed = (45 m/s) minus (zero)

Change in speed = 45 m/s .

Acceleration = (45 m/s) / (9 sec)

Acceleration = 5 m/s²

sasho [114]3 years ago

3 0

I believe the acceleration would be 5m/s

All you would need to do is divide the final speed by the time it took to get there. I am only about 80 sure this answer is correct, so take my advise only if you feel comfortable.

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A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the

Paha777 [63]

Answer:

c. V = 2 m/s

Explanation:

Using the conservation of energy:

$E_i =E_f$

so:

Mgh = $\frac{1}{2}IW^2 +\frac{1}{2}MV^2$

where M is the mass, g the gravity, h the altitude, I the moment of inertia of the pulley, W the angular velocity of the pulley and V the velocity of the mass.

Also we know that:

V = WR

Where R is the radius of the disk, so:

W = V/R

Also, the moment of inertia of the disk is equal to:

I = $\frac{1}{2}MR^2$

I = $\frac{1}{2}(5kg)(2m)^2$

I = 10 kg*m^2

so, we can write the initial equation as:

Mgh = $\frac{1}{2}IV^2/R^2 +\frac{1}{2}MV^2$

Replacing the data:

(5kg)(9.8)(0.3m) = $\frac{1}{2}(10)V^2/(2)^2 +\frac{1}{2}(5kg)V^2$

solving for V:

(5kg)(9.8)(0.3m) = $V^2(\frac{1}{2}(10)1/4 +\frac{1}{2}(5kg))$

V = 2 m/s

8 0

3 years ago

Which is a characteristic of the atom marked A

erastovalidia [21]

Its very dense. Hey, are you homeschooled?

6 0

3 years ago

A +1.0 nC charge is at x = 0 cm, a -1.0 nC charge is at x = 1.0 cm and a 4.0 nC at x= 2 cm. What is the electric potential energ

lesantik [10]

Answer:

- 2.7 x 10^-6 J

Explanation:

q1 = 1 nC at x = 0 cm

q2 = - 1 nC at x = 1 cm

q3 = 4 nC at x = 2 cm

The formula for the potential energy between the two charges is given by

$U=\frac{Kq_{1}q_{2}}{r}$

where r be the distance between the two charges

By use of superposition principle, the total energy of the system is given by

$U = U_{1,2}+U_{2,3}+U_{3,1}$

$U=\frac{Kq_{1}q_{2}}{0.01}+\frac{Kq_{2}q_{3}}{0.01}+\frac{Kq_{3}q_{1}}{0.02}$

$U=-\frac{9\times10^{9}\times 1\times10^{-9}\times 1\times10^{-9}}}{0.01}-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.01}+-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.02}$