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3 years ago

9

An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t

ravel during the time interval?

1 answer:

Gelneren [198K]3 years ago

6 0

Answer:

$1650\:\mathrm{m}$

Explanation:

We can use the following kinematics equations to solve this problem:

$v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x$ .

Using the first one to solve for acceleration:

$132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}$ .

Now we can use the second equation to solve for the distance travelled by the airplane:

$132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}$ (three significant figures).

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Question C) needs to be answered, please help (physics)

Zarrin [17]

(a) Differentiate the position vector to get the velocity vector:

<em>r</em><em>(t)</em> = (3.00 m/s) <em>t</em> <em>i</em> - (4.00 m/s²) <em>t</em>² <em>j</em> + (2.00 m) <em>k</em>

<em>v</em><em>(t)</em> = d<em>r</em>/d<em>t</em> = (3.00 m/s) <em>i</em> - (8.00 m/s²) <em>t</em> <em>j</em>

<em></em>

(b) The velocity at <em>t</em> = 2.00 s is

<em>v</em> (2.00 s) = (3.00 m/s) <em>i</em> - (16.0 m/s) <em>j</em>

<em></em>

(c) Compute the electron's position at <em>t</em> = 2.00 s:

<em>r</em> (2.00 s) = (6.00 m) <em>i</em> - (16.0 m) <em>j</em> + (2.00 m) <em>k</em>

The electron's distance from the origin at <em>t</em> = 2.00 is the magnitude of this vector:

||<em>r</em> (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m

(d) In the <em>x</em>-<em>y</em> plane, the velocity vector at <em>t</em> = 2.00 s makes an angle <em>θ</em> with the positive <em>x</em>-axis such that

tan(<em>θ</em>) = (-16.0 m/s) / (3.00 m/s) ==> <em>θ</em> ≈ -79.4º

or an angle of about 360º + <em>θ</em> ≈ 281º in the counter-clockwise direction.

3 0

3 years ago

When you view the pendulum’s swing, it shows that at the very top of the swing KE = 0. What does that tell you about the pendulu

nadezda [96]

it tells you that it has a lot of force

5 0

3 years ago

Read 2 more answers

A pair of narrow slits, separated by 1.8 mm, is illuminated by a monochromatic light source. Light waves arrive at the two slits

Nastasia [14]

Answer:

750 nm

Explanation:

$d$ = separation of the slits = 1.8 mm = 0.0018 m

λ = wavelength of monochromatic light

$D$ = screen distance = 4.8 m

$y$ = position of first bright fringe = $\frac{1cm}{5 fringe} = \frac{0.01}{5} = 0.002 m$

$n$ = order = 1

Position of first bright fringe is given as

$y = \frac{nD\lambda }{d}$

$0.002 = \frac{(1)(4.8)\lambda }{0.0018}$

λ = 7.5 x 10⁻⁷ m

λ = 750 nm

3 0

3 years ago

A. How many atoms of helium gas fill a spherical balloon of diameter 29.6 cm at 19.0°C and 1.00 atm? b. What is the average kine

Korolek [52]

Answer:

a) 3.39 × 10²³ atoms

b) 6.04 × 10⁻²¹ J

c) 1349.35 m/s

Explanation:

Given:

Diameter of the balloon, d = 29.6 cm = 0.296 m

Temperature, T = 19.0° C = 19 + 273 = 292 K

Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa

Volume of the balloon = $\frac{4}{3}\pi(\frac{d}{2})^3$

or

Volume of the balloon = $\frac{4}{3}\pi(\frac{0.296}{2})^3$

or

Volume of the balloon, V = 0.0135 m³

Now,

From the relation,

PV = nRT

where,

n is the number of moles

R is the ideal gas constant = 8.314 kg⋅m²/s²⋅K⋅mol

on substituting the respective values, we get

1.013 × 10⁵ × 0.0135 = n × 8.314 × 292

or

n = 0.563

1 mol = 6.022 × 10²³ atoms

Thus,

0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms

b) Average kinetic energy = $\frac{3}{2}\times K_BT$

where,

Boltzmann constant, $K_B=1.3807\times10^{-23}J/K$

Average kinetic energy = $\frac{3}{2}\times1.3807\times10^{-23}\times292$

or

Average kinetic energy = 6.04 × 10⁻²¹ J

c) rms speed = $\frac{3RT}{m}$

where, m is the molar mass of the Helium = 0.004 Kg

or

rms speed = $\frac{3\times8.314\times292}{0.004}$

or

rms speed = 1349.35 m/s

5 0

3 years ago

A spring with spring constant 33N/m is attached to the ceiling, and a 4.8-cm-diameter, 1.5kg metal cylinder is attached to its l

mylen [45]

Answer:

0.423m

Explanation:

Conversion to metric unit

d = 4.8 cm = 0.048m

Let water density be $\who_w = 1000 kg/m^3$

Let gravitational acceleration g = 9.8 m/s2

Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.

Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:

$F_s + F_b = W$

Where $F_s = kx$ N is the spring force, $F_b = W_w = m_wg = \rho_w V_s g$ is the buoyancy force, which equals to the weight $W_w$ of the water displaced by the submerged portion of the cylinder, which is the product of water density $\rho_w$ , submerged volume $V_s$ and gravitational constant g. W = mg is the weight of the metal cylinder.

$kx + \rho_w V_s g = mg$

The submerged volume would be the product of cross-section area and the submerged length x

$V_s = Ax = \pi(d/2)^2x$

Plug that into our force equation and we have

$kx + \rho_w \pi(d/2)^2x g = mg$

$x(k + \rho_w g \pi d^2/4) = mg$

$x = \frac{m}{(k/g) + (\rho_w\pi d^2/4)} = \frac{1.5}{(33/9.8) + (100*\pi * 0.048^2/4)} = 0.423 m$

6 0

2 years ago