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2 years ago

a jetliner goes from rest to a takeoff speed of 162 mi/h in 39.5 s. what is the magnitude of its average acceleration?

1 answer:

3 0

Acceleration = V/t

First set velocity to m/s

V = 72.42 m/s

Now we divide velocity by time

A = 72.42/39.5
A = 1.83 m/s^2

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A box weighs 100N and its base area of 2 m2. What pressure does it exert on the ground?

kotegsom [21]

Answer:

P = F/S = 100/2 =50 (N/m2)

5 0

2 years ago

Light hits a mirror at a 45Â° angle. It will be reflected at an angle _____.

Eddi Din [679]

It will be equal to 45Â°

6 0

2 years ago

An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference

kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

$K=qV$ (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

$K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J$

The kinetic energy of the particle is also:

$K=\frac{1}{2}mv^2$ (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}$

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

$|F_B|=qvBsin(\theta)$

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

$|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N$

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

$r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm$

the radius of the trajectory of the electron is 10.1 cm

6 0

3 years ago

The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve

Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

$\varepsilon _t=\dfrac{D-d}{d}$

$\varepsilon _t=\dfrac{32-30}{30}$

$\varepsilon _t=0.0667$

We know that

$n=-\dfrac{\epsilon _t}{\varepsilon _{long}}$

$\varepsilon _{long}=-\dfrac{\epsilon _t}{n}$

$\varepsilon _{long}=-\dfrac{0.0667}{0.45}$

$\varepsilon _{long}=-0.148$

So the axial pressure

$P=E\times \varepsilon _{long}$

$P=5\times 0.148$

P=740 KPa

The movement in the sleeve

$\Delta =\varepsilon _{long}\times L$

$\Delta =0.148\times 50$

Δ=7.4 mm

6 0

3 years ago

A ball is dropped from a building taking 3sec to fall to the ground. Calculate:

GenaCL600 [577]

Answer:

Vf = 29.4 m/s

h = 44.1 m

Explanation:

Data:

Initial Velocity (Vo) = 0 m/s
Gravity (g) = 9.8 m/s²
Time (t) = 3 s
Final Velocity (Vf) = ?
Height (h) = ?

==================================================================

Final Velocity

Use formula:

Vf = g * t

Replace:

Vf = 9.8 m/s² * 3s

Multiply:

Vf = 29.4 m/s

==================================================================

Height

Use formula:

$\boxed{h=\frac{g*(t)^{2}}{2}}$

Replace:

$\boxed{h=\frac{9.8\frac{m}{s^{2}}*(3s)^{2}}{2}}$

Multiply time squared:

$\boxed{h=\frac{9.8\frac{m}{s^{2}}*9s^{2}}{2}}$

Simplify the s², and multiply in the numerator:

$\boxed{h=\frac{88.2m}{2}}$

It divides:

$\boxed{h=44.1\ m}$

What is the velocity when falling to the ground?

The final velocity is <u>29.4 meters per seconds.</u>

How high is the building?

The height of the building is <u>44.1 meters.</u>

3 0

2 years ago