1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kumpel [21]
3 years ago
10

Plz help..During takeoff a plane accelerates at 4m/s^2 and takes 40s to reach takeoff speed.

Physics
1 answer:
matrenka [14]3 years ago
3 0

Answer:

The velocity of the plane at take off is 160 m/s.

The distance travel by the plane in that time is 3200 meter.

Explanation:

Given:

Acceleration, a = 4 m/s²

Time, t = 40 s

u = 0 i .e initial velocity

To Find:

velocity , v = ?

distance , s =?

Solution:

we have first Kinematic equation

v = u + at

∴ v = 0 + 4×40

∴ v = 160 m/s

Now by Third Kinematic equation

s = ut + \frac{1}{2}at^{2}

∴ s = 0 + 0.5 × 4× 40²

∴ s = 3200 meter

You might be interested in
According to the _______ the amount of energy in the universe doesn't change.
balandron [24]
The answer is B, Law of Kinetic Energy
6 0
3 years ago
Read 2 more answers
Why do astronauts need to wear pressurized suits in space?
SashulF [63]
The correct answer is C , because the space is vacuum and his body can explode and for this reason,  the astronaut need a special costum to be protected. It's the same on the moon, because there is no atmosphere 
6 0
3 years ago
A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar
Artist 52 [7]

Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = v_{oy} t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = v_{oy} t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

8 0
4 years ago
X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m

b) The change in photon energy is given by:

\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\

E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV

5 0
4 years ago
8) In a bumper car arena, two cars of equal mass are heading straight toward each other. The orange car is traveling at a veloci
Ghella [55]

The answer is D) Niether . I just took the test

3 0
4 years ago
Other questions:
  • Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of 45.0°
    12·1 answer
  • A worker on the roof of a house drops his 0.58 kg hammer, which slides down the roof at constant speed of 6.69 m/s. The roof mak
    15·1 answer
  • When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees
    14·1 answer
  • What are the similarities in the 3 types of waves
    6·1 answer
  • A football player kicks a 0.94 kg football with a force of 2.4 N. Calculate the acceleration of the football as the player kicks
    6·2 answers
  • Help please <br> hhjshwjsjejjenrhrhfhhfheisiw
    6·1 answer
  • Abigail runs one complete lap (400m) around the track, while Gabi runs a 50 meter dash in a straight line. Which runner had a gr
    6·2 answers
  • After watching the video below and based on your personal experiences, is there a difference
    6·1 answer
  • You can use a system of equations to graph and solve the polynomial equation 3 x cubed + x = 2 x squared + 1. Which statement is
    12·1 answer
  • Please, can somebody help me with this project? I'll give brainlest for the best answer! (Do not answer if you don't know or onl
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!