answer.
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
xxTIMURxx [149]
2 years ago
14

As a 15000 kg jet plane lands on an aircraft carrier, its tail hook snags a cable to slow it down. The cable is attached to a sp

ring with spring constant 60000 N/m.If the spring stretches 30 m to stop the plane, what was the plane's landing speed?
Physics
1 answer:
RideAnS [48]2 years ago
5 0

Answer:

60 m/s

Explanation:

From the law of conservation of energy,

The kinetic energy of the plane = Energy of store in the spring when the plane lands.

1/2mv²  = 1/2ke²

making v the subject of the equation.

v = √(ke²/m).................... Equation 1

Where  v = the plane landing speed, k = spring constant, e = extension. m = mass of the plane.

Given: m = 15000 kg, k = 60000 N/m, e = 30 m.

Substitute into equation 1

v = √(60000×30²/15000)

v = √(4×900)

v = √(3600)

v = 60 m/s.

Hence the plane's landing speed = 60 m/s

You might be interested in
The blue curve is the plot of the data. The straight orange line is tangent to the blue curve at t = 40 s. A plot has the concen
Sever21 [200]

Answer:

  0.00325 moles/liter/second

Explanation:

The tangent line has a slope of (y2 -y1)/(x2 -x1) = (0.35-0.48)/(40-0) = -0.00325.

The rate of the reaction is about 0.00325 moles/liter/second.

_____

This is the rate of decrease of the concentration of A.

5 0
2 years ago
Read 2 more answers
A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at 11.0
Serjik [45]

Answer:

The airspeed must be 7.78 m/s for the rectangular plate kept at 30°.

Explanation:

By looking at the images below wee see that the airspeed on one side of the rectangular plate decreases the statical pressure over this side. Since over the downside, the pressure still bein the atmospheric pressure. This difference in pressure produces a lift force in the plate. The list force is the net force obtained between the difference of the forces that produce the pressure over the upside and the downside:

F_{lift}=F_{up} - F_{dw}=0.5*p*V^2

Where up and down relate to what movement the forces produce. And p and V are the respective air density and velocity.

When the plate is kept horizontal the lift force balance the moment due to the weight of the plate and considering that both forces act at the same point:

F_{lift}=0.5*p*V^2=W

By replacing the known values it is possible to find the plate's weight:

F_{lift}=0.5*1.2 \frac{kg}{m^{3}}*(11 m/s)^2=W

W=72.6 N

When the plate kept to 30° from the vertical the moment equation balance is written as:

F_{lift}=0.5*p*V^2=W*sen(30\°)

The sine of 30° is due to the weight is 30° oriented, therefore the new value for the airspeed is:

V=\sqrt(W*sen(30\°)/0.5p)

V=\sqrt(\frac{72.6 N * 0.5}{0.5*1.2 kg/m^3})

V=\sqrt(60.5 \frac{N}{kg/m^3})

V=\sqrt(60.5 \frac{kg.m/s^2}{kg/m^3})

V=\sqrt(60.5 \frac{m^2}{s^2})

V= 7.78 m/s

7 0
2 years ago
You hang an object with mass m = 0.380 kg from a vertical spring that has negligible mass and force constant k = 60.0 N/m. You p
joja [24]

Answer:

a) 0.500 s

b) greater than 0.500 s

c) greater than 0.500 s

Explanation:

The time period of an oscillating spring-mass system is given by:

T=2\pi \sqrt{\frac{m}{k}}

where, m is the mass and k is the spring constant.

a) As the period of oscillation does not depend on the distance by which the mass is pulled, the period would remain same as 0.500 s for the given system.

b) As the period varies inversely with the square root of spring constant, so with the decrease in the spring constant, the period would increase. So, the new period would be greater than 0.500 s.

c) As the period varies directly with the square root of mass, so with the increase in mass, the period will also increase. The new period will be greater than 0.500 s.

8 0
2 years ago
The speed of light in benzene is 2.00×108 m/s. what is the index of refraction of benzene?
Klio2033 [76]
The index of refraction of a material is the ratio between the speed of light in vacuum, c, and the speed of light in that material, v:
n= \frac{c}{v}
where the speed of light in vacuum is c=3 \cdot 10^8 m/s. The speed of light in benzene is v=2.00 \cdot 10^8 m/s, so we can use the previous relationship to find the refractive index of benzene:
n= \frac{3 \cdot 10^8 m/s}{2.00 \cdot 10^8 m/s}=1.5
7 0
2 years ago
A major disturbance that caused the ecosystem to stabilize at a new equilibrium?
blsea [12.9K]

Answer: If you remove all of the hawks, it stabilizes an a new equilibrium.

Explanation: For the ecosystem to stabilize at a new equilibrium, the tertiary consumer (hawks) need to be totally removed from the ecosystem. Since there's no tertiary consumer to checkmate the density of other components of the food chain, it will make form a balanced ecosystem.

7 0
2 years ago
Other questions:
  • A bag is dropped from a hovering helicopter. The bag has fallen for 2.0 s. What is the bag’s velocity? How far has the bag falle
    7·1 answer
  • a bobsled has a momentum of 1500 kg*m/s to the south. Air resistance reduces its momentum to 750 kg*m/s to the south. What impul
    11·2 answers
  • a pebble is dropped down a well and hits the water 1.5 seconds later. using the equations for motion with constant acceleration,
    7·2 answers
  • A body A of mass 1.5kg, travelling along the positive x-axis with speed 4.5m/s, collides with another body B of mass 3.2kg which
    14·1 answer
  • The speed v of a sound wave traveling in a medium that has bulk modulus b and mass density ρ (mass divided by the volume) is v=b
    13·1 answer
  • Which description best explains a molecular bonding?
    5·1 answer
  • On a day when the barometer reads 75.23 cm, a reaction vessel holds 250 mL of ideal gas at 20 celsius. An oil manometer ( rho= 8
    12·1 answer
  • Suppose you are talking by interplanetary telephone to your friend, who lives on the Moon. He tells you that he has just won a n
    11·1 answer
  • You should have observed that there are some frequencies where the output is stronger than the input. Discuss how that is even p
    15·1 answer
  • 3. The expression 0.62 x10^3 is equivalent to...
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!