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GaryK [48]
3 years ago
14

An airplane flies at an altitude of 36,000 km and is traveling at a velocity of 300.0 km/h to the north, but the tailwind is 20.

0 km/h. What is the airplane's final velocity? (Remember that velocity is a vector.) vf =
Physics
2 answers:
Vitek1552 [10]3 years ago
8 0

Just a heads up, it's actually 320 km/h N. The verified answerer did not pay attention to the last sentence. Since it is about vectors, you must include the direction of the magnitude.

STALIN [3.7K]3 years ago
6 0

Vectors quantity have magnitude and direction, which means that they can be solved arithmetically with their corresponding signs. In this case, we can simply add the two velocities to get for the final velocity of the plane. But first let us determine the signs of each velocity value.

As a reference, we take the direction of the plane to be going the positive y-axis. Therefore it is going up and positive. Now take note that the wind is a “tailwind” which means that the wind is going WITH the direction of the plane, therefore it is also a positive. Now knowing that, we can add the two:

Final velocity = 300 km / hr + 20 km / hr

Final velocity = 320 km / hr

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The magnitude of the Normal Force on a
lbvjy [14]

Answer:

5. Is greater than mg, always

Explanation:

If the cone has an inclination of angle β, the sum of forces will be:

x-axis (centripetal axis):

N*sin β = m*ax  where ax is the centripetal acceleration

y-axis:

N*cos β - m*g = m*ay   where ay is the vertical acceleration. If the block starts falling down, ay will be negative. If the block starts sliding up, ay will be positive. If the block does not move up nor down, ay=0.

Solving for N:

N = \frac{m*g + m*ay}{cos \beta }

If ay is positive or zero, N will be greater than mg. If ay is negative, N will be less than mg.

If the block is sliding along a horizontal circular path (not up, nor down), ay = 0, so N will always be greater than mg.

7 0
3 years ago
What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0
3 years ago
Which of the following describes the efficiency of real machines?
Leokris [45]
Choice-B is the true one.
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Excess radiation can cause cancer. true or false
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It would be true. Excess radiation causes cancer.
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a moving billiard ball collides with an identical stationary billiard ball in an elastic collision. after the collision, the sec
MArishka [77]

A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.

<h3>Why does the first ball comes to rest after collision ?</h3>

Let m be the mass of the two identical balls.  

u1 = velocity before the collision of ball 1

u2 = 0 = velocity of second ball that is at rest

v1 and v2 are the velocities of the balls after the collision.

From the conservation of momentum,

∴ mu1 + mu2 = mv1 + mv2

∴ mu1 = mv1 + mv2

∴ u1 = v1 + v2

In an elastic collision, the kinetic energy of the system before and after collision remains same.

\frac{1}{2}  mu_1^2+0=\frac{1}{2}  mv_1^2+\frac{1}{2}  mv_2^2

∴  \frac{1}{2}  m(v_1+v_2 )^2=\frac{1}{2} mv_1^2+\frac{1}{2}mv_2^2

∴ \frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2+mv_1 v_2=\frac{1}{2}  mv_1^2+\frac{1}{2} mv_2^2

∴ mv₁v₂ = 0

  1. It is impossible for the mass to be zero.
  2. Because the second ball moves, velocity v2 cannot be zero.
  3. As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.
<h3>What is collision ?</h3>

An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.

Can learn more about elastic collision from brainly.com/question/12644900

#SPJ4

3 0
1 year ago
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