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Lena [83]
3 years ago
5

An airplane is flying from Dallas, Texas to Pensacola, Florida. Flying at maximum velocity, it encounters strong winds moving at

half the speed of the plane in the opposite direction. How long will it take the plane to arrive in Pensacola, relative to the original arrival time?
F. It will take half the time.
G. It will take twice as long.
H. It will take 1.5 times as long.
I. It will take three times as long.
Physics
1 answer:
jenyasd209 [6]3 years ago
4 0

Answer:

G. It will take twice as long.

Explanation:

Let's call v the original speed of the plane and d the distance between Dallas and Pensacola. The time the plane originally takes to complete the flight is

t=\frac{d}{v}

In this problem, we are told that the plane encounters wind moving at half of its speed: \frac{v}{2}, in the opposite direction. This means that the new speed of the plane is

v'=v-\frac{v}{2}=\frac{v}{2}

And so, the time the plane takes now to complete the flight is

t'=\frac{d}{v/2}=2\frac{d}{v}=2t

So, the plane takes twice the time as before.

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A hockey puck is sliding at a constant rate of 2m/s on a frictionless surface. How fast will the puck be moving after 10 sec
AnnyKZ [126]
2m/s because the hockey puck is traveling at a constant speed ( acceleration is 0 ). Unless something acts on the hockey puck it will travel 2 m/s forever.
5 0
3 years ago
If you were given distance and period of time, what can you calculate?
antiseptic1488 [7]

If you were given distance & period of time, you would be able to calculate the speed.

Hope this helps!

3 0
3 years ago
A bicycle rider has a speed of 19.0 m/s at a height of 55.0 m above sea level when he begins coasting down hill. The mass of the
lukranit [14]

Answer:

The mechanical energy of the rider at any height will be 6.34 × 10⁴ J.

Explanation:

Hi there!

The mechanical energy of the rider is calculated as the sum of the gravitational potential energy plus the kinetic energy. Since there are no dissipative forces (like friction), the mechanical energy of the rider at a height of 55.0 m above the sea level will be the same at a height of 25.0 m (or at any height), because the loss in potential energy will be compensated by a gain in kinetic energy, according to the law of conservation of energy.

Then, calculating the potential and kinetic energy at 55.0 m and 19 m/s, we can obtain the mechanical energy that will be constant:

Mechanical energy = PE + KE

Where:

PE = potential energy.

KE = kinetic energy.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the object.

g = acceleration due to gravity.

h = height.

Then, the potential energy of the rider will be:

PE = 88.0 kg · 9.81 m/s² · 55.0 m = 4.75 × 10⁴ J

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

Where "m" is the mass of the object and "v" its velocity. Then:

KE = 1/2 · 88.0 kg · (19.0 m/s)²

KE = 1.59 × 10⁴ J

The mechanical energy of the rider will be:

Mechanical energy = PE + KE = 4.75 × 10⁴ J + 1.59 × 10⁴ J = 6.34 × 10⁴ J

This mechanical energy is constant because when the rider coast down the hill, its potential energy is being converted into kinetic energy, so that the sum of potential energy plus kinetic energy remains constant.

5 0
3 years ago
You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an
a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, \mu=6\times 10^{-2}

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

KE=PE+W

\dfrac{1}{2}mv^2=mgh+W

W is the work done by the friction.

W=f\times d

f=\mu mg\ cos\theta

W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}

W=\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}

\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

8 0
3 years ago
A 70-kg ice skater is standing still on the ice, when a friend tosses him a 10-kg ball that is traveling at +8 m/s. If he catche
Misha Larkins [42]

Answer:+1.25 m/s

Explanation:

Given

mass of ice skater M=70 kg

mass of ball m=10 kg

the initial velocity of the ball  u_1=+8\ m/s

Conserving linear momentum

M\times0+m\timesu_1=(M+m)v\quad \quad [v=\text{combined velocity of skater and ball}]

v=\dfrac{10\times10}{80}=+1.25\ m/s

Therefore the velocity of the person holding the ball is 1.25 m/s

This collision represents the perfectly inelastic collision where particles stick together after the collision.

7 0
3 years ago
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