Ask question

Ask question

All categories

3 years ago

15

In 2 1/2 hours an airplane travels 1150 km against the wind. It takes 50 min to travel 450 km with the wind. Find the speed of t

he wind and the speed of the airplane in still air. g

1 answer:

Tamiku [17]3 years ago

5 0

Answer:

Velocity of airplane is 500 km/h

Velocity of wind is 40 km/h

Explanation:

$V_a$ = Velocity of airplane in still air

$V_w$ = Velocity of wind

Time taken by plane to travel 1150 km against the wind is 2.5 hours

$V_a-V_w=\frac {1150}{2.5}\\\Rightarrow V_a-V_w=460\quad (1)$

Time taken by plane to travel 450 km against the wind is 50 minutes = 50/60 hours

$V_a+V_w=\frac {450}{50}\times 60\\\Rightarrow V_a-V_w=540\quad (2)$

Subtracting the two equations we get

$V_a-V_w-V_a-V_w=460-540\\\Rightarrow -2V_w=-80\\\Rightarrow V_w=40\ km/h$

Applying the value of velocity of wind to the first equation

$V_a-40=460\\\Rightarrow V_a =500\ km/h$

∴ Velocity of airplane in still air is 500 km/h and Velocity of wind is 40 km/h

You might be interested in

A 5592 N piano is to be pushed up a(n) 3.79 m frictionless plank that makes an angle of 30.1 ◦ with the horizontal. Calculate th

otez555 [7]

Answer:

10628.87 J

Explanation:

We are given that

Force applied =F=5592 N

$\theta=30.1^{\circ}$

Displacement=D=3.79 m

We have to find the work done in sliding the piano up the plank at a slow constant rate.

Work done= $F\times displacement$

The perpendicular component of force= $FSin\theta=5592sin(30.1)=2804.45N$

Work done = $Fsin\theta\times D=2804.45\times 3.79=10628.87 J$

Hence, the work done in sliding the piano up the plank at a slow constant rate=10628.87 J

8 0

3 years ago

Read 2 more answers

The sidereal period of the moon around the Earth is 27.3 days. Suppose a satellite were placed in Earth orbit, halfway between E

icang [17]

Answer: 9.7 days

Explanation:

Applying Kepler's 3rd law, we can write the following proportion:

(Tm)² / (dem)³ = (Tsat)² / (dem/2)³

(As the satellite is placed in an orbit halfway between Erth's center and the moon's orbit).

Simplifyng common terms, and solving for Tsat, we have:

Tsat = √((27.3)²/8) = 9.7 days

7 0

3 years ago

Sarah observed that different kinds and amounts of fossils were present in a cliff behind her house. She wondered if changes in

MAXImum [283]

This is the answer: Fossil's found in Susan's yard are from prehistoric times.

6 0

3 years ago

A basketball player can jump 1.6 m off the hardwood floor. With what upward velocity did he leave the floor?

irinina [24]

First do 1.6 m (how far he jumps) 9.8 m/s (what gravity is measured at) then times 2

= 31.36

Sq root = 5.6

3 0

3 years ago

Read 2 more answers

Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

So

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

3 0

3 years ago