Ask question

Ask question

All categories

3 years ago

10

At take off, a plane flies 100 km north before turning to fly 200 km east. How far is its destination from where the plane took

off?
Question 3 options:

200 km

300 km

173 km

224 km

2 answers:

koban [17]3 years ago

7 0

224 km I did the quiz today

astra-53 [7]3 years ago

7 0

The destination is 224 km from where the plane took off. Therefore correct option is (d).

First the plane turns 100 km North, and than 200 km East. Since both the directions are perpendicular to each other, therefore we can apply the Pythagoras theorem to calculate the distance between the destination and the point where plane took off

$D^{2}$ =100^{2}+200^{2}

D=223.60 km=224 km

Therefore, The destination is 224 km from where the plane took off

You might be interested in

A proton moving at 3.90 106 m/s through a magnetic field of magnitude 1.80 T experiences a magnetic force of magnitude 7.20 10-1

VARVARA [1.3K]

Answer:

$\theta=40^0$

Explanation:

The magnitude of the magnetic force is

$F_m=evB\sin\theta$

To find the angle, we make $\sin\theta$ subject of the formula

$\implies \sin\theta=\frac{F_m}{evB}=\frac{7.20\times 10^{-13}}{1.6\times 10^{-19}\times 3.90\times 10^6\times 1.80}$

$\implies \sin\theta=0.641025641$

$\therefore \theta=\sin^{-1}=39.8683^0\\\implies \theta\approxeq 40^0$

8 0

3 years ago

A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.

andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows:

$2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}$

$13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}$

Let is clear $v_{1,f}$ in first equation:

$0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}$

$v_{1,f} = 5.86-2.25\cdot v_{2,f}$

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

$13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}$

$13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}$

$13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}$

$2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0$

$2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0$

There are two solutions:

$v_{2,f} = 0\,\frac{m}{s}$ or $v_{2,f} = 3.606\,\frac{m}{s}$

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: ( $v_{2,f} = 3.606\,\frac{m}{s}$ )

$v_{1,f} = 5.86-2.25\cdot (3.606)$

$v_{1,f} = -2.254\,\frac{m}{s}$

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

3 0

3 years ago

When does a ball that is thrown vertically upwards reach its maximum speed?

ladessa [460]

As soon as you let go of it it is at its max speed because gravity is constantly pulling it down

4 0

3 years ago

2. What do you understand by balanced and unbalanced force

labwork [276]

Answer:

forces that are equal in size and opposite in direction. Balanced forces do not result in any change in motion. unbalanced. forces: forces applied to an object in opposite directions that are not equal in size. Unbalanced forces result in a change in motion.

$\\$

hope helpful ~

8 0

2 years ago

Why do nations should establish a set of rules and principles for responsible lunar/moon explortions? ASAP pleaseeee :(

Marianna [84]

So that we do not contaminate it with microorganisms or garbage or other human stuff.

5 0

2 years ago