Ask question

Ask question

All categories

2 years ago

7

A plane travels 4.0 km at an angle of 25◦ to the ground, then changes direction and travels 10 km at an angle of 16◦ to the grou

nd. What is the magnitude of the plane's total displacement? Answer in units of km.

1 answer:

Maslowich2 years ago

5 0

Answer:

6.1 km

Explanation:

Given that a plane travels 4.0 km at an angle of 25◦ to the ground, then changes direction and travels 10 km at an angle of 16◦ to the ground. What is the magnitude of the plane's total displacement? Answer in units of km

The magnitude of the total displacement D can be calculated by using cosine formula

Ø = 25 - 16 = 9 degree

D^2 = 4^2 + 10^2 - 2 × 4 × 10 × cos 9

D^2 = 16 + 100 - 80cos9

D^2 = 116 - 79.02

D = sqrt( 36.98)

D = 6.1 m

Therefore, the magnitude of the plane's total displacement is 6.1 km

You might be interested in

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him

nataly862011 [7]

Answer:

a) $u_c=0\ m.s^{-1}$ & $m_c.v_c=m_b.v_b\times \cos\theta$

b) $v_c=0.0566\ m.s^{-1}$

c) $p_e=2.9218\ kg.m.s^{-1}$

Explanation:

Given:

mass of the book, $m_b=1.35\ kg$

combined mass of the student and the skateboard, $m_c=97\ kg$

initial velocity of the book, $v_b=4.61\ m.s^{-1}$

angle of projection of the book from the horizontal, $\theta=28^{\circ}$

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

$u_c=0\ m.s^{-1}$

where:

$u_c=$ initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

$m_c.v_c=m_b.v_b\times \cos\theta$

where:

$v_c=$ final velcotiy of the student after throwing the book

b)

$m_c.v_c=m_b.v_b\times \cos\theta$

$97\times v_c=1.35\times 4.61\cos28$

$v_c=0.0566\ m.s^{-1}$

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

$p_e=m_b.v_b\sin\theta$

$p_e=1.35\times 4.61\times \sin28^{\circ}$

$p_e=2.9218\ kg.m.s^{-1}$

6 0

2 years ago

High energy waves have what

GarryVolchara [31]

High energy waves have Gamma rays

3 0

2 years ago

Read 2 more answers

A 250–g piece of gold is at 19 °C. 5.192 kJ of energy is added to it by heat. The specific heat of gold is 129 J/(kg·°C). Calcul

algol [13]

Answer:

A. DT is given by Q= MCs DT

m = mass of the substances

Cs= is it's specific heat capacity

Ck= <u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>Q</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

Mk ×DTk

=<u>2</u><u>5</u><u>0</u><u> </u><u>×</u><u> </u><u>9</u><u> </u><u>×</u><u> </u><u>5</u><u> </u><u> </u>

129

=Dt = 180.1085271

answer is 180degree C.

Explanation:

B. = <u>2</u><u>5</u><u>×</u><u>1</u><u>0</u> ×100

1.082

=<u>2</u><u>5</u><u>0</u><u>0</u>

1.082

= 23105.360 g/kj.

7 0

2 years ago

a golfer is on the edge of a 12.5m bluff overlooking the 18th hole which is located 67.1m from the base of the bluff. she launch

Alja [10]

Answer: $26.359 m/s$

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

$x=V_{o}cos \theta t$ (1)

$y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}$ (2)

$V=V_{o}-gt$ (3)

Where:

$x=67.1 m$ is the horizontal distance traveled by the golf ball

$V_{o}$ is the golf ball's initial velocity

$\theta=0\°$ is the angle (it was a horizontal shot)

$t$ is the time

$y=0 m$ is the final height of the ball

$y_{o}=12.5 m$ is the initial height of the ball

$g=9.8 m/s^{2}$ is the acceleration due gravity

$V$ is the final velocity of the ball

Let's begin by finding $t$ from (2):

$t=\sqrt{\frac{2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}$ (5)

$t=1.597 s$ (6)

Substituting (6) in (1):

$67.1 m=V_{o} cos(0\°) 1.597 s$ (7)

Finding $V_{o}$ :

$V_{o}=42.01 m/s$ (8)

Substituting $V_{o}$ in (3):

$V=42.01 m/s-(9.8 m/s^{2})(1.597 s)$ (9)

Finally:

$V=26.359 m/s$

6 0

2 years ago

Our Moon a. was a separate small planet before it was captured by the Earth's gravity. b. spun off the fast-spinning Earth durin

Nadya [2.5K]

Answer: what do u need help with

Explanation:

The giant-impact hypothesis, sometimes called the Big Splash, or the Theia Impact, suggests that the Moon formed from the ejecta of a collision between the ...

5 0

2 years ago

Read 2 more answers