1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
natulia [17]
3 years ago
12

A bird watcher meanders through the woods, walking 1.46 km due east, 0.123 km due south, and 4.24 km in a direction 52.4° north

of west. The time required for this trip is 1.285 h. Determine the magnitudes of the bird watcher's (a)displacement and (b) average velocity.

Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
3 0

Answer:

(a). The displacement is 3.11 m.

(b). The average velocity is 2.42 km/hr.

Explanation:

Given that,

A bird watcher meanders through the woods, walking 1.46 km due east, 0.123 km due south, and 4.24 km in a direction 52.4° north of west.

(a). We need to calculate the displacement

Using Pythagorean theorem

D=\sqrt{(OA)^2+(AB)^2}

D=\sqrt{(1.46-4.24\sin52.4)^2+(0.123-4.24\cos52.4)^2}

D=3.11\ km

(b). We need to calculate the average velocity

Using formula of average velocity

v_{avg}=\dfrac{D}{T}

Where, D = displacement

T = time

Put the value into the formula

v_{avg}=\dfrac{3.11}{1.285}

v_{avg}=2.42\ km/hr

Hence, (a). The displacement is 3.11 m.

(b). The average velocity is 2.42 km/hr.

You might be interested in
An electron with a velocity of 10^7 m/s enter into a region of magnetic flux density of 0.10T,the angle between the direction of
wolverine [178]

Answer:

F = q v B sin Θ       describes force on the electron

F = m v^2 / R          describes force required to keep electron in circular path

q v B sin Θ = m v^2 / R

R = m v / (q B sin Θ)

R = 9.1E-31 * 1E7 / (.1 * 1.6E-19 * .423)

R = 9.1 / 6.77 * E-3 = .00134 m = .134 cm

6 0
2 years ago
PHYSICS PLEASE HELP 80 PTS!!!!!!
VladimirAG [237]

1.  B) 32 m/s

The speed of the baseball is given by

v=\frac{d}{t}

where

d = 48 m is the distance travelled

t = 1.5 s is the time taken

Substituting,

v=\frac{48}{1.5}=32 m/s

2. C) 12 m/s

The average velocity is given by

v=\frac{d}{t}

where

d = 60 - 0 = 60 m is the displacement

t = 5 s is the time interval

Substituting,

v=\frac{60}{5}=12 m/s

3. Missing diagram

4. D) velocity

In fact, velocity is defined as the ratio between the displacement (\Delta s) and the time interval (\Delta t) needed to achieve that change in position:

v=\frac{\Delta s}{\Delta t}

Similarly, acceleration is defined as the ratio between the change in velocity (\Delta v) and the time interval (\Delta t):

a=\frac{\Delta v}{\Delta t}

Comparing the two equations, we see that displacement is to velocity as velocity is to acceleration.

5. C) 5/1 cm/year

We know that the ridge moves 25 cm in 5 years, so we have:

- Displacement (d): d = 25 cm

- Time interval (t): t = 5 y

Using the equation for the velocity:

v=\frac{d}{t}=\frac{25 cm}{5 y}=5 cm/y

6.  D

The missing graph is attached. Each graph represents a distance (on the y-axis, measured in metres) versus a time (on the x-axis, measured in seconds). Therefore, the speed in each graph can be simply calculated from the slope of the curve, since speed is the ratio between distance and time:

v=\frac{\Delta y}{\Delta x}

For each graph:

A. \frac{\Delta y}{\Delta x}=\frac{8.6-2}{10}\sim 0.66

B. \frac{\Delta y}{\Delta x}=\frac{5.7-0}{10}\sim 0.57

C. \frac{\Delta y}{\Delta x}=\frac{9.6-3}{10}\sim 0.66

D. \frac{\Delta y}{\Delta x}=\frac{7.5-0}{5}\sim 1.5

So we can say that the correct graph must be graph D, the closest to 1.57.

7. B) The car has come to a stop and has zero velocity

The missing graph is attached.

The graph shows the position of the car versus time. From the graph, we can see that in segment B, the position of the car does not change: this means that its velocity is zero, since the velocity is defined as the ratio between the change in position (displacement) and the time interval:

v=\frac{\Delta s}{\Delta t}

And since the displacement is zero, the velocity is also zero.

8.  C) from 4.5 to 5.0 seconds

Missing graph is attached.

The graph represents the distance covered versus the time taken: therefore, the speed of the ball can be evaluated from the slope of the curve.

The only region in which the slope of the curve is zero is between 4.5 seconds and 5.0 seconds: therefore, this is the region where the soccer ball is not moving.

9. D) diagonal line with varying slope, from 3 to 4.

The table of data is:

Time (s)  0 1 2 3 4

Distance (m) 0 3 6 12 16

We want to know how the distance/time graph would like like. We have:

- At t = 1, the distance is d=3, so the slope is \frac{d}{t}=\frac{3}{1}=3

- Similarly, at t = 2:  \frac{d}{t}=\frac{6}{2}=3

- Instead, at t =3, the slope is \frac{d}{t}=\frac{12}{3}=4

- Similarly, at t=4,\frac{d}{t}=\frac{16}{4}=4

So the correct answer is D.

10. D) To move in a circle requires an acceleration and therefore a net force. This force is supplied by the slingshot.

For an object in circular motion, the velocity constantly changes (because the direction changes): this means that an acceleration is required (towards the centre of the circle), and therefore a force must be exerted (also towards the centre of the circle) to keep the object in the circular path.

As soon as this force is removed, the object is "free" to leave the circular trajectory and continue its motion following a straight path at constant velocity, according to Newton's first law.

11. B) The acceleration will become 1/4 as much.

Re-arranging Newton's second law formula,

F=ma \rightarrow a = \frac{F}{m} (1)

The force exerted on the satellite is actually the force of gravity:

F=\frac{Gm M}{d^2} (2)

Substituting (2) into (1),

a=\frac{F}{m}=\frac{GM}{d^2}

We see that there is no dependance on the mass of the satellite (m), but only on the distance. Since in this problem the distance is doubled (d'=2d), the new acceleration will be:

a'=\frac{GM}{(2d)^2}=\frac{1}{4}(\frac{GM}{d^2})=\frac{a}{4}

12. C) divide: distance : velocity

Velocity is distance divided by time:

v=\frac{d}{t}

Multiplying by t on both sides,

v\cdot t = \frac{d}{t} \cdot t \rightarrow vt = d

and dividing by v on both sides,

\frac{vt}{v}=\frac{d}{v} \rightarrow t = \frac{d}{v}

13. A) -8 km/h

Let's call:

v_k = +5 km/h the velocity of the kite

v_p = -3 km/h your velocity (opposite direction)

In your frame of reference, the velocity of the kite will be:

v'_k = v_p - v_k = -3 -(+5) = -8 km/h

3 0
4 years ago
Read 2 more answers
If you were part of a track and field team, what event would you like to participate in? And why? Please provide at least 3 comp
Serjik [45]

I wouldn't participate in any event, I'd leave the track and field team.

8 0
3 years ago
What distance will a truck trvael in 3 hours at an average speed of 50 per hour​
FromTheMoon [43]
The truck would of went 150 miles
5 0
3 years ago
Suppose a 60-kg boy and a 41-kg girl use a massless rope in a tug-of-war on an icy, resistance-free surface. If the acceleration
Digiron [165]

Answer:

Approximately 2.05\; {\rm m\cdot s^{-2}}.

Explanation:

The net force on the girl would be:

\begin{aligned}m(\text{girl}) \, a(\text{girl}) &= 41\; {\rm kg} \times 3.0\; {\rm m\cdot s^{-2}} \\ &= 123.0\; {\rm N} \end{aligned}.

Under the assumptions, the net force on this girl would be equal to the tension force in the rope. All other forces on the girl would be balanced.

In other words, the tension force that the rope exerted on the girl would be 123.0\; {\rm N}. The girl would exert a reaction force on the rope at the same magnitude (123.0\; {\rm N}\!) in the opposite direction. This force would translate to a 123.0\; {\rm N}\!\! force on the boy towards the girl.

Under similar assumptions, the net force on the boy would also be 123.0\; {\rm N}. Since the mass of the boy is m(\text{boy}) = 60\; {\rm kg}, the acceleration of the boy would be:

\begin{aligned}a(\text{boy}) &= \frac{(\text{net force})}{m(\text{boy})} \\ &= \frac{123.0\; {\rm N}}{60\; {\rm kg}} \\ &= 2.05\; {\rm m\cdot s^{-2}}\end{aligned}.

3 0
2 years ago
Other questions:
  • What is not an extrinsically motivated action
    5·1 answer
  • Why does sound move fastest in solid mediums?
    6·2 answers
  • What is the chemical formula for the binary compound nitrogen dioxide? A) (NO)2 B) 1N2Ox C) NO2 D) nitro1oxide2
    6·1 answer
  • An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/
    13·1 answer
  • In the visible spectra of stars, absorption lines of hydrogen are produced when atoms are excited from n = 2 to higher levels (t
    5·1 answer
  • What are the general tasks that a person can perform if he or she is considered functional? (4 points)
    6·2 answers
  • Which of the following is the most appropriate unit to describe the time you can drive a car based on the amount of gas?
    9·2 answers
  • A piece of aluminium with mass 1 kg and density 2700 kg/m3 is suspended from a string and then completely immersed in a containe
    8·1 answer
  • Definition of measurements?​
    14·1 answer
  • A stone is thrown vertically upward with an initial speed of 24.7 m/s. Neglect air resistance. The speed of the stone when it is
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!