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
Kitty [74]
3 years ago
15

If you walk 1.4 km north and then 4.8 km east, what are the magnitude and direction of your resultant displacement?

Physics
1 answer:
olchik [2.2K]3 years ago
8 0

A


North and east are perpendicular directions hence we can use them as the basis of our coordinate system. The coordinates of this point can then simply be represented as

P = 1.4 N + 4.8 E (km)

The magnitude can be given by the right triangle formula :

P = \sqrt{N^2 + E^2}

= 5 km

Direction  = atan(N/E) = 16.3  deg north of east = 73.7 deg east of north


You might be interested in
Is metal denting a physical change
Scilla [17]
No it is a chemical change
7 0
3 years ago
A force of 10N acts on a car for 3 seconds. The impulse imparted on the car is
ss7ja [257]

Answer:

30N*s

Explanation:

Given the following data;

Force = 10N

Time = 3 seconds

To find the impulse;

Impulse = force * time

Substituting into the equation, we have;

Impulse = 10 * 3

Impulse = 30Ns

7 0
3 years ago
If a sled has a mass of 4kg what is the force of gravity on the sled?
ale4655 [162]

Answer:

Explanation:

Gravity pulls everything down at the same rate of 9.8 m/s/s. If you're looking for the normal force, which is the same as the weight of the object, we'll find that, just in case.

w = mg which says that the normal force/weight of an object is equal to its mass times the pull of gravity:

w = 4.0(9.8) so

w = 39N

7 0
3 years ago
After a collision between two different massed objects; the larger objects accelerate at a faster rate than the smaller object?
Nitella [24]

Answer: Things continue doing what they are doing unless a force is applied to it. Objects have a natural tendency to resist change. This is INERTIA. Heavier objects (objects with more mass) are more difficult to move and stop. Heavier objects (greater mass) resist change more than lighter objects, so true

Explanation:

Pushing a bicycle or a Cadillac, or stopping them once moving. The more massive the object (more inertia) the harder it is to start or stop. The Cadillac has more of a tendency to stay stationary (or continue moving), and resist a change in motion than a bicycle.

6 0
3 years ago
Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W
jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

T^{2}=\frac{4\pi^{2}}{GM}r^{3}    (1)

Where;

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24}kg is the mass of the Earth

r  is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period T_{X} is:

T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}    (2)

Where r_{X}=1.2(10)^{6}m

T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}    (3)

T_{X}=413.712 s    (4)

For satellite Y, the orbital period T_{Y} is:

T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}    (5)

Where r_{Y}=1.9(10)^{5}m

T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}    (6)

T_{Y}=26.064 s    (7)

This means T_{X}>T_{Y}

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

V_{X}=\sqrt{\frac{GM}{r_{X}}} (8)

V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}

V_{X}=18224.783 m/s (9)

<u>For Satellite Y:</u>

V_{Y}=\sqrt{\frac{GM}{r_{Y}}} (10)

V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}

V_{Y}= 45801.13 m/s (11)

This means V_{Y}>V_{X}

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0
3 years ago
Read 2 more answers
Other questions:
  • The time needed for a wave to make one complete cycle is its
    10·1 answer
  • What deep ocean feature forms at subduction zones
    9·1 answer
  • Can someone help me PLEASE? I need to finish this because I have more homework and I need to go to bed
    7·1 answer
  • 50 points and brainleist
    7·2 answers
  • A circular flat coil that has N turns, encloses an area A, and carries a current i, has its central axis parallel to a uniform m
    9·1 answer
  • When you push a toy car it eventually stops this is due to something called
    15·1 answer
  • True or false: Increasing the Young’s modulus of a beam in bending will cause it to deflect less.
    13·2 answers
  • А masd<br>Of 500kg a raised to a height of 6m In 30s<br>Find (a) Workdone .​
    13·1 answer
  • In this photograph, a soccer player is about to kick the ball. Use the situation to explain that when two objects interact, the
    12·1 answer
  • What is the speed of an object at rest?<br> a. 0.0 m/s<br> c. 9.8 m/s<br> b. 1.0 m/s<br> d. 9.81 m/s
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!