20 POINTS 20 POINTS On a bright sunny day you decide to take a walk. You begin at your home and walk 1000 meters to an ice cream shop in 10 minutes. You spend 15 minutes ordering your ice cream and then return home. Since you have an ice cream cone in your hand, it takes 20 minutes to walk home. 1. Find the total displacement and total distance traveled. 2. Find your average speed and average velocity in meters/min. 3. Find your average speed and average velocity in meters/sec.
Answer:
A) Total Distance = 2000 m and Total displacement = 0 m
B) Average Speed = 44.44 m/min and Average Velocity = 0 m/min
C) Average Speed = 0.7407 m/s and Average velocity = 0 m/s
Explanation:
A) Distance to reach ice cream shop from home = 1000 meters
Therefore distance to get back home would also be 1000 meters.
Total distance traveled = 1000 + 1000 = 2000 metres
Since journey started at home and ended at home, then total displacement = 0 metres.
B) Average speed = Total distance/total time.
Total time = 10 + 15 + 20 = 45 minutes
Since total distance = 2000 m
Then;
Average speed = 2000/45
Average speed = 44.44 m/min
Average velocity = Total displacement/total time
Average velocity = 0/45 = 0 m/min
C) We now want answers in B to be in m/s.
Total time = 45 minutes.
From conversion, 60 seconds make 1 minute. Thus, 45 minutes = 45 × 60 = 2700 seconds
Thus;
Average Speed = 2000/2700
Average Speed = 0.7407 m/s
Average displacement = 0/2700 = 0 m/s
I point
Is the car's speed increasing or decreasing with time?
not enough information
decrease
increase
constant
Answer:
It's increasing with time
Find the magnitude of the magnetic field ∣∣B⃗ (r)∣∣ inside the cylindrical resistor, where r is the distance from the axis of the cylinder, in terms of i, r, r0, l, and other given variables. You will also need π and μ0. Ignore fringing effects at the ends of the cylinder.
Answer:
The magnetic field inside the cylindrical resistor is [tex]\dfrac{\mu_{0}ir}{2\pir_{0}^2}[/tex]
Explanation:
Given that,
Distance from the axis of the cylinder = r
We need to calculate the magnetic field inside the cylindrical resistor
Using formula of magnetic field
[tex]\oint{\vec{B}\cdot\vec{dl}}=\mu_{0}i_{encl}[/tex]
[tex]B\cdot(2\pi r)=\mu_{0}\dfrac{i\pir^2}{\pi r_{0}^2}[/tex]
Where, r₀ = radius
r = distance
i = current
[tex]|\vec{B}(r)|=\dfrac{\mu_{0}ir}{2\pir_{0}^2}[/tex]
Hence, The magnetic field inside the cylindrical resistor is [tex]\dfrac{\mu_{0}ir}{2\pir_{0}^2}[/tex]
From the gravitational law, calculate the weight W (gravitational force with respect to the earth) of a 70 kg spacecraft traveling in a circular orbit 275 km above the earth's surface. Express W in Newtons and pounds.
Answer:
The value in Newton is [tex]W = 631.92 \ N[/tex]
The value in pounds is [tex]W = 142 \ lb[/tex]
Explanation:
From the question we are told that
The mass of the spacecraft is [tex]m = 70 \ kg[/tex]
The distance above the earth is [tex]d = 275 \ km = 275000 \ m[/tex]
Generally the gravitational force with respect to the earth is mathematically represented as
[tex]W = \frac{G * m * m_e}{ (d + r_e)^2}[/tex]
Here [tex]m_e[/tex] is the mass of earth with value [tex]m_e = 5.978 *10^{24} \ kg[/tex]
[tex]r_e[/tex] is the radius of the earth with value [tex]r_e = 6371 \ km = 6371000 \ m[/tex]
G is the gravitational constant with value [tex]G = 6.67 *10^{-11} \ m^3/ kg\cdot s^2[/tex]
So
[tex]W = \frac{ 6.67 *10^{-11} * 70 * 5.978 *10^{24}}{ (275000 + 6371000)^2}[/tex]
[tex]W = 631.92 \ N[/tex]
Converting to pounds
[tex]W = \frac{631.92 }{4.45}[/tex]
[tex]W = 142 \ lb[/tex]
How long did it take me to make this?
Answer:
Its according from what it is made with if wax it takes longer but if with a craft paper it takes lesser time if paper approximately 20 to 25 minutes
What line on a wheather map indicates áreas where the temperature is the same?
Answer:Isobars and isotherms
Explanation:
Check Concepts
4.
35. Which of the following do you calculate
when
you
divide the total distance trav-
eled by the total travel time?
A) average speed
B) constant speed
C) variable speed
D) instantaneous speed
Answer:
I think its A.........
Is the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth?
Answer:
Yes
Explanation:
This is because the interaction between piece of paper and earth.is gravitational while that of piece of paper and rod is electrostatic
Yes, the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth.
Gravitational force is the force by which earth attracts other objects by mass.The electrostatic force is the force of an object due to charge.Electrostatic forces are much stronger than gravitational forces. because gravity depends on mass, atoms have less masses so that the gravitational forces between them is close to zero. Whereas, the electrostatic force related to charges is bigger.Therefore, the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth.
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Write all the different ways you can think of that describe what it means to be healthy
Answer:
Eat more healthy foods. Workout and build your immune system.
Explanation:
Eat Healthy foods like Carrots, Apples, Bannas, Pears, and anything that deals with not much of any sugar. An example of unhealthy foods is Cakes, Chocolates, Candy, and more. Drink a lot of water.
The main force(s) acting on the puck after receiving the kick is (are):_________.A) a downward force of gravity and an upward force exerted by the surfaceB) a downward force of gravity, and a horizontal force in the direction of motionC) a downward force of gravity, an upward force exerted by the surface, and a horizontal force in the direction of motionD) a downward force of gravityA) a downward force of gravity and an upward force exerted by the surface
Answer:
the statements, the correct one is A
a downward force of gravity and an upward force exerted by the surface
Explanation:
When the disc is hit, a thrust force is exerted in the direction of movement, at the moment the disc moves this force loses contact and becomes zero.
When the movement is already established there are two main forces: gravity that acts downwards and the reaction force to the support of the disk called normal that acts upwards.
As it is not mentioned that there is friction, this force that opposes the movement is zero.
Analyzing the statements, the correct one is A
What are the benefits of testing your heart rate during exercise
Answer: Heart rate provides an objective measurement of how hard your body is working. The higher the exercise intensity, the higher your heart rate will be.
Explanation:
Answer:
You're getting exercise and training your heart!
Explanation:
Which of the following is not a valid use of your driver's license?
O proof of your ability to operate a motor vehicle
O proof of your age
proof of your residency
O proof that you have liability insurance
NEXT QUESTION
In the state where I live, your driver's license is not a proof that you have liability insurance. You don't need liability insurance to get a driver's license, but you need it in order to operate a car that you own.
It may be different in the state where YOU live.
NEXT QUESTION
A bird flutters around in a tree in a path described by the dark line: Which vector represents the average velocity of the bird? Question 2 options:
Answer:
Displacement vector represents the average velocity of the bird.
Explanation:
Given that,
A bird flutters around in a tree in a path described by the dark line.
Suppose, given vectors
(a). Time, (b). Displacement, (c). speed, (d). distance
We know that,
Vector quantity :
Vector quantity has direction and magnitude.
Average velocity :
Average velocity is equal to the displacement divided by time.
In mathematically,
[tex]v=\dfrac{D}{t}[/tex]
Where, D = displacement
t = time
We need to find the vector which is represents the average velocity of the bird
Using given data
Average velocity of the bird shows the displacement over the time.
Displacement is the vector quantity.
Hence, Displacement vector represents the average velocity of the bird.
Consider two balls in motion at the same time. Joe drops the first ball from rest at height h. Directly below Joe, on the ground, Hayley simultaneously tosses a second ball upward with speed v0.a. If the two balls collide at the moment the second ball is instantaneously at rest, what is the height of the collision?b. What is the relative speed of the balls when they collide?
Answer:
A. To find height of collision let's find
Speed of collision first
Using
Vf= √2g(h-g)
But for Harvey's ball we have
V²-2gy= 0
y=( v²/2g) so this is height of collision
B.
To find relative speed of the ball
Using V at y to find Vsr
Vs= √2g(h-v²/2g)
= √2gh- v²
You are standing on top of the Empire State Building (height 330 [m]) when Superman flies past. He is headed straight down with a steady speed of 35 [m/s]. (Superman can do things like this.) At the instant he goes past, you drop a 10 [kg] lead ball over the edge. At what height above the sidewalk does the ball pass Superman?
Answer:
The ball will be overtake superman after travelled distance 248.5 m.
Explanation:
Given that,
Speed of superman = 35 m/s
Height = 330 m
Mass of ball = 10 kg
Let the height at which they are at same position be S.
For superman,
We need to calculate the time
Using equation of motion
[tex]S=ut[/tex]...(I)
For ball,
[tex]S=u't+\dfrac{1}{2}gt^2[/tex]
From equation (I) and (II)
[tex]ut=u't+\dfrac{1}{2}gt^2[/tex]
[tex]ut=0+\dfrac{1}{2}\times g\times t^2[/tex]
[tex]u=\dfrac{1}{2}\times g\times t[/tex]
[tex]t=\dfrac{2u}{g}[/tex]
Put the value into the formula
[tex]t=\dfrac{2\times35}{9.8}[/tex]
[tex]t=7.1\ sec[/tex]
We need to calculate the distance travelled
Using formula of distance
[tex]d=v\times t[/tex]
Put the value into the formula
[tex]d=35\times 7.1[/tex]
[tex]d=248.5\ m[/tex]
Hence, The ball will be overtake superman after travelled distance 248.5 m.
Two equal charges are 2m2m apart. If the charges and the distance are divided by two, how is the force between the charges affected? g
Answer:
The force between the charges are not affected.
Explanation:
Given;
distance between two equal charges, R = 2m
The force between the charges is given by;
[tex]F = \frac{kq^2}{R^2}\\\\F_1 = \frac{kq_1^2}{R_1^2}\\\\When\ the \ charges \ and \ the \ distance \ are \ divided \ by \ two \ (q_2 = \frac{q_1}{2}, \ R_2 = \frac{R_1}{2} )\\\\ F_2 = \frac{kq_2^2}{R_2^2}\\\\F_2 = \frac{k(q_1/2)^2}{(R_1/2)^2}\\\\F_2= \frac{4k*q_1^2}{4*R_1^2}\\\\F_2 = \frac{k*q_1^2}{R_1^2}\\\\F_2 = F_1[/tex]
Therefore, the force between the charges are not affected.
A pilot flies her route in two straight-line segments. The displacement vector A for the first segment has a magnitude of 232 km and a direction 30.0o north of east. The displacement vector B for the second segment has a magnitude of 168 km and a direction due west. The resultant displacement vector is R = A + B and makes an angle θ with the direction due east. Using the component method, find (a) the magnitude of R and (b) the directional angle θ.
Answer:
a) 121 km
b) 74°
Explanation:
To start with, we assume that there exist two components, the East and the North. We'd be representing the East, by "e" and the North, by "n"
Now we start with the
First vector:
east1 = 232 cos 30 = 201
north1 = 232 sin 30 = 116
Now, that of the second vector will be
east 2 = - 168
north2 = 0
Next, we add the two together and get
East components
201 - 168 = 33 east
North components
116 + 0 = 116
Therefore, the magnitude has to be
magnitude = √(33² + 116²)
Magnitude = √14545
Magnitude = 121
tanθ = 116/33
Tanθ = 3.51
θ = tan^-1 3.51
θ = 74° North East
2 differences between calorimeter and thermometer ?
Answer:
Calorimeter is used to measure heat in and represents that in units of joules per kelvin units J/˚C or kJ/K
A calorimeter is can be used to measure the amount of heat released or involved in a chemical reaction.
Whereas thermometer can only measures temperature or hotness of a substance. It cannot be used to measure the thermal rate or amount of heat energy of a reaction. Unit measurement used by thermometer is Celsius (°C).
Explanation:
A ship sets out to sail to a point 116 km due north. An unexpected storm blows the ship to a point 121 km due east of its starting point. (a) How far (in km) and (b) in what direction (as an angle from due east, where north of east is a positive angle) must it now sail to reach its original destination?
Explanation:
In this problem, we are meant to slove for the resultant and the direction of the the vectors given
Given data
let the sail to a point due north be y= 116km
and the point due east be x= 121 km
(a) How far (in km)
The resultant between the two points is the distance between them
[tex]r=\sqrt{x^2+y^2} \\\\r=\sqrt{121^2+116^2} \\\\r=\sqrt{14641+13456} \\\\r=\sqrt{28097} \\\\r=167.62[/tex]
The distance between the points is 167.62
(b) in what direction (as an angle from due east, where north of east is a positive angle) must it now sail to reach its original destination
the direction can be gotten using
tan∅= y/x
∅= tan-1 (y/x)
∅= tan-1(116/121)
∅= tan-1(0.958)
∅= 43.77°
The direction is 43.77°
A car originally traveling at 30.0 m/s manages to brake for 5.0 seconds while traveling 125 m along a road. After those first 5.0 seconds, the brakes fail. After an additional 5.0 seconds it travels an additional 150 m further down the road. What was the magnitude of the acceleration of the car after the brakes failed
Answer:
The magnitude of the acceleration of the car after the brakes failed is 4 m/s²
Explanation:
The car was originally traveling at 30.0 m/s, that is
The initial velocity, [tex]u[/tex] = 30.0 m/s
The time spent while the car manages to brake is 5.0 seconds, that is
time, [tex]t[/tex] = 5.0 secs
and the distance traveled during this time is
distance, [tex]s[/tex] = 125 m
From one of the equations of kinematics for linear motion,
[tex]s = ut + \frac{1}{2}at^{2} \\[/tex]
Where [tex]a[/tex] is the acceleration
We can determine the deceleration of the car during the first 5.0 seconds
Hence,
From,
[tex]s = ut + \frac{1}{2}at^{2} \\[/tex]
[tex]125 = 30.0(5.0) + \frac{1}{2}(a)(5.0)^{2}[/tex]
[tex]125 =150.0 + 12.5a[/tex]
[tex]12.5a = 125 - 150.0[/tex]
[tex]12.5a = -25\\a = \frac{-25}{12.5}\[/tex]
[tex]a = - 2.0 m/s^{2}[/tex] (Negative sign indicates deceleration)
Now we will calculate the final velocity reached at this time
From,
[tex]v^{2} = u^{2} + 2as[/tex]
Where [tex]v[/tex] is the final velocity
[tex]v^{2} = 30.0^{2} + 2(-2.0)(125)\\v^{2} = 400\\v = \sqrt{400} \\v = 20 m/s \\[/tex]
This is the final velocity reached by the car during the first 5.0 seconds
Now, for the magnitude of the acceleration of the car after the brakes failed,
After the brakes failed,
it travels an additional 150 m further down the road, that is
s = 150m
an additional 5.0 seconds, that is
t = 5.0 seconds
Also, from
[tex]s = ut + \frac{1}{2}at^{2} \\[/tex]
The initial velocity here will be the final velocity for the first 5.0 seconds, that is,
u = 20 m/s
Hence,
[tex]s = ut + \frac{1}{2}at^{2} \\[/tex] becomes
[tex]150 = 20(5.0) + \frac{1}{2}(a)(5.0)^{2}[/tex]
[tex]150 = 100 + 12.5a\\12.5a = 150 - 100\\12.5a = 50\\a = \frac{50}{12.5} \\a = 4m/s^{2}[/tex]
Hence, the magnitude of the acceleration of the car after the brakes failed is 4 m/s²
A sidewalk has a length of 75.00m. How many inches is this? (Hint: you need to use two unit conversion fraction. 1 cm equals about 0.3937 inches)
Length = (75.00 m)
Length = (75 meter) x (3.28084 foot/meter) x (12 inch/foot)
Length = (75 x 3.28084 x 12) (meter-foot-inch / meter-foot)
Length = 2,952.76 inches
A rod attracts a positively charged hanging ball. the rod is?a) negativeb) positivec) neutrald) either negative or neutrale) either positive or neutral
Answer:
The correct option is a
Explanation:
This question seeks to test a general rule in physics (on charges) which states that like charges repel but unlike charges attract. This means that, a negatively charged substance will repel or not attract another negatively charged material and the same applies to a positively charged substance also. However, a negatively charged substance will attract a positively charged material and vice versa, hence only a negatively charged rod will attract a positively charged hanging ball.
Peter left Town A at 13:30 and travelled towards Town B at an
average speed of 40 mph. At 13:45, Philip left Town A for Town
B at an average speed of 30 mph. What was the distance
between them at 15:00?
Answer:
Explanation: From 13:30 to 15:00, it past: 1 h 30 mins = 1.5
Then, the distance covered by Peter: 40x1.5= 60 miles
From 13:45 to 15:00, it pasts; 1 h 15min =1.25
Then, the distance covered by Philip. 30 x 1.25 = 37.5 miles
Lastly, the distance between them: 60-37.5= 22.5 miles
So the answer is 22.5
Common benefits of lower body endurance include improved
Answer:
Hearing, Vision and Metabolism.
chemical reaction to make a foam?
Answer: Hydrogen Peroxide
Explanation: Hydrogen peroxide breaks down into oxygen and water. As a small amount of hydrogen peroxide generates a large volume of oxygen, the oxygen quickly pushes out of the container. The soapy water traps the oxygen, creating bubbles, and turns into foam.
You are driving your car at a speed of 19.0 m/s and you hit the brakes. The car accelerates at -3.50 m/s2. (a) How long does it take the car to cover 10.0 m? (b) What is the final velocity of the car?
Recall the formulas,
[tex]x_f=x_i+v_it+\dfrac12at^2[/tex]
[tex]v_f=v_i+at[/tex]
(a)
[tex]10.0\,\mathrm m=\left(19.0\dfrac{\rm m}{\rm s}\right)+\dfrac12\left(-3.50\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
[tex]\implies \boxed{t\approx0.555\,\mathrm s}[/tex]
(b)
[tex]v_f=19.0\dfrac{\rm m}{\rm s}+\left(-3.50\dfrac{\rm m}{\mathrm s^2}\right)(0.555\,\mathrm s)[/tex]
[tex]\implies \boxed{v_f\approx17.1\dfrac{\rm m}{\rm s}}[/tex]
What is the value of the radius of the following circle with an area of 154 cm2?
The area of ANY circle is (π) · (radius²).
So ...
(π) · (radius²) = 154 cm²
radius² = (154 cm²) / (π)
radius² = 49.02 cm²
radius = √(49.02 cm²)
radius = 7 cm
Answer:
[tex] \boxed{\sf Radius \ of \ circle \ (r) = 7 \ cm} [/tex]
Given:
Area of circle = 154 cm²
To Find:
Radius of circle (r)
Explanation:
[tex] \bold{Area \: of \: circle = \pi r^2}[/tex]
[tex] \sf \implies \pi {r}^{2} = 154 \\ \\ \sf \implies {r}^{2} = \frac{154}{\pi} \\ \\ \sf \implies {r}^{2} = \frac{154}{ \frac{22}{7} } \\ \\ \sf \implies {r}^{2} = \frac{154 \times 7}{22} \\ \\ \sf \implies {r}^{2} = \frac{1078}{22} \\ \\ \sf \implies {r}^{2} = 49 \\ \\ \sf \implies {r}^{2} = {7}^{2} \\ \\ \sf \implies r = \sqrt{ {7}^{2} } \\ \\ \sf \implies r = 7 \: cm[/tex]
A car going initially with a velocity 13.5 m/s accelerates at a rate of 1.9 m/s 2 for 6.2 s. It then accelerates at a rate of -1.2 m/s2 until it stops. a) Find the car’s maximum speed b) Find the total time from the start of the first acceleration until the car is stopped c) What’s the total distance the car travels?
(a) For the first 6.2 s, the car has velocity at time t given by
[tex]v(t)=13.5\dfrac{\rm m}{\rm s}+\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]
so that after 6.2 s, it attains a velocity of
[tex]v(6.2\,\mathrm s)=25.28\dfrac{\rm m}{\rm s}[/tex]
For any time t after 6.2 s, its velocity is given by
[tex]v(t)=25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]
which tells us the velocity only falls from this point onward. This means the maximum speed is 25.28 m/s, or about 25.3 m/s.
(b) Solve for t (after 6.2 s) that makes v(t) = 0 :
[tex]25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t=0[/tex]
[tex]\implies t\approx21.067\,\mathrm s[/tex]
It takes the car about 21.2 s to come to a rest, so the car travels a total of about 6.2 s + 21.2 s = 27.4 s.
(c) For the first 6.2 s, the car undergoes a displacement at time t of
[tex]x(t)=\left(13.5\dfrac{\rm m}{\rm s}\right)(6.2\,\mathrm s)+\dfrac12\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)(6.2\,\mathrm s)^2[/tex]
[tex]\implies x\approx120.218\,\mathrm m[/tex]
For time t beyond 6.2 s, its displacement is
[tex]x(t)=120.218\,\mathrm m+\left(25.28\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
The car comes to a rest after 21.2 s (accelerating at a rate of -1.2 m/s^2), so that its total displacement is
[tex]x(21.2\,\mathrm s)\approx386.49\,\mathrm m[/tex]
so the car travels a total distance of about 387 m.
hows a map of Olivia's trip to a coffee shop. She gets on her bike at Loomis and then rides south 0.9mi to Broadway. She turns east onto Broadway, rides 0.8 mi to where Broadway turns, and then continues another 1.4mi to the shop.
The question is incomplete. Here is the complete question.
The map (in the attachment) shows Olivia's trip to the coffee shop. She gets on her bike at Loomis and then rides south 0.9mi to Broadway. She turns east onto Broadway, rides 0.8mi to where Broadway turns, and then continues another 1.4mi to the shop.
What is the magnitude of the total displacement of her trip?
Whta is the direction of the total displacement of her trip?
Answer: Magnitude = 2.6mi
Direction: 54.65° east
Explanation: Displacement is the change in postition of a moving object.
There are a few ways to determine total displacement. For this case, the Perpendicular Components of a Vector method will be used.
For this method, total displacement is given by:
[tex]\Delta d_{t}=\sqrt{(\Delta d_{x})^{2}+(\Delta d_{y})^{2}}[/tex]
[tex]\theta=tan^{-1}(\frac{\Delta d_{y}}{\Delta d_{x}})[/tex]
[tex]\Delta d_{x}[/tex] is the x-component of total displacement and it is the sum of each individual x-components;
[tex]\Delta d_{y}[/tex] is the y-component of total displacement and it is the sum of each individual y-components;
θ is the angle the resulting displacement;
For Olivia's trip, there are no x-component of the first part and for the third part, the path she bikes is a hypotenuse of a right triangle. So, that right triangle's x-component is:
[tex]sin30=\frac{x}{1.4}[/tex]
[tex]\frac{1}{2} =\frac{x}{1.4}[/tex]
x = 0.7
Then,
[tex]\Delta d_{x}[/tex] = 0 + 0.8 + 0.7
[tex]\Delta d_{x}[/tex] = 1.5
Related to y, there are no y-component in the second part of Olivia's trip and for the third part:
[tex]cos30=\frac{y}{1.4}[/tex]
[tex]\frac{\sqrt{3} }{2} =\frac{y}{1.4}[/tex]
y = 1.21
Then,
[tex]\Delta d_{y}[/tex] = 0.9 + 0 + 1.21
[tex]\Delta d_{y}[/tex] = 2.11
Total displacement is
[tex]\Delta d_{t}=\sqrt{(1.5)^{2}+(2.11)^{2}}[/tex]
[tex]\Delta d_{t}=\sqrt{6.7021}[/tex]
[tex]\Delta d_{t}=[/tex] 2.6
Magnitude of Olivia's total displacement is 2.6mi
On the map, joining the initial and final points gives a vector pointing towards east at angle:
[tex]\theta=tan^{-1}(\frac{2.11}{1.5})[/tex]
[tex]\theta=tan^{-1}(1.41)[/tex]
θ = 54.65°
Direction of total displacement is 54.65° East.
Answer:
1.9 mi, 350.7°
Explanation:
If continued another 1.2 mi*
solving for x
0 + 0.8 + 1.2(cos30) = 1.83923048 --> 1.84
solving for y
-0.9 + 0 + 1.2(sin30) = -0.3
^negative because it is going downward
a) solving for magnitude
[tex]\sqrt{(1.84)^2+(-0.3)^2} = \sqrt{3.4756} = 1.86429611... = 1.9 mi[/tex]
b) solving for direction of total displacement
[tex]tan^-1 = (\frac{x}{y} ) \\tan^-1 = (\frac{-0.3}{1.84}) = -0.16 \\[/tex]
∘, measured counterclockwise from the eastward direction
360 - 0.16 = 359.84°
*replace any of the needed values in the equation, such as 1.2 mi to 1.4 mi
A typical atom has a diameter of about 1.0×10−10m.
Approximately how many atoms are there along a 2.0 −cm line?
Express your answer using two significant figures.
Answer:
10m
Explanation:
if it was at the 2.0 line it would be 10 m
If a typical atom has a diameter of about 1.0×10⁻¹⁰ m, then there are approximately atoms are there along a 2.0-centimeter line.
What are significant figures?In positional notation, significant figures refer to the digits in a number that is trustworthy and required to denote the amount of something, also known as the significant digits, precision, or resolution.
As given in the problem If a typical atom has a diameter of about 1.0×10⁻¹⁰ m, then we have to find out approximately how many atoms are there along a 2.0-centimeter line,
diameter of the one atom = 1.0×10⁻¹⁰
approximate number of atoms in 2 cm line = 2 ×10⁻² /( 1.0×10⁻¹⁰ )
=2 ×10⁸ atoms
Thus, there are approximately 2 ×10⁸ atoms are there along a 2.0-centimeter line.
Learn more about significant figures here,
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