Answer:
The energy of the wave is 1.435 x 10⁻⁴ J
Explanation:
Given;
area of the window, A = 0.5 m²
the rms value of the field, E = 0.06 V/m
The peak value of electric field is given by;
[tex]E_o = \sqrt{2} *E_{rms}\\\\E_o = \sqrt{2}*0.06\\\\E_o = 0.0849 \ V/m[/tex]
The average intensity of the wave is given by;
[tex]I_{avg} = \frac{c \epsilon_o E_o^2 }{2}\\\\I_{avg} = \frac{(3*10^8)( 8.85*10^{-12}) (0.0849)^2 }{2}\\\\I_{avg} = 9.569*10^{-6} \ W/m^2[/tex]
The average power of the wave is given by;
P = I x A
P = (9.569 x 10⁻⁶ W/m²) (0.5 m²)
P = 4.784 x 10⁻⁶ W
The energy of the wave is given by;
E = P x t
E = (4.784 x 10⁻⁶ W)(30 s)
E = 1.435 x 10⁻⁴ J
Therefore, the energy of the wave is 1.435 x 10⁻⁴ J
A stone is thrown vertically upward with a speed of 28.0 m/s how much time is required to reach this height
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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Your friend says, “chemical changes are caused by an input in energy. In physical changes, there is no transfer of energy” is your friend correct? Why or why not?
Answer:
Ok, let's suppose the simplest of the physical changes:
We have an object that is not moving (so it is not accelerated)
and there is change, now the object moves.
Because there was a change, means that there was an acceleration, and by the second Newton's law.
Force equals mass times acceleration:
F = m*a
There must be a force.
So suppose that you pushed the object, then some energy that you had, you transferred it to the object, that now is moving and now has kinetic energy.
Now, is kinda true that in a closed system the total energy is always constant, but it depends on what is our system.
So if we think in our system as you and the object, then in the whole system the energy does not change because the energy that you lost is now on the object, but again, there was a transfer of energy.
So no, your friend is not correct.
Both chemical and physical changes involve energy transfer, so your friend is not correct.
A chemical change occurs when a new substance is formed or created through a process of chemical reaction which is reversible.
The addition or removal of heat energy can affect the rate of chemical reaction. This addition or removal is known as energy transfer process.
Examples of chemical changes include;
Acid-base reaction.Rusting of iron in presence of moisture and oxygen. Cooking any food.A physical change on the other hand doesn't involve formation of new substance and it is can be reversible or irreversible.
The addition or removal of heat energy can affect the rate of physical changes.
Examples of such physical changes include;
vaporization of liquid (liquid to gas),freezing of liquid (liquid to solid), and condensation of gas (gas to liquid).Thus, both process (chemical and physical changes) involve energy transfer, so your friend is not correct.
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Common benefits of lower body endurance include improved
Answer:
Hearing, Vision and Metabolism.
A 10 kg object is dropped from rest. a. How far will it drop in 2s? b. How long will it take a 5 kg object to drop the same distance?
Answer:
Explanation:
Using the equation of motion S = ut + 1/2at² to get the height of drop where
u is the initial velocity of the object = 0m/s
a is the acceleration due to gravity = +9.81m/s² (downwards motion of object)
t is the time it takes the object to drop = 2secs
Substituting the given parameters into the formula to get the height of drop S, we will have;
S = 0(2)+1/2(9.81)(2)²
S = 0+9.81*2
S = 19.62m
Hence the object will drop at a distance of 19.62m
b.) To determine the time it will take a 5 kg object to drop the same distance, we will use the same formula S = ut+1/2at²
Since the 5kg object also drops at the same distance, then S = 19.62m
Substituting this values into the equation we will have;
19.62 = 0(t) + 1/2(9.81)t²
19.62 = 4.905t²
t² = 19.62/4.905
t² = 4
t =2secs
This shows that it will take 5kg object 2secs to fall from the same distance. This means that no matter the mass of the object, it will take them the same time to fall at the same distance because they are all falling under the same influence of gravity.
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
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
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]
Difference between calorimeter and thermometer ?
Answer:
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.
Explanation:
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.
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:
What quantity do units represent in a value?
A. size
Question:
What quantity do units represent in a value? A. Size B. Direction C. Magnitude D. Dimension
Answer:
D. Dimension
Explanation:
A unit is simply the measurement given to a dimension. In other words, units represent dimensions. For example, consider a ruler that is 20m long.
i. The value is 20m.
ii. The magnitude of the measurement is 20
iii. The unit of the measurement is meters(m) and that essentially represents the dimension (length) of the ruler.
Which statement best defines inertia? An object's motion is unaffected by any external forces acting upon it. An object responds to a force by tending to move in the direction of that force. An object opposes any motion, naturally returning to a state of rest on its own. An object opposes any change in its velocity, either to its direction or to its speed.
Answer:
An object opposes any change in its velocity, either to its direction or to its speed.
Explanation:
Edmentum Answer
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.
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°
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]
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.
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
Assuming it is a van der Waals gas, calculate the critical temperature, pressure and volume for CO2. (a = 3.610 atm L2 mol-2, b = 0.0429 L mol-1)
pc = ___ atm
Tc = ___ K
Vc = ___ L/mol
Answer
To get critical pressure
We use
Pc = a/(27b²)
So
= 3.610/(27 X 0.0429²)
We have
= 72.7 atm
Critical temperaturewe
We use
Tc = 8a/27Rb
= 8 x 3.610/(27 x 0.0812 x 0.0429)
= 307 K
Critical volume
We use
Vc =3b =
3 x 0.0429
= 0.129L/mol
Put Newton’s 1st Law in your own words
Answer:Newton’s law also states that larger bodies with heavier masses exert more gravitational pull, which is why those who walked on the much smaller moon experienced a sense of weightlessness, as it had a smaller gravitational pull. To help explain his theories of gravity and motion, Newton helped create a new, specialized form of mathematics.
Explanation:Here I dont know
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 technical machinist is asked to build a cubical steel tank that will hold 310L of water. Calculate in meters the smallest possible inside length of the tank.
Answer:
0.68 m
Explanation:
Since the volume of water is 310 L, and we know that 1 litre = 1 dm³. So the volume of water is V = 310 dm³. Since this volume of water is the volume of water the cubical steel tank can contain, it equals the volume of the cubical steel tank.
We know that the volume of the cubical steel tank V = L³ where L is the length of side of the cube on the inside.
So the length of side of the cube L = ∛V = ∛310 dm³ = 6.77 dm = 6.77 dm × 1m/10 dm = 0.677 m ≅ 0.68 m
So, the smallest possible inside length of the tank is 0.68 m
What is the binary number system?
Explanation:
binary number is a number expressed in the base-2 numeral system or binary numeral system, which uses only two symbols: typically "0" and "1".
A ball with a mass of 3.7 kg is thrown downward with an initial velocity of 8 m/s from a high building. How fast will it be moving after 3 seconds?
Answer:
v=37.4 m/s
Explanation:
It is given that,
Mass of a ball, m = 3.7 kg
Initial velocity of the ball is u = 8 m/s
We need to find its velocity after 3 seconds. It is moving downwards. The equation of motion is this case is
v=u+gt
[tex]v=8+9.8\times 3\\\\v=37.4\ m/s[/tex]
So, the velocity of the ball after 3 seconds is 37.4 m/s.
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.
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.........
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
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:
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