Answer: A ([tex]\sqrt{61}[/tex],309.8°)
B (2[tex]\sqrt{2}[/tex], 315°)
C ([tex]3\sqrt{5}[/tex], 26.56°)
Explanation: To transform rectangular coordinates into polar coordinates use:
[tex]r=\sqrt{x^{2}+y^{2}}[/tex] and [tex]\theta=tan^{-1}(\frac{y}{x})[/tex]
For point A:
[tex]r=\sqrt{(-5)^{2}+6^{2}}[/tex]
[tex]r=\sqrt{61}[/tex]
[tex]\theta=tan^{-1}(\frac{6}{-5})[/tex]
[tex]\theta=tan^{-1}(-1.2)[/tex]
[tex]\theta=-50.2[/tex]°
Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:
[tex]\theta=360-50.2[/tex]
[tex]\theta=[/tex] 309.8°
Polar coordinates for point A is ([tex]\sqrt{61}[/tex], 309.8°)
For point B:
[tex]r=\sqrt{2^{2}+(-2)^{2}}[/tex]
[tex]r=\sqrt{8}[/tex]
[tex]r=2\sqrt{2}[/tex]
[tex]\theta=tan^{-1}(\frac{-2}{2} )[/tex]
[tex]\theta=tan^{-1}(1)[/tex]
[tex]\theta=-45[/tex]°
Point B is in IV quadrant, so:
[tex]\theta=360-45[/tex]
[tex]\theta=[/tex] 315°
Polar coordinates for point B is ([tex]2\sqrt{2}[/tex], 315°)
For point C:
[tex]r=\sqrt{(-6)^{2}+(-3)^{2}}[/tex]
[tex]r=\sqrt{45}[/tex]
[tex]r=3\sqrt{5}[/tex]
[tex]\theta=tan^{-1}(\frac{-3}{-6} )[/tex]
[tex]\theta=tan^{-1}(0.5)[/tex]
[tex]\theta=[/tex] 26.56°
Polar coordinates for point C is ([tex]3\sqrt{5}[/tex], 26.56°)
A social movement dedicated to sustaining Earth's systems that support life
Answer:
A social movement dedicated to sustaining Earth's systems that support life
Explanation:
That would be environmentalism.
Hope that helped:)
When air is heated, it will MOST likely
A.condense and rise
B.expand and rise
C.condense and fall
D.expand and fall
Answer:
B
Explanation:
When air is heated, it expands and rises.
Answer:
B
Explanation:
hot air is less dense than cold air, meaning its particles are farther apart. this causes it to expand and rise, since it is less dense. if you know about convection currents, you'll know that hot air rises and cold air falls.
A plane travelling at 63 m/s[S] down a runway begins accelerating uniformly at 2.8 m/s?[S]. How far does it travel
in 4.0 s?
Answer:
270 m
Explanation:
Given:
v₀ = 63 m/s
a = 2.8 m/s²
t = 4.0 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (63 m/s) (4.0 s) + ½ (2.8 m/s²) (4.0 s)²
Δx = 274.4 m
Rounded to two significant figures, the displacement is 270 meters.
A plane traveling at 63 meters/seconds down a runway begins accelerating uniformly at 2.8 meters/seconds, then the total distance traveled by plane would be 274.4 meters.
What is speed?The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object.
As given in the problem a plane traveling at 63 meters/seconds down a runway begins accelerating uniformly at 2.8 meters/seconds, then we have to find the total distance traveled in 4.0 seconds,
By using the second equation of the motion,
S = ut + 1/2 at²
=63×4 +0.5×2.8×4²
=274.4 meters
Thus, the total distance traveled by plane would be 274.4 meters.
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if you toss a ball upward, is the momentum of the moving ball conserved? is the momentum of the system consisting of ball + earth conserved?
Please help I will give 20 points.
Antarctica has received increased snow pack in recent years but has also had greater
melting of ice. In 1-2 sentences, explain why scientists would use more than one
computational model to make predictions about future melting.
2 point
Question 9
An object is shot from the ground at 75m/s at an angle of 45° above the horizontal. How high does the object get before beginning its descent?
Answer:
Before beginning its descent, the object gets 143.5 meters high
Explanation:
Projectile Motion
It's known as the type of motion that experiences an object that is projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
Being vo the initial speed of the object, θ the initial launch angle, and [tex]g=9.8m/s^2[/tex] the acceleration of gravity, then the maximum height reached by the object is:
[tex]\displaystyle y_m=\frac{v_o^2\sin^2\theta}{2g}[/tex]
The object is shot at vo=75 m/s at an angle of θ=45°. The maximum height is calculated below:
[tex]\displaystyle y_m=\frac{75^2\sin^2 45^\circ}{2\cdot 9.8}[/tex]
[tex]\displaystyle y_m=\frac{5625\cdot 0.707^2}{19.6}[/tex]
[tex]y_m=143.5\ m[/tex]
Before beginning its descent, the object gets 143.5 meters high
what is a plane mirror ?
Answer:
A plain mirrior is a mirrior with flat reflective surface.
hope it is helpful for you.
In ancient wars, heavy rocks are thrown horizontally from the wall as weapons. f the initial speed of a rock is 3 m/s , how many meters will it travel horizontally from a wall 25 meters abov the ground after 2 seconds(neglect air resistance, g=9.8m/s2)
Answer:
The horizontal distance traveled by the rock from the wall after 2 seconds is 6m.
Explanation:
Given;
initial speed of the rock, u = 3 m/s
height of the wall, h = 25 m
time of travel, t = 2 seconds
The horizontal distance is determined as follows;
X = ut + ¹/₂gt²
where;
g is acceleration due to gravity, horizontal distance is not affected by gravity, g = 0
X = ut
X = 3 x 2
X = 6 m
Therefore, the horizontal distance traveled by the rock from the wall after 2 seconds is 6m.
A object spins through an angular displacement of 10 radians and has an angular acceleration of 2.3 rad/sec-squared. Assuming it began spinning from rest, over what time interval did the acceleration occur?
Answer:
The acceleration of the object occurred at 2.95 s
Explanation:
Given;
initial angular velocity of the object, ω = 0
angular acceleration, α = 2.3 rad/s²
angular displacement of the object, θ = 10 radians
The time of the motion is given by the following kinematic equation;
θ = ω + ¹/₂αt²
θ = 0 + ¹/₂αt²
θ = ¹/₂αt²
[tex]t^2 = \frac{2 \theta}{\alpha}\\\\t = \sqrt{\frac{2 \theta}{\alpha}}\\\\t = \sqrt{\frac{2 *10}{2.3}}\\\\t = 2.95 \ s[/tex]
Therefore, the acceleration of the object occurred at 2.95 s
A lady driver of car applied brakes and barely avoided hitting the truck parked due to technical problem on the road way.
The vehicle left skid marks of 25 m. Assuming, f = 0.6 and the braking efficiency of 95%, determine whether the drive
was in violation of the 60 Kmph speed limit at that location if she was traveling,
(a) Uphill on 5º slope. (b) Downhill on 5º slope and (c) On the level roadway.
Answer:
do no answer pls tell sorry
Two bodies A and B with 8 Kg and 24 kg respectively move
directly to one another for head
of
a speed of 15 m/s
and 15 m/s respectively If e=0.5 for colliding
bodies, determine the following
(1) Their velocities after collision
(2) The loss of K.E of colliding bodies.
Which of the following statements about air masses is true
Answer:
Air masses are created by patterns of heating and cooling of the Earth's surface.
One of the difficulties in studying mechanics is that many common words are used with highly specific technical meanings, among them velocity, acceleration, position, speed, and displacement. Answer the question in this problem using words from the following list:
a. position
b. direction
c. displacement
d. coordinates
e. velocity
f. acceleration
g. distance
h. magnitude
i. vector
j. scalar components
Identify the following physical quantities as scalars or vectors.
a. Position
b. Velocity
c. Displacement
d. Speed
e. Average velocity
f. Distance
Answer:
Explanation:
Physical quantities in mathematics are divided into two:
1) Scalar quantity: This are quantities that gas only magnitude but doesn't specify direction. They only talk about the position of an object in space. For example, Distance, position, speed at all scaler quantities because they didn't specify direction. Distance only talk about how far a body has traveled while speed talks about how far a body has traveled in a specified period of time.
Scalar quantities are:
- Position: coordinate of point on a plane.
- Speed: Speed is the distance covered in a specified time.
- Distance: is how far a body has travelled.
2) Vector quantity: This is a quantity that has both MAGNITUDE and DIRECTION. They specify both position of an object as well as the direction of travel. Example of vector quantities are:
- Velocity: This is the change in displacement of a body with respect to time. Since displacement specifies direction, hence velocity is classified as a vector quantity.
- Displacement: This is the distance traveled by a body in a specified direction.
- Average velocity: distance travelled in a specified time.
When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 75-kg man just before contact with the ground has a speed of 4.6 m/s. (a) In a stiff-legged landing he comes to a halt in 2.1 ms. Find the average net force that acts on him during this time. N (b) When he bends his knees, he comes to a halt in 0.09 s. Find the average force now. N (c) During the landing, the force of the ground on the man points upward, while the force due to gravity points downward. The average net force acting on the man includes both of these forces. Taking into account the directions of these forces, find the force of the ground on the man in parts (a) and (b). stiff legged landing N bent legged landing
Answer:
a) 1.725*10^5 N
b) 3.83*10^3 N
c) i) 173.24 kN
c) ii) 4.57 kN
Explanation:
See the attachment for calculations
Compare and contrast groups and periods of the periodic table.
Answer:
Wheres the table?
Explanation:
How much work is done to push a car 20m using a 200 N force?
W =
J
How much power did you developed in pushing the car in 30 s?
P =
W
Answer:
F=200N d=20m t=30s
W=F*d
W=200*20
W=4000J
P=W/t
P=4000/30
P=400/3
P=133.333333334 W
Given this measurement 2642 ft, how many significant figures are there?
A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the ball was hit. The ball lands with a velocity of 20.5 m/s at an angle of 38 degrees below horizontal. Ignoring air resistance
(a) Find the initial velocity and the angle above horizontal with which the ball leaves the bat
(b) Find the height of the ball relatively to the ground.
(a) The ball has a final velocity vector
[tex]\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j[/tex]
with horizontal and vertical components, respectively,
[tex]v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}[/tex]
[tex]v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}[/tex]
The horizontal component of the ball's velocity is constant throughout its trajectory, so [tex]v_{x,i}=v_{x,f}[/tex], and the horizontal distance x that it covers after time t is
[tex]x=v_{x,i}t=v_{x,f}t[/tex]
It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:
[tex]103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s[/tex]
The vertical component of the ball's velocity at time t is
[tex]v_{y,f}=v_{y,i}-gt[/tex]
where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:
[tex]-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}[/tex]
So, the initial velocity vector is
[tex]\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j[/tex]
which carries an initial speed of
[tex]\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}[/tex]
and direction θ such that
[tex]\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}[/tex]
(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height y at time t is
[tex]y=v_{y,i}t-\dfrac12gt^2[/tex]
so that when it lands in the seats at t ≈ 6.38 s, it has a height of
[tex]y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}[/tex]
something that can not be used up or depleted
Answer:
oxygen
Explanation:
as it is in the air it can't be depleted or used up
Answer:
Deplete definition is - to empty of a principal substance. How to use deplete in a ... to reduce in amount by using up The soil was depleted of minerals. deplete.
Explanation:
Two students grab a slinky and start waving it up and down. A third student counts the number of waves that pass by every second and measures the distance between the wave peaks. This data is recorded and a graph is made to show the results.
In looking at the graph, what wave property is remaining constant over the experiment?
A) amplitude
B) energy
C) period
D) velocity
Identify the characteristic of the transverse wave that halved from wave A (black) to wave B (green).
A) amplitude
B) crest
C) trough
D) wavelength
Two point charges of −1.0 × 10−6 and +2.0 × 10−6 are separated
by a distance of 0.30 . What is the electrostatic force on each
particle? Hint ( = 9.0 × 1092
−2)
the circuit consist of three resistors R1=5, R2=3, R3=4 and are connected in series to each other. A voltmeter connected in parallel to resister R2 measures voltage of 6v. what is the net in the current?
Answer:
Current in the circuit= 2 A
Explanation:
Current in a Series Connection
When two or more resistors are connected in series, all of them have the same current, and the sum of their individual voltages is the total voltage applied to the circuit.
We know a voltmeter connected to R2=3Ω measures 6V. Applying Ohm's law:
[tex]V=R.I[/tex]
We can calculate the current through R2:
[tex]\displaystyle I=\frac{6}{3}=2\ A[/tex]
Since the current in R2 is the same in the rest of the resistors, the current of R1 and R3 is also 2 A.
In case of collision of objects in two dimensions, which statement is true after the collision?
A.
The horizontal momentum is not conserved.
B.
The vertical momentum is not conserved.
C.
The horizontal momentum and the vertical momentum are both conserved.
D.
The horizontal momentum and the vertical momentum are not conserved.
Answer:
C. The horizontal momentum and the vertical momentum are both conserved.
Explanation:
Answer:
The horizontal momentum and the vertical momentum are both conserved.
Explanation:
Just took it
PLZZ ANSWER THE QUESTION
Answer:
I would say a but it could be b too. It’s definetly not the last two.
I say a because no reason to go up the hill and that’s not gonna give her speed going down so. Good luck!
Explanation:
A train is accelerating at a rate of 1.8 m/s2. If its initial velocity is 18 m/s,
what is its velocity after 28.1 sec?
Answer:
velocity= 68.58m/s
Explanation:
formula: v = initial velocity + (acceleration)(time)
plug in your values: v= 18m/s + (1.8 m/s2)(28.1s)
solve the parenthesis first: v = 18 m/s + 58.58m/s
add: v = 68.58 m/s
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 21 kg and the larger bottom crate has a mass of m2 = 90 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.8 and the coefficient of kinetic friction between the two crates is μk = 0.64. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem). 1)The rope is pulled with a tension T = 261 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate? m
Answer:
2.35 m/s²
Explanation:
Given that
Mass of the smaller crate, m₁ = 21 kg
Mass of the larger crate, m₂ = 90 kg
Tensión of the rope, T = 261 N
We know that the sum of all forces for the two objects with a force of friction F and a tension T are:
(i) m₁a₁ = F
(ii) m₂a₂ = T - F, where m and a are the masses and accelerations respectively.
1) no sliding can also mean that:
a₁ = a₂ = a
This makes us merge the two equations written above together as:
m₂a = T - m₁a
If we then solve for a, we would have something like this
a = T / (m₁+m₂)
a = 261 / (21 + 90)
a = 261 / 111
a = 2.35 m/s²
Therefore, the needed acceleration of the small crate is 2.35 m/s²
analogy
Teacher:Educate. student:____
,
Answer:
Study
Explanation:
From the given analogy, we understand that.
Teachers educates
If this is true,
The students are expected to learn from what the teacher teaches.
This can also be said that, the students studies what the teachers teach
Hence, to complete the analogy.
We have;
(1) Students, Study.
We can also make use of
(2) Students, learn.
(1) & (2) answer the question completely.
Answer:
STUDENT:STUDY OR LEARN
Explanation:
A blue shark accelerates at a rate of 16m/s for a time of 0.8s. During this time it travels a
distance of 11.52m. Calculate its initial speed.
Answer:
The initial velocity of this shark is [tex]8.0\; \rm m \cdot s^{-1}[/tex]. (Assuming that the unit of the acceleration in this question is [tex]\rm m\cdot s^{-2}[/tex].)
Explanation:
Let [tex]a[/tex] denote the acceleration of this shark.
Let [tex]v_0[/tex] denote the initial velocity of this shark.
Assume that the acceleration [tex]a[/tex] of this shark is constant (as it is in this question.) Over a period of time [tex]t[/tex], the shark would have travelled a distance of:
[tex]\displaystyle x = \frac{1}{2}\, a\, t^2 + v_0\, t[/tex].
This question states that:
[tex]x = 11.52\; \rm m[/tex], [tex]t = 0.8\; \rm s[/tex]. (That is: this shark travelled a distance of [tex]11.52\; \rm m[/tex] in [tex]0.8\; \rm s[/tex].)[tex]a = 16\; \rm m \cdot s^{-2}[/tex] (the acceleration of this shark is indeed a constant.)This question is asking for [tex]v_0[/tex], the initial velocity of this shark at the beginning of this [tex]0.8[/tex]-second period. Substitute the three known values into the equation:
[tex]\displaystyle 11.52 = \frac{1}{2}\times 16\times (0.8)^2 + 0.8\, v_0[/tex].
Solve for [tex]v_0[/tex]:
[tex]v_0 = \displaystyle \frac{11.52 - (1/2) \times 16 \times (0.8)^2}{0.8} = 8.0\; \rm m \cdot s^{-1}[/tex].
When a charge is placed on a metal sphere, it ends up in equilibrium at the outer surface. Use this information to determine the electric field of +3.0 µC charge put on a 5.0-cm aluminum spherical ball at the following two points in space:
a. A point 1.0 cm from the center of the ball (an inside point).
b. A point 10 cm from the center of the ball (an outside point)
Answer:
a
The value at an inside point is Zero
b
The electric field is [tex]E = 2.7*10^{6} \ N/C[/tex]
Explanation:
From the question we are told that
The magnitude of the charge is [tex]q = 3.0 \mu C = 3.0 *10^{-6} \ C[/tex]
The radius of the spherical ball is [tex]r = 5.0 \ mm = 0.005 \ m[/tex]
Generally according to law postulated by Gauss the magnitude of charge enclosed inside a conducting material is zero which implies that the electric field inside the spherical ball is zero
Generally the electric field out side the spherical ball is mathematically represented as
[tex]E = \frac{kq}{ a^2}[/tex]
Here a is the position outside the spherical ball that is been considered and the value is [tex]a = 10 \ cm = \frac{10}{100} = 0.1 \ m[/tex]
and k is the coulombs constant with value
[tex]k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
=> [tex]E = \frac{ 3 *10^{-6} * 9*10^9 }{ (0.1)^2}[/tex]
=> [tex]E = 2.7*10^{6} \ N/C[/tex]
An object with mass m is given initial velocity to slide across a horizontal plane AB towards a semi circle BCD with radius R.
Between the object and the plane exists a kinetic friction u_k=0.5, but only between the section FB with length R. F is in the middle of A and B. The inside of the circle is smooth.
When the object reached to point C, it applies a force of 3mg on the semi circle.
The object is going to the left.
1. Write an expression for the initial velocity at point A.
The expression for the initial velocity at point A is:
0 = (velocity at point A - 0) / time
Simplifying the equation, we find:
Velocity at point A = 0
The initial velocity at point A is zero, indicating that the object starts from rest before sliding on the horizontal plane AB.
To write an expression for the initial velocity at point A, we need to analyze the forces acting on the object and apply the principles of motion.
Given:
Mass of the object, m
Radius of the semi circle, R
Coefficient of kinetic friction, μ[tex]_k[/tex] = 0.5
Force applied at point C, F = 3mg
The object is initially at rest.
Let's break down the motion into two parts: the motion on the horizontal plane AB and the motion along the semi circle BCD.
1. Motion on the horizontal plane AB:
The only force acting on the object on the horizontal plane is the force of kinetic friction. The frictional force can be calculated using:
Frictional force, f = μ[tex]_k[/tex]* Normal force
The normal force is equal to the weight of the object, which is mg.
Normal force, N = mg
Frictional force, f = μ[tex]_k[/tex] * mg
The frictional force acts in the opposite direction to the motion, so its magnitude is negative. Thus, the net force on the object on the horizontal plane is:
Net force = -f = -μ[tex]_k[/tex]* mg
Using Newton's second law, we can relate the net force to the acceleration:
Net force = mass * acceleration
-μ[tex]_k[/tex] * mg = m * acceleration
The acceleration can be expressed as the rate of change of velocity:
Acceleration = (final velocity - initial velocity) / time
Since the object is initially at rest, the initial velocity is zero.
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