it is meters and ounces
Two velocity vectors, one is twice of the other, and separated by 90 Degree angle. If their resultant is calculated 90 m/s, what the magnitude of bigger vector?
Answer:
Vy = 80.5 [m/s]
Explanation:
In order to solve this problem we must use the Pythagorean theorem.
V = 90 [m/s]
The components are Vx and Vy:
Therefore:
[tex]v=\sqrt{v_{x}^{2} + v_{y}^{2} }[/tex]
where:
Vy = 2*Vx ; because one is twice of the other.
[tex]90 = \sqrt{v_{x}^{2} +(2*v_{x})^{2} }\\ 90 =\sqrt{v_{x}^{2}+4*v_{x}^{2}} \\90 =\sqrt{5v_{x}^{2}} \\90=2.23*v_{x} \\v_{x}=40.25[m/s][/tex]
and the bigger vector is:
Vy = 40.25*2
Vy = 80.5 [m/s]
what is power?
a- the magnitude of a force needed to move an object
b- how much work can be done in a given time
c- the distance over time that an object moves
d- the energy needed to create work
Answer:
b- how much work can be done in a given time
how much work can be done in a given time
Julie takes a cold glass of dear, doness houd out the register and pusi on the babe hathewwad warme up to room temperature, Julie sees buthles of gas form in the und and me to the face Which conclusion is best supported by Jules onervations? The liquid is a pure sutrance because it is closers O The liquid is a mixture because its temperature res The liquid is a pure substance because it remains lowd when The liquid is a mixture because it has a gas dissolved in a loud
Answer:
D. The liquid is a mixture because it has a gas dissolved in a liquid.
A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand in the second trial 4H. Compare the FINAL VELOCITY'S for the packages right as they hit the
ground (or right before)?
Answer:
The final speed of the second package is twice as much as the final speed of the first package.
Explanation:
Free Fall Motion
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:
[tex]v=gt[/tex]
And the distance traveled downwards is:
[tex]\displaystyle y=\frac{gt^2}{2}[/tex]
If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:
[tex]\displaystyle t=\sqrt{\frac{2y}{g}}[/tex]
Replacing into the first equation:
[tex]\displaystyle v=g\sqrt{\frac{2y}{g}}[/tex]
Rationalizing:
[tex]\displaystyle v=\sqrt{2gy}[/tex]
Let's call v1 the final speed of the package dropped from a height H. Thus:
[tex]\displaystyle v_1=\sqrt{2gH}[/tex]
Let v2 be the final speed of the package dropped from a height 4H. Thus:
[tex]\displaystyle v_2=\sqrt{2g(4H)}[/tex]
Taking out the square root of 4:
[tex]\displaystyle v_2=2\sqrt{2gH}[/tex]
Dividing v2/v1 we can compare the final speeds:
[tex]\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}[/tex]
Simplifying:
[tex]\displaystyle v_2/v_1=2[/tex]
The final speed of the second package is twice as much as the final speed of the first package.
A man climbs on to a wall that is 3.6 m high and gains 2222.64 J of potential
energy. What is his mass?
63 kg
17.5 kg
405 kg
617.4 kg
43. The particle consists of fast moving electrons is.
Answer:
Explanation: someone help me please you want free points right
What do virtually all daily task require ?
Answer:muscle strength
Explanation: bc well you would need it
9. A plane starts at rest & accelerates along the ground before takeoff. It
moves 1600m in 18s. Calculate the acceleration rate of the plane. *
Answer:
9.877 m/s^2
Explanation:
The acceleration can be computed from ...
d = (1/2)at^2
(1600 m) = (1/2)a(18 s)^2
a = (1600/162) m/s^2 ≈ 9.877 m/s^2
What is the displacement from the forest to the doctor’s office?
Look up "Everything You Need To Know About Math In One Big Fat Notebook pdf." It's the best thing I've ever been given, I have it with me in math class all the time and I've aced every test. I have it with me right now and it has everything I've ever been taught about math in it so it might help you.
When you shock your self because of a door knob this is an example of
A.the law of conservation of charge
B. Force field
C.static discharge
D.torture lol
Answer:
C
Explanation:
Static electricity is what makes your hair stand up when you rub a balloon against it or gives you a shock from your doorknob. In static electricity, electrons are moved around mechanically (i.e. by someone rubbing two things together).
HOPE THIS HELPED AND LETTER D MADE ME LAUGH
How much distance did this object travel in meters between 0 and 10 seconds
Answer:
5m
Explanation:
Answer:
5
Explanation:
because that's the average of how far it when the most
blank refers to the method of spreading fertilizer evenly over the entire field by hand it is done at the blank stage
Answer:
Broadcasting is the method, not sure about the stage it is done in
Explanation:
Answer:
Broadcasting is the first (blank) second (blank) is Cultivation.
Explanation:
I took the test & got this answer correct.
The human circulatory system is closed-that is, the blood pumped out of the left ventricle of the heart into the arteries is constrained to a series of continuous, branching vessels as it passes through the capillaries and then into the veins as it returns to the heart. The blood in each of the heart’s four chambers comes briefly to rest before it is ejected by contraction of the heart muscle. If the aorta (diameter dad a) branches into two equal-sized arteries with a combined area equal to that of the aorta, what is the diameter of one of the branches? (a) da−−√ d a ; (b) da/2–√d a / 2 ; (c) 2da2d a ; (d) da/2d a /2.
The question is missing parts. The complete question is here.
The human circulatory system is closed, that is, the blood pumped out of the left ventricle of the heart into the arteries is constrained to a series of continuoous, branching vessels as it passes through the capillaries and then into the veins as it returns to the heart. The blood in each of the heart's four chambers comes briefly to rest before it is ejected by cotraction of the heart muscle. If the aorta (diameter [tex]d_{a}[/tex]) branches into two equal-sized arteries with a combined area equal to that of the aorta, what is the diameter of one of the branches?
(a) [tex]\sqrt{d_{a}}[/tex]
(b) [tex]\frac{d_{a}}{\sqrt{2} }[/tex]
(c) [tex]2d_{a}[/tex]
(d) [tex]\frac{d_{a}}{2}[/tex]
Answer: (b) [tex]\frac{d_{a}}{\sqrt{2} }[/tex]
Explanation: The cross-sectional area of a vessel is a circle. So area is:
[tex]A=\pi.r^{2}[/tex]
Radius is half of a diameter, i.e.:
[tex]r=\frac{d}{2}[/tex]
Suppose [tex]d_{b}[/tex] is diameter of the branches, radius of one the branches is:
[tex]r_{b}=\frac{d_{b}}{2}[/tex]
Both branches are equal sized, which means, both have the same radius. So, combined area of both branches is:
[tex]A_{b}=\pi\frac{d_{b}^{2}}{4} +\pi\frac{d_{b}^{2}}{4}[/tex]
[tex]A_{b}=2\frac{d_{b}^{2}}{4}[/tex]
Area of aorta is
[tex]A_{a}=\pi.(\frac{d_{a}}{2})^{2}[/tex]
[tex]A_{a}=\pi.\frac{d_{a}^{2}}{4}[/tex]
Area of aorta is equal the combined area of the branches, then:
[tex]2\pi\frac{d_{b}^{2}}{4}= \pi\frac{d_{a}^{2}}{4}[/tex]
Rearraging:
[tex]d_{b}^{2}=\frac{d_{a}^{2}}{2}[/tex]
[tex]d_{b}=\frac{d_{a}}{\sqrt{2} }[/tex]
The diameter of one of the branches is [tex]\frac{d_{a}}{\sqrt{2}}[/tex].
5 13. If a train going 60 m/s hits the brakes, and it takes the train 2 minute 25 seconds to stop, what is the train's acceleration?
Answer:
Acceleration = -12/17 m/s ^2
Explanation:
Name the variables
V = final velocity( velocity after the train stops) u= intial velocity(velocity when the train was moving at constant speed, before brakes are applied)
A = acceleration( in this case, it is deceleration)
T = time ( time taken to fully stop)
Second Step: What equation should we use?
The 3 Big Motion in One Dimension Formulas are:
V = u+at
v^2 = u^2 + 2as
S = ut+ (1/2) at^2
For this question, we will be using the first one.
Third Step: Solve
V = 0 (the final velocity is 0, because the train stopped)
U = 60 (the train was going at 60 mps)
T = 85 seconds( 1 min and 25 sec)
0 = 60+ 85a
Sand is made of different types of rocks. The shapes and sizes are not all the same. Sand is a ----?
Answer:
A soil
Explanation:
Sand is a granular material composed of finely divided rock and mineral particles. It is defined by size, being finer than gravel and coarser than silt. Sand can also refer to a textural class of soil or soil type; i.e., a soil containing more than 85 percent sand-sized particles by mass.
Answer:
mixture
sand is a mixture
how can you observe the law of conservation of energy in action at the skatepark?
Explanation:
When the skater is dropped onto the ramp from above, the potential energy decreases and the kinetic energy increases.
Every time the skater bounces from the impact, thermal energy is gained, and both potential and kinetic energy are lost.
How long does it take a cheetah that runs with a velocity of 34m/s to run 750m?
Answer:
It would take 5 seconds
Explanation:
I can't find the 'delta' sign nor the vector sign so just pretend that displacement, time and velocity has them.
V = d / t
34 = 170 / t
34 x t = 170
34t = 170
t = 170 / 34
t = 5
What is the acceleration of a ball with a mass of 0.40 kg is hit with a force of 16N?
Answer:
40 m/s^2
Explanation:
Mass= 0.40 kg
Force= 16 N
Therefore the acceleration can be calculated as follows
F = ma
16= 0.40 × a
16= 0.40 a
a= 16/0.40
a= 40 m/s^2
Hence the acceleration is 40 m/s^2
Which statement describes the interaction between the north and south poles of two magents?
Answer:
A south pole attracts a north pole :)
Explanation:
11. What is the momentum of a truck with a mass of 3000 kg traveling north at a
speed of 25 m/s?
Answer:
We are given:
mass of the truck (m) = 3000 kg
velocity of the truck (v) = 25 m/s
Calculating the Momentum:
We know that the momentum of an object is equal to its mass times it's velocity
P = mv (where P is the momentum of the truck)
Momentum of the truck:
replacing with the values of the truck
P = mv
P = 3000 * 25
P = 75000 kg m/s
Therefore, the momentum of the truck is 75000 kg m/s
A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular acceleration of -4.46 rad/s2. Because of this acceleration, the angular velocity of the wheel changes from its initial value to a final value of -31.4 rad/s. While this change occurs, the angular displacement of the wheel is zero. (Note the similarity to that of a ball being thrown vertically upward, coming to a momentary halt, and then falling downward to its initial position.) Find the time required for the change in the angular velocity to occur.
Answer:
The time for the change in the angular velocity to occur is 14.08 secs
Explanation:
From the question,
the angular acceleration is - 4.46 rad/s²
Angular acceleration is given by the formula below
[tex]\alpha =\frac{\omega -\omega _{o} }{t - t_{o} }[/tex]
Where [tex]\alpha[/tex] is the angular acceleration
[tex]\omega[/tex] is the final angular velocity
[tex]\omega _{o}[/tex] is the initial angular velocity
[tex]t[/tex] is the final time
[tex]t_{o}[/tex] is the initial time
From the question
[tex]\alpha[/tex] = - 4.46 rad/s²
[tex]\omega _{o}[/tex] = 0 rad/s (starting from rest)
[tex]\omega[/tex] = -31.4 rad/s
[tex]t_{o}[/tex] = 0 s
Now, we will determine t
From [tex]\alpha =\frac{\omega -\omega _{o} }{t - t_{o} }[/tex], then
[tex]-4.46 = \frac{-31.4 - 0}{t - 0}[/tex]
[tex]-4.46 = \frac{-31.4}{t}[/tex]
[tex]t = \frac{-31.4}{-4.46}[/tex]
t = 7.04 secs
This is the time spent in one direction,
Since the angular displacement of the wheel is zero ( it returned to its initial position), then the time required for the change in the angular velocity will be twice this time, that is 2t
Hence,
The time is 2×7.04 secs = 14.08 secs
This is the time for the change in the angular velocity to occur.
Chen is testing the friction of three surfaces. He pushes the same ball across three different surfaces with the same force and measures the distance the ball rolls over each surface. The ball moved ten inches across Surface 1, six inches across Surface 2, and fifteen inches across Surface 3. Which could most likely describe the three surfaces?
Answer:Surface 1 is blacktop, Surface 2 is gravel, and Surface 3 is ice.
Explanation:
Answer: its C
Explanation:
distinguish between current and current density.
Answer:
1- the rate of flow of charge through a conductor is called current .whereas ,current density is the current per unit area of conductor
Electric charges are either positive or ____
Answer:
Negative
Explanation:
duh
Answer:
:)
Explanation:
negative.
go add the snap carmel.bratz
A car rounds a banked curve as we will discuss in class on Tuesday. The radius of curvature of the road is R and the banking angle is θ. (a) In the absence of friction, what is the safe speed for the car to take this curve? (b) Now assume the coefficient of friction between the car’s tires and the road is µs. Determine the range of speeds the car can have without slipping up or down the road. (c) What is the minimum value of µs that makes the minimum speed zero? (d) If θ = 25.0 ◦ , for what values of µs can the curve be taken at any speed? Note: The upper limit of µs you will find is practically impossible to achieve for the car’s tires and the road.
Answer:
A) v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]
B)√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]
C) µ_s = tan θ
D) µ_s = 0.4663
Explanation:
A) The forces acting on the car will be;
Force due to friction; F_f
Force due to Gravity; F_g
Normal Force; F_n
Now, let us take the vertical direction to be j^ and the direction approaching the centre to be i^ downwards and parallel to the road surface by k^.
Also, we will assume that initially, F_n is in the negative k^ direction and that it will have a maximum possible value of; F_f = µ_s × F_n
Thus, sum of forces about the vertical j^ direction gives;
ΣF_j^ = F_n•cos θ − mg + F_f•sin θ = 0
Since F_f = µ_s × F_n ;
F_n•cos θ − mg + (µ_s × F_n × sin θ) =0
F_n = mg/[cos θ + (µ_s•sin θ)]
Also, sum of forces about the centre i^ direction gives;
ΣF_i^ = F_n(sin θ + (µ_s•cos θ)) = mv²/r
Plugging in formula for F_n gives;
ΣF_i^ = [mg/[cos θ + (µ_s•sin θ)]] × (sin θ + (µ_s•cos θ)) = mv²/r
Making v the subject gives;
v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]
B) What we got in a above is the minimum speed the car can have while going round the turn.
The maximum speed will be gotten by making the frictional force(F_f) to point in the positive k^ direction. This means that F_f will be negative.
Now, if we change the sign in front of F_f in the equation in part a that led to the minimum velocity, we will have the maximum as;
v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]
Thus the range is;
√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]
C) For the minimum speed to be 0, it implies that F_f will be in the negative k^ direction. Thus, Sum of the forces in the k^ direction gives;
ΣF_k^ = mg(sin θ - µ_s•cos θ) = 0
Thus;
mg(sin θ - µ_s•cos θ) = 0
Making µ_s the subject gives;
µ_s = sin θ/cos θ
µ_s = tan θ
D) If θ = 25.0°;
Thus;
µ_s = tan 25
µ_s = 0.4663
A hawk is flying due west at 10.1 m/s carrying a field mouse in its talons. The mouse manages to break free at a height of 15 m. What is the magnitude of the mouse's velocity as it reaches the ground? Note: neglect air resistance
Answer:
When the mouse broke free from the hawk, its vertical velocity is zero since the mouse just fell from the grips of the hawk
We are given:
initial vertical velocity (v) = 0 m/s
initial horizontal velocity (u) = 10.1 m/s
height from the ground (h) = 15 m
acceleration due to gravity (a) = 10 m/s/s
final vertical velocity (vf) = v m/s
Solving for Final vertical velocity:
from the third equation of motion
(vf)² - (v)² = 2ah
replacing the variables
v² - 0² = 2(10)(15)
v² = 300
v = √300
v = 10√3 OR 17.3 m/s
A 55-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 40.0o above the horizontal. If the tension in the rope is 125 N, how much work is done on the crate to move it 5.0 m?
Answer:
478.75 J
Explanation:
W=force* displacement
constant speed= (a=0) net F=0
Horizontal component of tension
Tcosx
125Ncos40= 95.76 N
W= (95.76 N)(5 m)
=478.75 J
The work done in moving the crate across the given distance is 478.75 J.
The given parameters;
Mass of the packing create, m = 55 kgAngle of inclination of the rope, Ф = 40°Tension on the rope, T = 125 NDistance through which the crate is the moved, d = 5 mThe work done in moving the crate is the product of the horizontal component of the tension and the distance through which the crate is moved.
The work-done in moving the crate is calculated as;
W = Tcos(Ф) x d
W = 125cos(40) x 5
W = 478.75 J.
Thus, the work done in moving the crate across the given distance is 478.75 J.
Learn more here: https://brainly.com/question/19498865
PLEASEEEEE!!!!
A cube has a mass of 100 g and a volume of 50 mL. What is the density of the cube?
Answer:
The answer is 2.0 g/mLExplanation:
The density of a substance can be found by using the formula
[tex]density = \frac{mass}{volume} \\[/tex]
From the question
mass = 100 g
volume = 50 mL
We have
[tex] density = \frac{100}{50} = \frac{10}{5} \\ [/tex]
We have the final answer as
2.0 g/mLHope this helps you
What feature does not require a planet to have any particular characteristics?
Stream Beds
Dunes
Impact Craters
Volcanic Lava Flows
What is the correct answer?
Answer:
Impact Craters.
Explanation:
An impact crater can be defined as a circular depression that is caused by impact on any planet or asteroids or any other celestial body's surface. When smaller body in galaxy impacts these larger bodies, they form a circular depression on it's surface.
This is a major feature found in solid object such as the Moon, Mercury, etc.
Therefore, the feature that a planet does not require is an impact crater. As other features are important to define a planet. Thus correct option is C.
A black widow spider hangs motionless from a web that extends vertically from the ceiling above. If the spider has a mass of 1.81 g, what is the tension in the web?
N=
Answer:
The tension in the web is 0.017738 N
Explanation:
Net Force
The net force exerted on an object is the sum of the vectors of each individual force applied to an object.
If the net force equals 0, then the object is at rest or moving at a constant speed.
The spider described in the question is hanging at rest. It means the sum of the forces it's receiving is 0.
A hanging object has only two forces: The tension of the supporting string (in our case, the web) and its weight. If the object is in equilibrium, the tension is numerically equal to the weight:
T=W=m.g
The mass of the spider is m=1.81 gr = 0.00181 Kg, thus the tension is:
[tex]T = 0.00181\ Kg\cdot 9.8\ m/s^2[/tex]
[tex]T=0.017738\ N[/tex]
The tension in the web is 0.017738 N