Answer:
0.25s
Explanation:
5/20 = 0.25s
This might be correct
Find the transformation matrix that rotates a rectangular coordinate system through an angle of 60 about axes equal angels with original three coordinate axes
Answer:
[tex]M = \left[\begin{array}{ccc}cos \ 60&0\\0&-sin \ 60\end{array}\right][/tex]
Explanation:
To find the matrix, let's decompose the vectors, the rotated angle is (-60C) for the prime system
x ’= x cos (-60)
y ’= y sin (-60)
we use
cos 60 = cos (-60)
sin 60 = - sin (-60)
we substitute
x ’= x cos 60
y ’= - y sin 60
the transformation system is
[tex]\left[\begin{array}{ccc}x'\\y'\end{array}\right] = \left[\begin{array}{ccc}cos 60&0\\0&-sin60\end{array}\right] \ \left[\begin{array}{ccc}x\\y\end{array}\right][/tex]x '
the transformation matrix is
[tex]M = \left[\begin{array}{ccc}cos \ 60&0\\0&-sin \ 60\end{array}\right][/tex]
As a main sequence star exhausts hydrogen in its core, its surface becomes ___ and its energy output per second (luminosity) becomes ____.
Answer:
b
Explanation:
bc
A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the mass of the body
Answer:
the mass of the body is 0.02 kg.
Explanation:
Given;
relative density of the oil, [tex]\gamma _0[/tex] = 0.875
mass of the object in oil, [tex]M_o[/tex] = 0.013 kg
mass of the object in water, [tex]M_w[/tex] = 0.012 kg
let the mass of the object in air = [tex]M_a[/tex]
weight of the oil, [tex]W_0 = M_a - 0.013[/tex]
weight of the water, [tex]W_w = M_a - 0.012[/tex]
The relative density of the oil is given as;
[tex]\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg[/tex]
Therefore, the mass of the body is 0.02 kg.
Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m each. If Margy is walking at 1.4 m/s and accelerates at 0.20 m/s2 during
Complete question is;
Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m each. If Margy is walking at 1.4 m/s and accelerates at 0.20 m/s² during one of the running portions, what is her final velocity at the end of the 100.0 m? Round your answer to the nearest tenth.
Answer:
6.5 m/s
Explanation:
We are told that she is walking at 1.4 m/s and accelerates at 0.20 m/s².
Thus;
Initial velocity; u = 1.4 m/s
Acceleration; a = 0.2 m/s²
Distance; s = 100 m
From Newton's equation of motion, we know that;
v² = u² + 2as
Where v is final velocity.
Thus;
v² = 1.4² + 2(0.2 × 100)
v² = 41.96
v = √41.96
v ≈ 6.5 m/s
difine scalar quantity
Scalar quantity are physical quantities that have just magnitude, not direction.
It is always positive.Examples: Speed, distanceLight strikes a smooth wooden tabletop.
What happens to the light after it is reflected?
The light rays bounce off the table and all move in the same direction.
The light rays bounce off the table and move in different directions.
The light rays pass through the table and all move in the same direction.
The light rays pass through the table and move in different directions.
Answer:
For smooth surface:The light rays bounce off the table and all move in the same direction.
what is acceleration
[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]
[tex]\sf What \: is \: acceleration? \\ \\ \sf The \: rate \: of \: change \: of \: velocity \: of \: an \\ \sf object \: with \: respect \: to \: time \\ \sf is \: known \: as \: acceleration. [/tex]
Classes. frequency
0-20 2
20-40. 2
40-60. 3
60-80 12
80-100 18
100-120 5
120-140. 2
Find mean, median and mode
Answer:
The mean is 79.[tex]\overline {54}[/tex]
The median is 80 - 100
The mode is 80 - 100
Explanation:
The given table is presented as follows;
[tex]\begin{array}{lcrc}Classes&Mid \ point &Frequency &Frequency \times Mid \ point\\0 - 20&10& 2&20\\20-40&30&2&60\\40-60&50&3&150\\60-80&70&12&840\\80-100&90&18&1620\\100-120&110&5&550\\120-140&130&2&260\end{array}[/tex]
The mean of a class of values, [tex]\overline x[/tex] = ∑(Frequency × Midpoint)/∑(Frequency)
Therefore, we get;
[tex]\overline x[/tex] = (20+60+150+840+1620+550+260)/(2+2+3+12+18+5+2) = 79.[tex]\overline {54}[/tex]
The mean, [tex]\overline x[/tex] =79.[tex]\overline {54}[/tex]
The median class = The middle value lass = The class at the 22 nd value = 80 - 100
The median = 80 - 100
The modal class = The class with the highest frequency = 80 - 100
The mode = 80 - 100
The Displacement is 5m. We found that using the
Pythagorean Theorem.
Vector Quantities require both a Displacement and a
Direction.
What direction is this Vector?
South
Northeast
West
Answer:
A vector can be written as:
(R, θ)
Where R is the magnitude, in this case, we know that the magnitude of the displacement is 5m
Then:
R = 5m
and θ defines the direction, it's an angle measured from the positive x-axis.
(In the image, θ would be the angle located at the point A)
Now, if you look at the image, you can see a triangle rectangle.
Where the adjacent cathetus has a length of 4,
the opposite cathetus has a length of 3 units
the hypotenuse has a length of 5 units.
So we can use any trigonometric rule to find the value of θ, like:
sin(θ) = (opposite cathetus)/hypotenuse
Then:
sin(θ) = 3m/5m
Now we can use the inverse sin function, Asin(x), in both sides
Asin( sin(θ)) = θ = Asin( 3/5) = 36.87°
then the vector is:
(5m, 36.87°)
Now, if we define the positive y-axis as the North, and the positive x-axis as the East.
This vector would point at 36.87° North of East.
(or almost Northeast)
Calculate the distance travelled by the car in part Q use the equation distance travelled= average speed x time
Explanation:
distance travelled = average speed x time
=30m/s*100s
=3000m
Answer:
3000m
Explanation:
30m/s*100s
3000m
Why are the largest craters we find on the Moon and Mercury so much larger than the largest craters we find on the Earth
Answer:
Because Moon and Mars has no atmosphere.
Explanation:
Moon and Mars has no atmosphere, so there is no friction on the falling object due to the atmosphere. The speed of the falling object is more at Moon and Mars.
When a small object impact on the surface of moon or Mars with high speed, the size of crater is large than the earth as out earth has atmosphere.
A jet plane lands at a speed of 100 m/s and can accelerate at a maximum rate of -5.00 m/s^2 as it comes to a rest.
(a from the instant the plane touches the runaway, what is the minimum time needed before it can come to a rest?
(b Can this plane land on a runaway that is only 0.800 km long?
shown work pls will reward alot of points
Answer:
a) t = 20 s, b) x = 1000 m, As the runway is only 800 m long, the plane cannot land at this distance
Explanation:
This is a kinematics exercise
a) in minimum time to stop,
v = vo + at
v = 0
t = -v0 / a
we calculate
t = -100 / (5.00)
t = 20 s
b) Let's find the length you need to stop
v² = vo² + 2 a x
x = -v0 ^ 2 / 2a
x = - 100² / 2 (-5.00)
x = 1000 m
As the runway is only 800 m long, the plane cannot land at this distance.
SINGLE CORRECT OBJECTIVE
Question 9
speed of a moving object is said to be uniform if it covers equal distances in equal interva
time.
о O
ghte
O false
Your Answer:
Next
Answer:
True
Explanation:
Formula for speed is;
Speed = distance/time
Now, if an object covers an equal distance in equal time intervals, it means the speed will remain the same.
For example if an object covers 3 m every 1 second it means speed will always be; 3/1 = 3 m/s.
Thus the statement is correct.
While using a digital radiography system, suppose a radiographer uses exposure factors of 10 mAs and 70 kVp with an 8:1 grid for an AP shoulder radiograph with acceptable anatomical part penetration and detector element (DEL) exposure. If the radiographer desires to increase scatter absorption using a 12:1 grid, what new exposure factors should be used to maintain the same DEL exposure
Answer:
b. 12.5 mAs, 70 kVp
Explanation:
The given parameter are;
The initial exposure factors := 10 mAs and 70 kVp
The initial Grid Ratio, G.R.₁ = 8:1
The Grid Ratio with which the radiographer desires to increase the scatter absorption, G.R.₂ = 12:1
Given that the lead content in the 12:1 grid, is higher than the lead content in 8:1 grid and that 12:1 grid needs more mAs to compensate, and provides a higher image contrast, the amount of extra mAs is given by the Grid Conversion Factors, GCF, as follows;
The GCF for G.R. 8:1 = 4
The GCF for G.R. 12:1 = 5
Therefore, given that the mAs used by the radiographer for 8:1 Grid Ratio is 10 mAs, the mAs required for a G.R. of 12:1 in order to maintain the same exposure is given as follows;
mAs for G.R. of 12:1 = 10 mAs × 5/4 = 12.5 mAs
Therefore the new exposure factors are;
12.5 mAs, 70 kVp
1. 20kg of water is ejected horizontally in 10s; the speed of the water leaving the nozzle is 30m/s. Calculate the force experienced by a fire-fighter holding the hose.
Answer 60 NEWTON
Explanation:
FORCE = MASS * acceleration
acceleration= VELOCITY / TIME
acceleration= 30 / 10 = 3 M/S2
FORCE = MASS * acceleration
FORCE = 20 *3 = 60 NEWTON
the bodies in this universe attract one another name the scientist who propounded this statement
Answer:
It was proposed by Isaac Newton
Explanation:
The law of universal attraction of expression
F = [tex]G \ \frac{m_1m_2}{ r^2}[/tex]G m1m2 / r ^ 2
where G is a constant, m₁ and m₂ are the masses of the bodies and r the distance between them.
It was proposed by Isaac Newton
With this law Newton explained that the force that pulls the moon towards the earth is the same as that which attracts an apple towards the earth
A thin rod of length 1.4 m and mass 180 g is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed 1.80 rad/s. Neglecting friction and air resistance, find (a) the rod's kinetic energy at its lowest position and (b) how far above that position the center of mass rises.
Answer:
[tex]K.E = 0.1905 J[/tex]
Explanation:
From the question we are told that:
Length [tex]L=1.4m[/tex]
Mass [tex]m=180g[/tex]
Angular Velocity [tex]\omega=1.80rads/s[/tex]
Generally the equation for Kinetic energy K.E is mathematically given by
[tex]K.E =0.5 (1/3 ML^2 )w^2[/tex]
[tex]K.E =0.5 ( 1/3 * 0.18 * 1.4^2 ) 1.8^2[/tex]
[tex]K.E = 0.1905 J[/tex]
A system has a pressure of 5 N/m2
If a force of 2000N is applied, what is the area that the force is applied to?
Give the units.
400 m2
Explanation:
Pressure = Force ÷ Area
5 N/m2 = 2000 N ÷ A
A = 2000 N ÷ 5
= 400 m2
If a force of 2000N is applied, the area that the force is applied to is 400 m²
What is force?The word "force" has a specific meaning in science. At this level, calling a force a push or a pull is entirely appropriate. A force is not something an object "has in it" or that it "contains."
One thing experiences a force from another. There are both living things and non-living objects in the concept of a force.
The amount of force applied to a certain region is referred to as pressure. The force per unit area is called pressure. F in this condensed version of the equation stands in for the force, which is expressed in newtons.
Given that the pressure of 5 N/m²
Force is 2000N
Pressure = Force ÷ Area
5 N/m² = 2000 N ÷ A
A = 2000 N ÷ 5 = 400 m²
Therefore, the area that the force is applied to is 400 m².
To learn more about force, refer to the link:
https://brainly.com/question/19529052
#SPJ2
The work function for silver is 4.73 eV. (a) Convert the value of the work function from electron volts to joules.
Answer:
[tex]W=7.56\times 10^{-19}\ J[/tex]
Explanation:
Given that,
The work function for silver is 4.73 eV.
We need to find the value of the work function from electron volts to joules.
We know that,
[tex]1\ eV=1.6\times 10^{-19}\ J[/tex]
For 4.73 eV,
[tex]4.73\ eV=1.6\times 10^{-19}\times 4.73\\\\=7.56\times 10^{-19}\ J[/tex]
So, the work function for silver is [tex]7.56\times 10^{-19}\ J[/tex].
A 72 kg swimmer dives horizontally off a raft floating in a lake. The diver's speed immediately after leaving the raft is 3.8 m/s. If the time interval of the interaction between the diver and the raft is 0.25 s, what is the magnitude of the average horizontal force by diver on the raft?
Answer:
F = 1094.4 N
Explanation:
From impulse - momentum theorem, we now that ;
Impulse = momentum
Where;
Formula for impulse = force (F) × time(t)
Momentum = mass(m) × velocity(v)
Now, we are given;
Mass of swimmer; m = 72 kg
Speed; v = 3.8 m/s
Time; t = 0.25 s
Thus;
F × t = mv
F = mv/t
F = (72 × 3.8)/0.25
F = 1094.4 N
This value of force is the magnitude of the average horizontal force by diver on the raft.
The most successful types of plants on Earth are
Answer:
The angiosperms dominate Earth's surface and vegetation in more environments, particularly terrestrial habitats, than any other group of plants. As a result, angiosperms are the most important ultimate source of food for birds and mammals, including humans.
Explanation:
plz mark brainlest
Suponga que la pelota de la figura se proyecta desde una altura de 35.0 m sobre el suelo y se le imprime una velocidad horizontal inicial de 8.25 m/s. a) ¿Cuánto tiempo tardará la pelota en golpear el suelo? b) ¿A qué distancia del edificio tocará el suelo la pelota?
Answer:
Explanation:
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Internal energy of a diatomic gas consists of:
OA. kinetic energy due to vibration and rotation.
B. kinetic energy due to translation, vibration, and rotation.
C. potential energy due to intermolecular forces.
D. kinetic energy due to translation only.
Answer:
C) Potential energy due to intermolecular forces.
100 POINTS !!! PLEASE HELP !!!!
What is the independent variable?
What is the dependent variable?
I
Materials
1. 4 antacid tablets
2. 2 clear cups labeled Hot and Cold
3. Water
4. Timing device
Answer:
dependent: the outcome of the experience
independent variable: everything literaly.
Independent is where you change some variables and see the result
Dependent is literaly the result, or the outcome dependent on the exprience.
Explanation:
I got u.
A ball of mass 0.3 kg is released from rest at a height of 8 m. How fast is it going when it hits the ground? (Gravity being equal to 9.8)
Answer:
Explanation:
Mass doesn't matter here because when something is falling, gravity plays fairly; an elephant falls at the same rate of acceleration as does a feather. What DOES matter is everything pertinent to the y-dimension of free-fall:
a = -9.8 m/s/s
v₀ = 0 (since the ball was held before it was dropped)
v = ??
Δx = -8 m (negative because the ball drops this far below the point from which it was released).
Putting all this together in one equation:
v² = v₀² + 2aΔx and filling in this equation:
v² = (0)² + 2(-9.8)(-8) and
v² = 156.8 so
v = 12.5 which rounds to 13 if you're using 2 sig figs, and rounds to 10 if you're only using 1 (which you should be, according to the way the numbers have been given in this problem)
A student walks 350 m [S], then 400 m [E20°N], and finally 550 m [N10°W]. Using the component method, find the resultant (total) displacement). Round your answer to the appropriate significant figures. Round your angle to the nearest degree.
In component form, the displacement vectors become
• 350 m [S] ==> (0, -350) m
• 400 m [E 20° N] ==> (400 cos(20°), 400 sin(20°)) m
(which I interpret to mean 20° north of east]
• 550 m [N 10° W] ==> (550 cos(100°), 550 sin(100°)) m
Then the student's total displacement is the sum of these:
(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m
≈ (280.371, 328.452) m
which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].
An ammeter with a resistance of 5.0 ohm is connected in series with a 3.0V cell and a lamp rated at 300 mA, 3V. Calculate the current that the ammeter will measure
Answer:
I = 0.2 A
Explanation:
Lamp is rated at 300 mA
I_lamp = 0.3 A
Voltage is; V = 3V
Thus; Resistance is given by;
R = V/I
R = 3/0.3
R = 10 ohms
Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;
R_eq = 10 + 5
R_eq = 15 ohms
Ammeter current will be;
I = V/R_eq
I = 3/15
I = 0.2 A
A 5.0 kg block of ice is at rest at the top of a smooth inclined plane. The block is released and slides 2.0 m down the plane. Assuming there is no friction between the block and the surface, calculate
a) the gravitational potential energy at the top of the plane
b) the component of the weight parallel to the plane
c) the acceleration of the block
d) the velocity of the block at the bottom of the plane
e) the kinetic energy at the bottom of the plane.
Answer:
a) 98.1 Joules
b) 49.05 N × sin(θ)
c) 9.81 × sin(θ)
d) The velocity of the block at the bottom of the plane, v is approximately 6.264 m/s
e) 98.1 Joules
Explanation:
The given parameters of the block are;
The mass of the block, m = 5.0 kg
The distance down the plane the block slides, h = 2.0 m
The friction between the block and the surface = 0
Let θ represent the angle of inclination oof the plane
a) The gravitational potential energy, P.E. = m·g·h
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
∴ P.E. ≈ 5.0 kg × 9.81 m/s² × 2.0 m = 98.1 Joules
The gravitational potential energy, P.E. ≈ 98.1 Joules
b) The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex], is given as follows;
[tex]w_{\parallel}[/tex] = w × sin(θ) = m·g·sin(θ)
∴ [tex]w_{\parallel}[/tex] ≈ 5.0 kg × 9.81 m/s² × sin(θ) = 49.05 × sin(θ) N
The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex] ≈ 49.05 N × sin(θ)
c) The component of the weight along the inclined plane = The force with which the block moves along the inclined plane, therefore;
[tex]w_{\parallel}[/tex] = m·g·sin(θ) = m·a
Where a represents the acceleration of the block along the plane
Therefore, by comparison, we have;
g·sin(θ) = a
∴ a ≈ 9.81 × sin(θ)
d) Given that the motion of the block is 2.0 m downwards, we have;
The velocity of the block at the bottom of the plane, v² = 2·g·h
Therefore, v² ≈ 2 × 9.81 m/s²× 2.0 m = 39.24 m²/s²
v = √(39.24 m²/s²) ≈ 6.264 m/s
e) The kinetic energy at the bottom of the plane, K.E. = (1/2)·m·v²
∴ K.E. = (1/2) × 5.0 kg × 39.24 m²/s² = 98.1 J
Quickly pls!!! A wave with a wavelength of 0.5 m moves with a speed of 1.5 m/s. What is the frequency of the wave?
A. 2.0 Hz
B. 1.0 Hz
C. 0.33 Hz
D. 3.0 Hz
suppose the pilot starting again from rest opens the throttle part.way at constant acceleration the airboat then covers a distance of 60.0m in 10.0s find the net force action on the boat
Answer:
Acceleration is 1.2 m/s^2.
Explanation:
initial velocity, u = 0
distance, d = 60 m
time, t = 10 s
Let the acceleration is a.
use second equation of motion
[tex]s= u t +0.5 at^2\\\\60 = 0 + 0.5 \times a \times 10\times 10\\\\a = 1.2 m/s^2[/tex]
Now according to the Newton's second law
Force = mass x acceleration
Let the mass is m.
F = m x 1.2 = 1.2 m Newton
