Answer:
2.1m/s towards your grandmother's house
Explanation:
Given parameters:
Time taken = 204s
Distance = 430m
Unknown:
Velocity = ?
Solution:
The velocity is determined by:
Velocity = [tex]\frac{displacement}{time}[/tex]
Velocity = [tex]\frac{430}{204}[/tex] = 2.1m/s towards your grandmother's house
Match the measurements with the proper SI unit.
Acceleration:
A. Meters
B. Meters per second
C. Meters per second squared
Velocity:
A. Meters
B. Meters per second
C. Meters per second squared
Distance:
A. Meters
B. Meters per second
C. Meters per second squared
Explanation:
C. meter per second squared
B. meter per second
A. meter
Answer:
b. meters per second
c.meters per second squared
c.meters
Explanation:
ginawa ko na rin KC toh
The height (in centimeters) at time t (in seconds) of a small mass oscillating at the end of a spring is h(t)=5sin(2πt). Estimate its instantaneous velocity at t=3
Answer:
h (3) = 0
Explanation:
In this exercise they give us the expression that governs the movement
h (t) = 5 sin (2πt)
remember that the angles are in radians.
To calculate the instantaneous velocity we substitute
h (3) = 5 sin (2π 3)
h (3) = 0
therefore the body this is its position of equilibrium
Two ropes are connected to a 200 kg dinghy. Two cousins each take one rope and pull. When the cousins pull in the same direction, the dinghy accelerates at a rate of 1.31 m/s2 to the east. If they pull in opposite directions, the dinghy has an acceleration of 0.526 m/s2 to the west. Assume the ropes are horizontal, and ignore any other horizontal forces acting on the dinghy. What is the magnitude of the force each cousin exerts on the dinghy
Answer:
The magnitude of the force each cousin exerts on the dinghy is 183.6 N and 78.4 N.
Explanation:
When the cousins pull in the same direction we have:
[tex] F_{1} + F_{2} = ma_{e} [/tex] (1)
Where:
F₁ and F₂ are the forces exerted by the two boys.
m: is the mass of the dinghy = 200 kg
[tex]a_{e}[/tex]: is the acceleration in the east direction
When the cousins pull in opposite directions we have:
[tex] F_{1} - F_{2} = ma_{w} [/tex] (2)
By adding equation (1) and (2):
[tex] 2F_{1} = m(a_{e} + a_{w}) [/tex]
[tex] F_{1} = \frac{200 kg(1.31 m/s^{2} + 0.526 m/s^{2})}{2} = 183.6 N [/tex]
Now, by entering F₁ into equation (1) we can find F₂:
[tex] F_{2} = ma_{e} - F_{1} = 200kg*1.31 m/s^{2} - 183.6 N = 78.4 N [/tex]
Therefore, the magnitude of the force each cousin exerts on the dinghy is 183.6 N and 78.4 N.
I hope it helps you!
You are inside the Great Hall, 15 m from the north wall with the doors to the RMC, and centered between two open doors that are 3 m apart. Someone is blairing a 200 Hz tone outside the Great Hall so that it enters the doors as a plane wave. You hear a maximum intensity in your current position. As you walk along the direction of the wall with the doors (but maintain a distance 15 m from the wall), how far will you walk (in m) to hear a minimum in the sound intensity
Answer:
Δr = 0.425 m
Explanation:
This is a sound interference exercise, the expression for destructive interference is
Δr = (2n + 1) λ / 2
in this case the movement is in the same direction as the sound, therefore the movement is one-dimensional
let's use the relationship between the speed of sound and its frequency and wavelength
v = λ f
λ = v / f
the first minium occurs for n = 0
Δr = λ / 2
Δr = v / 2f
Δr = [tex]\frac{340}{2 \ 400}[/tex]
Δr = 0.425 m
this is the distance from the current position that we assume in the center of the room
6th grade science I mark as brainliest !
Answer:
first is Atoms
4) is True
Identify the independent variable(s) in Asch's original experiment.
Answer:
How do I answer this I don't understand the question
Explanation:
Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant (and quite modest) acceleration. A train travels through a congested part of town at 4.0 m/s . Once free of this area, it speeds up to 11 m/s in 8.0 s. At the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed. Part A What is the final speed
Answer:
25 m/s
Explanation:
Given that:
Initial speed, u = 4 m/s
Final velocity, V = 11 m/s
Time, t = 8 seconds
t2, = 16 seconds
Acceleration, a= (change in velocity) / time interval
a = (11 - 4) / 8
a = 7 / 8 = 0.875m/s²
Final velocity, v2 ;
Acceleration * t2
0.875 * 16 = 14
V2 = 14 m/s
Final speed : v + v2 = (11 + 14)m/s = 25m/s
How does Doppler ultrasound technology differ from ultrasound technology
that does not use the Doppler effect?
A. Doppler ultrasound collects data from moving objects.
B. Other ultrasound technology creates images, but Doppler
ultrasound does not.
C. Doppler ultrasound creates images, but other ultrasound
technology does not.
D. Doppler ultrasound is based on absorption of sound, and other
ultrasound technology is based on reflection.
Answer:
A. Doppler ultrasound collects data from moving objects
Explanation:
Did the test !!
Answer:A. Doppler ultrasound collects data from moving objects.
Explanation: just got it right
on my test
what is the mathematical definition of momentum? what is a more conceptual or descriptive definition of momentum?
Answer:
Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion. The amount of momentum that an object has is dependent upon two variables: how much stuff is moving and how fast the stuff is moving.
Explanation:
5. An astronaut has a mass of 65kg where the gravitational field strength is 10N/kg
a. Calculate the weight of the astronaut on earth
[3]
Answer: a) weight on Earth = mass of the object and gravity n the Earth. = 65*10 = 650 kg.
Explanation:
An astronaut has a mass of 65 kg on Earth where the gravitational field strength is 10 N kg A calculate the astronaut's weight on Earth
hope this helps :)
Answer:
650
Explanation:
use the equation
weight = gm
The standard test to determine the maximum lateral acceleration of a car is to drive it around a 200-ft diameter circle painted on a level asphalt surface. The driver slowly increases the vehicle speed until he is no longer able to keep both wheel pairs straddling the line. If the maximum speed is 35 mi/hr for a 3000-lb car, compute the magnitude F of the total friction force exerted by the pavement on the car tires.
Answer:
the magnitude F of the total friction force is 2456.7 lb
Explanation:
Given the data in the question;
maximum speed = 35 mi/hr = ( 35×5280 / 60×60) = 51.3333 ft/s
diameter = 200ft
radius = 200/2 = 100 ft
First we calculate the normal component of the acceleration;
[tex]a_{n}[/tex] = v² / p
where v is the velocity of the car( 51.3333 ft/s)
p is the radius of the curvature( 100 ft)
so we substitute
[tex]a_{n}[/tex] = (51.3333 ft/s)² / 100ft
[tex]a_{n}[/tex] = (2635.1076 ft²/s²) / 100ft
[tex]a_{n}[/tex] = 26.35 ft/s²
we convert Feet Per Second Squared (ft/s²) to Standard Gravity (g)
1 ft/s² = 0.0310809502 g
[tex]a_{n}[/tex] = 26.35 ft/s² × 0.0310809502 g
[tex]a_{n}[/tex] = 0.8189g
Now consider the dynamic equilibrium of forces in the Normal Direction;
∑[tex]F_{n}[/tex] = m[tex]a_{n}[/tex]
F = m[tex]a_{n}[/tex]
we know that mass of the car is 3000-lb = 3000lb([tex]\frac{1}{g}slug[/tex]/1 lb)
so
we substitute
F = 3000lb([tex]\frac{1}{g}slug[/tex]/1 lb) × 0.8189g
F = 2456.7 lb
Therefore; the magnitude F of the total friction force is 2456.7 lb
fertilization that takes place when the union of the sex cells happens outside the body
Answer:external
Explanation:EDGE 2021
A plane wishes to fly due north to an airport which is 205 km away. The plane can fly at a speed in still air of 220 km/h. A wind of 43 km/h blows from east to west.
a. In which direction,relative to north, should the plane head to reach it’s destination?
b. How long does this take?
Answer:
nique ta mama
Explanation:
A fan is set on a desk next to a stack of paper. The fan is turned on and then turned
on HIGH SPEED. Which of the following would best apply Newton's First Law to this
example.
The papers accelerated due to the force of the fan.
The papers are acted upon by an unbalanced force from the fan.
O The papers exerted an equal force on the air blown by the fan.
O The papers that were the heaviest were blown the closest.
Answer:
The second option - the papers are acted upon by an unbalanced force from the fan.
Explanation:
3. What were some of the materials use in ancient history times in the
making of the ball?
Answer: Fur stuffed with feathers.
Explanation:
Ball games were popular throughout ancient history but also in the Middle Ages. Many of these games are the forerunners of today's American football and soccer. There are data on almost all continents on how peoples practiced some ball sports in antiquity. The game balls' look was not like today, it was mostly egg-shaped, and the materials for making it were different. The most common material for making the ball was leather, which was filled with feathers.
If a 500-pound object is moved 200 feet how much work is being done?
a. 200 FT LB
b. 500 FT LB
c. 1000 FT LB
d. 100,000 FT LB
Answer:
D
Explanation:
Work = Distance x Mass
work done = 100,000 FT LB
What is work done ?
Work is done whenever a force moves something over a distance or The work done by a force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement.
Work done = force * displacement
given :
force = 500 pound
displacement = 200 feet
work done = 500 * 200 = 100,000 FT LB
correct option is d. 100,000 FT LB
learn more about work done
https://brainly.com/question/13662169?referrer=searchResults
#SPJ2
6th grade science I mark as brainliest !
Answer:
The answer is B. molecules, atoms, elements, compund, cells, tissues, organs, organ systems, and organism
A motorcycle reaches the speed of 40 m / s, how far does it travel in 10 seconds?
Answer:
d = 200 m
Explanation:
Data:
Initial Velocity (Vo) = 0 m/s Final Velocity (Vf) = 40 m/s Time (t) = 10 s Distance (d) = ?Use formula:
[tex]\boxed{d=\frac{Vf+Vo}{2}*t}[/tex]Replace:
[tex]\boxed{d=\frac{40\frac{m}{s}+0\frac{m}{s}}{2}*10s}[/tex]Sum in the numerator:
[tex]\boxed{d=\frac{40\frac{m}{s}}{2}*10s}[/tex]It divides:
[tex]\boxed{d=20\frac{m}{s}*10s}[/tex]Simplify the seconds (s), and multiply:
[tex]\boxed{d=200\ m}[/tex]How far does it go?
Travel a distance of 200 meters.
Which option tells the forces that influence the movement of earths plates
Answer:
Gravity
Explanation:
A wall clock has a minute hand with a length of 0.46 m and an hour hand with a length of 0.24 m. Take the center of the clock as the origin, and use a Cartesian coordinate system with the positive x axis pointing to 3 o'clock and the positive y axis pointing to 12 o'clock. Write the vector that describes the displacement of a fly if it quickly goes from the tip of the minute hand to the tip of the hour hand at 3:00 P.M. (Let vector D represents the displacement of the fly.)
Answer:
the vector that describes the displacement of a fly if it quickly goes from the tip of the minute hand to the tip of the hour hand at 3:00 P.M is ; { 0.24 m(i) - 0.46 m(j) }
Explanation:
Given the data in the question;
as illustrated in the image below,
3:00 pm means
the hour hand is on 3 i.e along x-axis
while the minute hand is on 12 i.e along y-axis
so Displacement will be;
D = ( 0.24 + 0i) - ( 0 + 0.46j )
D = { 0.24 m(i) - 0.46 m(j) }
Therefore, the vector that describes the displacement of a fly if it quickly goes from the tip of the minute hand to the tip of the hour hand at 3:00 P.M is ; { 0.24 m(i) - 0.46 m(j) }
What do you call the height of a wave?
a. wavelength
b. frequency
c. amplitude
d. resonance
Answer:
amplitude is the answer
WILL GIVE BRAINLIEST!!
What is a way to transfer charge in which an object becomes polarized?
Answer:
I answered Number 4 (Solids and Elasticity)
Explanation:
solids and elasticity
Determine one way you can contribute to water in the atmosphere in your day-to-day activities pleaseeeee helppp
Answer:
agricultural production of food
Explanation:
On a level test track, a car with antilock brakes and 90% braking efficiency is determined to have a theoretical stopping distance (ignoring aerodynamic resistance) of 408 ft (after the brakes are applied) from 100 mi/h. The car is rear-wheel drive with a 110-inch wheelbase, weighs 3200 lb, and has a 50/50 weight distribution (front and back), a center of gravity that is 22 inches above the road surface, an engine that generates 300 ft-lb of torque, and overall gear reduction of 8.5 to 1 (in first gear), a wheel radius of 15 inches and a driveline efficiency of 95%. What is the maximum acceleration from the rest of this car on this test track
Answer:
a = 30.832 ft/s²
Explanation:
To solve this problem let's start by finding the braking acceleration using kinematics, where the distance is x = 408 ft, the initial velocity vo = 100 mi / h and the final velocity is zero v = 0
v² = v₀² - 2 a x
0 = v₀² - 2ax
a = [tex]\frac{v_o^2}{2x}[/tex]
Let's start by reducing the magnitudes to ft / s
v₀ = 100 mi / h (5280 foot / 1 mile) (1h / 3600 s) = 146.666 ft / s
let's calculate
a = [tex]\frac{146.66^2}{2 \ 408}[/tex]
a = 26.36 ft / s²
Let's call this acceleration a_effective, this acceleration is in the opposite direction to the speed of the vehicle.
Let's use a rule of three (direct proportions) to find the acceleration applied by the brake system (a1) which has an efficiency of 95%. or 0.95
a₁ = [tex]\frac{a_e}{0.95}[/tex]
Let's use another direct proportion rule If the acceleration of the brake system (a₁) for an applied acceleration (a) with an efficiency of 0.90
a = [tex]\frac{a_1}{0.90}[/tex]
we substitute
a = [tex]\frac{a_e}{0.95 \ 0.90}[/tex]
let's calculate
a = [tex]\frac{26.36}{ 0.95 \ 0.90}[/tex]
a = 30.832 ft/s²
This is the maximum relationship that the vehicle can have for when it brakes to stop at the given distance
Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the power measured at the detector when only the vertical arm is blocked is 2.25 mW, while the power measured at the detector when only the horizontal arm is blocked is only 2.025 mW. Assume initially the intensity is at its maximum. How much would we need to translate the perfect mirror to the right to get a minimum intensity at detector, and what is that minimum intensity
This question is incomplete, the complete question;
you make an interferometer using 50-50 beam splitter and two mirrors, one being a perfect mirror and one which does not reflect all light. The wavelength of the 9 mW incident laser is 400 nm.
Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the power measured at the detector when only the vertical arm is blocked is 2.25 mW, while the power measured at the detector when only the horizontal arm is blocked is only 2.025 mW. Assume initially the intensity is at its maximum. How much would we need to translate the perfect mirror to the right to get a minimum intensity at detector, and what is that minimum intensity
Options;
a) 200 nm; 0.9 mW
b) 100 nm, 0.0059 mW
c) 200 nm; 0 mW
d) 100 nm; 0.9 mW
e) 200 nm; 0.0059 mW
Answer:
the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;
100 nm; 0.0059 mW
Option b) 100 nm, 0.0059 mW is the correct answer
Explanation:
Given that the instrument here is an interferometer.
Maximum intensity is obtained when the two waves are exactly in phase.
that is the peaks (crusts and troughs) and nodes (zero value points) of the two waves will be at the exact same point when the wave falls on the detector.
The phase factor of this point is taken as ∅ = 0
Now, to get a minimum point, the phase difference between the two waves should be should be ∅ = π
This corresponds to a path difference between the two waves as half of the wavelength. λ/2
The light gets reflected from the mirror.
Hence, when we move the mirror by a length l, the extra/less path the light has to travel is 2l (light is going and coming back)
hence, to get a path difference of λ/2 the mirror should move half of this distance only
so, the mirror should move;
[tex]l[/tex] = λ/4
here, wavelength is 400nm
the length moved by the mirror = 400/4 = 100 nm
The intensity is given by the equation;
[tex]l[/tex] = [tex]l[/tex]1 + [tex]l[/tex]2 + 2√[tex]l[/tex]1[tex]l[/tex]2cos(∅)
where
[tex]l[/tex]1 = 2.25 mW
[tex]l[/tex]2 = 2.025 mW
∅ = π
so we substitute
[tex]l[/tex] = 2.25 + 2.025 - 2√(2.25 × 2.025)
[tex]l[/tex] = 4.275 - 4.26907
[tex]l[/tex] = 0.0059
Therefore; the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;
100 nm; 0.0059 mW
Option b) 100 nm, 0.0059 mW is the correct answer
A wave in which the movement of the wave is perpendicular to the movement of the wave traveling through the medium-compression wave (longitudinal wave)
True or false?
Answer:
I'm rusty sorry if I'm wrong but true?
A car initially traveling at 15 m/s speeds up at a constant rate of 2.0 m/s2 for 3 seconds. The velocity of the car at the end of the 3 second interval is _________ m/s.
Answer:
Vf = 21 m/s
Explanation:
Data:
Initial Velocity (Vo) = 15 m/sAcceleration (a) = 2.0 m/s²Time (t) = 3 sFinal Velocity (Vf) = ?Use formula:
Vf = Vo + a * tReplace:
Vf = 15 m/s + 2.0 m/s² * 3sMultiply the acceleration with time:
Vf = 15 m/s + 6 m/sSolve the sum:
Vf = 21 m/sThe velocity of the car at the end of the 3 second interval is 21 meters per second.
A warm hockey puck has a coefficient of restitution of 0.50, while a frozen hockey puck has a coefficient of restitution of only 0.35. In the NHL, the pucks to be used in games are kept frozen. During a game, the referee retrieves a puck from the cooler to restart play but is told by the equipment manager that several warm pucks were just put into the cooler. To check to make sure he has a game-ready puck, the referee drops the puck on its side from a height of 2 m. How high should the puck bounce if it is a frozen puck
Answer:
the required height is 0.2449 m only
Explanation:
Given the data in the question;
Initial height = 2m
so speed of the puck before hitting the ground will be;
u² = 2gh
Initial speed u_ball = √2gh
u_ball = √( 2 × 9.8 × 2 )
u_ball = √39.2
u_ball = 6.26 m/s
given that; FOR THE FROZEN PUCK, coefficient of restitution = 0.35 only
R = - (v_ball - v_ground / u_ball - u_ ground)
so
0.35 = - (v_ball - 0 / 6.26 - 0)
0.35 = -v_ball / - 6.26
-v_ball = 0.35 × (- 6.26)
-v_ball = -2.191 m/s
v_ball = 2.191 m/s
to get the height;
v² = 2gh
h = v² / 2g
we substitute
h = (2.191)² / 2×9.8
h = 4.800481 / 19.6
h = 0.2449 m
Therefore, the required height is 0.2449 m only
A 4 m long metal bar tapped by a hammer on one end. A microphone recording the sound at the other end picks up two sound pulses; one which traveled through the metal bar and other which was traveling through room temperature air. These two pulses are separated by 11 msec. What is the speed of sound in the metal
Answer:
Explanation:
Speed of sound in air at room temperature = 346 m /s
Let speed of sound in metal given = v m /s
Time take by sound wave travelling through air = 4 / 346 = .01156 s
Time taken by sound wave travelling through metal = 4 / v s
Given ,
.01156 - 4 / v = 11 x 10⁻³ s = .011 s ( time taken by sound travelling in air is more )
4 /v = .01156 - .011 = .00056
v = 4 / .00056 = 7142.85 m /s
You are removing branches from your roof after a big storm. You throw a branch horizontally from your roof, which is a height 3.00 m above the ground. The branch lands a horizontal distance 8.00 m away from where you threw it (assuming you are the 0 position in x, and the branch traveled in the x direction). You can assume there is no air resistance. You can assume that the upwards direction is positive. What is the initial velocity in x of the branch (how fast did you throw the branch)
Answer:
The initial velocity in the x-direction with which the branch was thrown is approximately 10.224 m/s
Explanation:
The given parameters of the motion of the branch are;
The height from which the branch is thrown = 3.00 m
The horizontal distance the branch lands from where it was thrown, x = 8.00 m
The direction in which the branch is thrown = Horizontally
Therefore, the initial vertical velocity of the branch, [tex]u_y[/tex] = 0 m/s
The time it takes an object in free fall (zero initial downward vertical velocity) to reach the ground is given as follows;
s = [tex]u_y[/tex]·t + 1/2·g·t²
Where;
[tex]u_y[/tex] = 0 m/s
s = The initial height of the object = 3.00 m
g = The acceleration due to gravity = 9.8 m/s²
∴ s = 0·t + 1/2·g·t² = 0 × t + 1/2·g·t² = 1/2·g·t²
t = √(2·s/g) = √(2 × 3/9.8) = (√30)/7 ≈ 0.78246
The horizontal distance covered before the branch touches the ground, x = 8.00 m
Therefore, the initial velocity in the horizontal, x-direction with which the branch was thrown, 'uₓ', is given as follows;
uₓ = x/t = 8.00 m/((√30)/7 s)
Using a graphing calculator, we get;
uₓ = 8.00 m/((√30)/7 s) = (28/15)·√30 m/s ≈ 10.224 m/s
The initial velocity in the horizontal, x-direction with which the branch was thrown, uₓ ≈ 10.224 m/s.
