Answer:
a = 5 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}=v_{o}+a*t[/tex]
where:
Vf = final velocity = 20 [m/s]
Vo = initial velocity = 10 [m/s]
t = time = 2 [s]
a = acceleration [m/s²]
Now replacing:
[tex]20 =10 +a*2\\10=2*a\\a=5[m/s^{2} ][/tex]
200. newton·meters of work is put into a machine over a distance of 20. meters. The machine does 150. newton·meters of work as it lifts a load 10. meters high. What is the mechanical advantage of the machine?
Answer:
Answer:
Mechanical advantage of the machine is 1.5
Explanation:
The formula for mechanical advantage is:
MA = OutputForce/InputForce
To calculate this, we need the force of input and the force of output.
The formula to calculate the force given the work (W) and the distance (d) is:
F = W/d
Calculating the input force Fi:
Fi = 200Nm/20m = 10 N
Calculating the output force Fo:
Fo = 150Nm/10m = 15 N
Thus, the mechanical advantage MA is:
MA = 15N / 10N = 1.5
"When the ball leaves the ramp at Point B, students measure the horizontal" distance traveled. They repeat the experiment five times, being careful to release the ball from the same starting Point A and find the average horizontal distance traveled to be 2.0 m. One student suggests they use a stopwatch to find the time the ball is in the air whereas another student suggests they use a meter stick. Given these materials, describe the procedure students should follow to minimize error and calculate the speed of the ball as it leaves the ramp.
Answer:
Speed of ball as it leaves ramp is 2 m/s
Explanation:
From the question it is given that the average distance traveled by ball from point A to B is 2.0 m and the time taken is 1sec
one student is using stopwatch to calculate the time taken by ball and another student is using meter stick to calculate the distance traveled
since speed of ball is given by v = [tex]\frac{distance}{time}[/tex] thus,
the speed = 2 m/s
now for minimizing the error of speed it is necessary to record the readings by single students at-least 5 times and take average
by doing this, the error of speed calculation will be minimum as the it decreases the error due to random error of system caused by taking the reading by different students
state two uses of total internal reflection
Answer:
The phenomenon of total internal reflection of light is used in many optical instruments like telescopes, microscopes, binoculars, spectroscopes, periscopes
A train travels at 100 km/h heading east to reach a town in 4 hours. The train then reversed ad heads west at 50 km/h for 4 hours
Which of the following are car safety features that rely on increasing the
time of a crash? *
seat belts
crumple zones
air bags
all of the above
What is her velocity?
O 1.5 m/s
O 2 m/s
O 2.5 m/s
O 5 m/s
Answer:
C. 2.5
Explanation:
It said it was right, so that's cool.
why iron-rims are heated before putting them on the wheels of Bullock-carts?
Answer:
Explanation:
The iron ring to be put on the rim of a cart wheel is always of slightly smaller diameter than that of the wheel. When the iron ring is heated to become red hot, it expands and slips on to the wheel easily. When it is cooled, it contracts and grips the wheel firmly.
Explanation:
Iron rims are made slightly smaller than the wheels. They are heated red hot before fixing them on the cart wheels to expand them so that they can be easily fixed on the wheels and water is poured on them to cool them. As the rims cool they contract and take the shape of the wheel and get firmly fixed on the wheels.
So iron-rims are heated before putting them on the wheels of Bullock-carts
Hope it will help :)
A new prototype cup has been designed to keep liquids, such as hot coffee or cold annk, near their original temperature for long periods of time Which statement describes the function of the newly designed
cup
Answer:
hope this was good for u and I believe it would be solid
1. Which is not an example of vaporization?
Bubbles form as water boils on the stove.
Water droplets form on a mirror during a shower.
Air gains moisture as it moves over the ocean.
Wet pavement dries after a rain shower.
"Wet pavement dries after a rain shower" is not an example of vaporization. The correct answer is D.
What is vaporization?
Vaporization is the process by which a liquid is converted into a gas or vapor. The two main types of vaporization are boiling and evaporation.
Here in the Question,
Option A: Bubbles form as water boils on the stove. This is an example of vaporization by boiling, which occurs when a liquid is heated to its boiling point and its vapor pressure becomes equal to the atmospheric pressure.
Option B: Water droplets form on a mirror during a shower. This is an example of vaporization by evaporation, which occurs when a liquid changes into a gas at temperatures below its boiling point. In this case, the water on the mirror evaporates due to the heat and humidity in the shower.
Option C: Air gains moisture as it moves over the ocean. This is also an example of vaporization by evaporation. The warm air over the ocean absorbs moisture from the water surface, causing the water to evaporate and form water vapor in the air.
Option D: Wet pavement dries after a rain shower. This is not an example of vaporization. The water on the pavement may evaporate due to the heat and dryness of the surrounding air, but this process does not involve a liquid changing into a gas or vapor. Rather, the water on the pavement may be absorbed by the ground, run off into nearby drainage systems, or be removed by physical means like squeegees.
Therefore, among the given options, option D is not an example of vaporization.
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7. An outfielder throws a baseball to the first baseman at a speed of 19.6
m/s and an angle of 30° above the horizontal. If the ball in the described
situation is caught at the same height from which it was thrown, calculate
the amount of time the ball was in the air. (use -9.8 m/s2 for g in this
problem] [include the numerical answer and correct unit] *
The amount of time that the ball took in air is 1 second.
We must recall that the ball was thrown as a projectile and its time of flight must be obtained from the formula for the time of flight of a projectile as follows;
T = 2usinθ/g
u = initial velocity = 19.6 m/s
θ = angle of projection = 30°
g = acceleration due to gravity = 9.8 m/s2
Substituting values;
T = 2 × 19.6 m/s sin (30°)/9.8 m/s2
T = 1 second
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____ is the perceived frequency of a sound wave.
Answer:
Pitch is the correct answer!
Explanation:
~Hope this helps! :)
ANSWER FAST PLSSSSSS Use this diagram of the wave below to answer any questions that follow
Answer:
1. Letter A indicates the crest.
2. Letter E indicates the wavelength.
3. Letter C indicates the trough.
Explanation:
1. Determination of the the crest.
A crest is simply defined as a point on a wave where the displacement is maximum (i.e highest).
Considering the diagram given above, the point where the displacement is maximum is A. Therefore, A is the crest of the wave.
2. Determination of the wavelength.
The wavelength of a wave is simply defined as the distance between two successive crest or trough.
Obeserving the diagram above, we can see that E gives the distance between two successive crest. Therefore, E is the wavelength of the wave.
3. Determination of the trough.
A trough is simply defined as a point on a wave where the displacement is minimum (i.e lowest). Thus we can say that the trough is the opposite of the crest.
Considering the diagram given above, the point where the displacement is minimum is C. Therefore, C is the trough of the wave.
A plane starting from rest accelerates
to takeoff velocity of 75 m/s in 15
seconds. What was the plane’s
acceleration and how far did it travel
before takeoff?
Answer:
The acceleration is 5 m/s² and the distance is 562.5 m.
Explanation:
Given that,
Initial velocity of the plane, u = 0 (at rest)
Final speed, v = 75 m/s
Time, t = 15 s
We need to find the acceleration of the plane and distance it travel before takeoff.
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{75-0}{15}\\\\a=5\ m/s^2[/tex]
Let the distance is d.
[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{(75)^2-(0)^2}{2\times 5}\\\\d=562.5\ m[/tex]
So, the acceleration is 5 m/s² and the distance is 562.5 m.
A truck pushes a pile of dirt horizontally on a frictionless road with a net force of 20\, \text N20N20, start text, N, end text for 15.0\,\text m15.0m15, point, 0, start text, m, end text. How much kinetic energy does the dirt gain
Answer:
300 Nm ; 300 J
Explanation:
Given that:
Force (F) = 20 N
Distance (d) = 15 m
The kinetic energy (Workdone) = Force * Distance
Kinetic Energy = 20N * 15m
Kinetic Energy = 300Nm
K. E = 1/2
Answer:
77.0k/M
Explanation:
Today basketball is played with nine players on the court at a time.
True
False
Answer:
yes basketball is played with nine players
why should a rain gauge be raised 30cm above the ground
To avoid splashing, rain gauges are positioned 30 cm above the surface of the ground.
What is the reason behind a rain gauge be raised 30cm above the ground?The rainfall may be precisely measured by a rain gauge without any loss from evaporation. It is a tool for measuring the amount of precipitation that falls over a certain area. As a result, it gauges rainfall. A millimeter of measured precipitation is equal to one liter of rainfall per square meter.
Standard rain gauges have a 150–170 cm channeled aperture. They are made to be a straightforward passive collector. For measuring the amount of precipitation, use a graduated measuring glass. To avoid rainwater splashing while measuring rainfall, the rain gauze should be positioned 30 cm above the ground.
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A guitar player can change the frequency of a string by "bending" it-pushing it along a fret that is perpendicular to its length. This stretches the string, increasing its tension and its frequency. The B string on a guitar is 64 cm long and has a tension of 74 N. The guitarist pushes this string down against a fret located at the center of the string, which gives it a frequency of 494 Hz. He then bends the string, pushing with a force of 4.0 N so that it moves 8.0 mm along the fret.
* What is the new frequency?
Answer:
[tex]f'=504hz[/tex]
Explanation:
From the question we are told that
The B string on a guitar is 64 cm long
The B string tension tension of 74 N.
Frequency of 494 Hz
Pushed with a Force of 4.0 N
It moves 8.0 mm along the fret.
Generally the equation for frequency of ring under tension is mathematically given as
[tex]2Lf=\sqrt{x\frac{T}{\mu} }[/tex]
[tex]2*(64/100)*494=\sqrt{\frac{74}{m/0.64}[/tex]
[tex](632.32)^2={\frac{74}{m/0.64}[/tex]
[tex](632.32)^2=74*{\frac{0.64}{m}[/tex]
[tex](632.32)^2={\frac{47.36}{m}[/tex]
[tex]m=1.18450761*10^-^4[/tex]
Therefore finding the New frequency f'
[tex]f'=\frac{(\sqrt{\frac{74+11}{(\frac{1.18450761*10^-^4}{0.642})}})}{2*0.642}[/tex]
[tex]f'=\frac{(\sqrt{(\frac{74+11}{1})*(\frac{0.642}{1.18450761*10^-^4}})}{2*0.642}[/tex]
[tex]f'=\frac{(\sqrt{(\frac{85}{1})*(\frac{0.642}{1.18450761*10^-^4}})}{2*0.642}[/tex]
[tex]f'=\frac{(\sqrt{(\frac{54.57}{1.18450761*10^-^4}})}{2*0.642}[/tex]
[tex]f'=\frac{678.7471973}{2*0.642}[/tex]
[tex]f'=572.9111096hz[/tex]
Assuming the density of water is 62.4 lbm/ft3, and using standard gravity, how much does 1 gallon of water weigh?
Answer:
The weight of the water is 37.14 N.
Explanation:
Given;
density of water, ρ = 62.4 lbm/ft³
volume of water, V = 1 gallon = 0.1338 ft³
The mass of the water is calculated as;
m = ρV
[tex]m = 62.4 \ \frac{lbm}{ft^3} \ \times \ 0.1338 \ ft^3\\\\m = 8.349 \ lbm[/tex]
1 lbm = 0.454 kg
8.349 lbm = ?
= 3.79 kg
The weight of the water is calculated as;
W = mg
where;
g is acceleration due to gravity = 9.8 m/s²
W = (3.79)(9.8)
W = 37.14 N.
Therefore, the weight of the water is 37.14 N.
A particle moves along the x axis so that its velocity at time t is given by v(t)=
−(t + 1) sin(t2/2). At time t = 0, the particle is at position x = 1. Find the acceleration of the particle at time t = 4. Is the speed of the particle increasing at t = 4? Why or why not?
.
Answer:
a = 0.7267 , acceleration is positive therefore the speed is increasing
Explanation:
The definition of acceleration is
a = dv / dt
they give us the function of speed
v = - (t-1) sin (t² / 2)
a = - sin (t²/2) - (t-1) cos (t²/2) 2t / 2
a = - sin (t²/2) - t (t-1) cos (t²/2)
the acceleration for t = 4 s
a = - sin (4²/2) - 4 (4-1) cos (4²/2)
a = -sin 8 - 12 cos 8
remember that the angles are in radians
a = 0.7267
the problem does not indicate the units, but to be correct they must be m/s²
We see that the acceleration is positive therefore the speed is increasing
If the balloon can barely lift an additional 3500 N of passengers, breakfast, and champagne when the outside air density is 1.23 kg/m3, what is the average density of the heated gases in the envelope
The complete question is :
A hot-air balloon has a volume of 2100 m3 . The balloon fabric (the envelope) weighs 860 N . The basket with gear and full propane tanks weighs 1300 N .
If the balloon can barely lift an additional 3400 N of passengers, breakfast, and champagne when the outside air density is 1.23kg/m3, what is the average density of the heated gases in the envelope?
Solution :
Given volume of the hot air balloon [tex]$=2100 \ m^3$[/tex]
The balloon fabric weights = 860 N
The weight of the basket with the gear and propane tank = 1300 N
Density of outside air [tex]$= 1.23 \ kg/m^3$[/tex]
∴ Total pay load = Weight of the air displaced - weight of gas inside the balloon
Total pay load = 860 + 1300 + 3400
= 5560 N
Mass = density x volume
Weight [tex]$= \text{mass} \times g$[/tex]
Weight = volume x density [tex]$\times \text{ acceleration due to gravity (g)}$[/tex]
Weight of the displaced air = 2100 x 1.23 x 9.8
= 25313 N
Weight of the gas inside the balloon = density [tex]$\times \text{ acceleration due to gravity (g)}$[/tex] x volume
= density x 9.8 x 2100
= density x 20580 N
Therefore substituting the values, we get
⇒ 25313 - (density x 20580) = 5560
⇒ density [tex]$=\frac{19753}{20580}$[/tex]
[tex]$= 0.96 \ kg/m^3$[/tex]
So the density of the heated gas [tex]$= 0.96 \ kg/m^3$[/tex]
Consider two positive charges separated by a distance ((A 20)/100) meters. One charge has a value of (B 5) mC (milli-coulombs) and is located to the right of the other (C 5) mC charge. Assume the left charge is the origin and the other charge is located with a positive horizontal position. How large and in what direction is the electrostatic field at the mid-point, directly between the two charges
Answer:
E_total = 0
Explanation:
In this exercise we are asked for the electric field created by two charges at the midpoint between them, since the electric field is a vector magnitude it must be added as vectors. The equation that describes the electric field for a charge is
[tex]E = k \frac{q}{r^{2} }[/tex]
in this case the two charges are positive, therefore the field is salient, in the adjoint we can see a diagram of the system
E_total = E₁ -E₂
we calculate the electric field of charge 1 with q = + 5 10⁻³ C and a distance r = L / 2 = 100/2 = 50 m
E₁ = [tex]\frac{9 10^{9} \ 5 10^{-3} }{ 50^{2} }[/tex]
E₁ = 1.8 10⁴ N / C
the magnitude of the electric field of charge 2 is equal, but the sense is opposite.
Since the two charges have the same sign, the addition of vectors gives zero
E_total = 0
if the charges were of different signs, the fields would be added
E_total = 3.6 10⁴ N / A
A bowling ball hits two standing bowling pins at the same time. Which of the following is true? Assume that all collisions are elastic. A. The sum of the momentum of both pins after the collision will equal the momentum of the ball before the collision. B. The momentum of the ball after the collision will equal the momentum of the ball before the collision. C. The sum of the momentum gained by both pins will equal the amount of momentum lost by the ball. D. The momentum of each pin after the collision will equal the momentum of the ball before the collision.
Answer:
C. The sum of the momentum gained by both pins will equal the amount of momentum lost by the ball.
Explanation:
The law of conservation of momentum states that total momentum before collision must be equal to the total momentum after collision.
Momentum could be transferred from one object to another during collision. When the ball hits the stationary pins, momentum is transferred from the ball to the pins.
Since the collision is elastic, momentum is conserved hence total momentum gained by the pins equals the total momentum lost by the ball.
Paola conducts an investigation to determine how temperature affects plant growth. Each plant started at the same height, received the same amount of water and was planted in the same type of soil She
recorded her results every seven days for three weeks.
Pls help
Answer:
HE IS WON
Explanation:
JESUS
At a certain distance from the center of the Earth, a 0.4-kg object has a weight of 2.0 N. (a) Find this distance. (b) If the object is released at this location and allowed to fall toward the Earth, what is its initial acceleration
Answer:
a) The distance of the object from the center of the Earth is 8.92x10⁶ m.
b) The initial acceleration of the object is 5 m/s².
Explanation:
a) The distance can be found using the equation of gravitational force:
[tex]F = \frac{GMm}{r^{2}}[/tex]
Where:
G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²
M: is the Earth's mass = 5.97x10²⁴ kg
m: is the object's mass = 0.4 kg
F: is the force or the weight = 2.0 N
r: is the distance =?
The distance is:
[tex]r = \sqrt{\frac{GMm}{F}} = \sqrt{\frac{6.67 \cdot 10^{-11} Nm^{2}/kg^{2}*5.97 \cdot 10^{24} kg*0.4 kg}{2.0 N}} = 8.92 \cdot 10^{6} m[/tex]
Hence, the distance of the object from the center of the Earth is 8.92x10⁶ m.
b) The initial acceleration of the object can be calculated knowing the weight:
[tex] W = ma [/tex]
Where:
W: is the weight = 2 N
a: is the initial acceleration =?
[tex] a = \frac{W}{m} = \frac{2 N}{0.4 kg} = 5 m/s^{2} [/tex]
Therefore, the initial acceleration of the object is 5 m/s².
I hope it helps you!
A rock weighing 6 Newton’s is lifted 2 meters. How much work is done?
Answer:
12 Joules
Explanation:
I believe a Joule is a Newton meter, and work is represented in joules, so 6 N * 2m = 12 Nm (joules)
In Requiem for a Dream, the DP uses a device that straps the camera to the actor's chest to achieve a very subjective shot. What is this apparatus called
Answer:
A snorricam
Explanation:
The apparatus used here is a snorricam. A snorricam can be described to be a camera device which is used during the making of films. The snorricam is rigged to the actor. It faces him or her directly in such a way that as they walk, they would not appear to make any movements but the things around them would. It is also called the bodymount camera.
Which of the following is true about a field goal?
A shot from half court.
It is worth two points.
It is shot from behind the 3-point line.
It is a free shot given when a player is fouled.
Answer:
which sport are you referring to?
Explanation:
You throw a ball into the air. Which two forces cause the ball to gradually stop moving upward and then fall back to Earth?
A.
Balanced forces
B.
Friction
C.
Normal force
D.
Gravitational force
will mark brainliest
Answer:
I'm pretty sure the answer is D
Explanation:
Honestly it's just a guess so let me know if it's right :3
Answer:
it should be A and B
Explanation:
because obviously gravity causes it to fall down and the second force acted upon it is Friction from the air and ball coming in contact with eachother. I hope this helps
PLEASE HELP The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the wave shown?
A.6 Hz
B.0.3 Hz
C.0.7 Hz
D.3 Hz
Answer:
[tex]F = 0.3\ Hz[/tex]
Explanation:
Given
See attachment for the graph
Required
Determine the frequency
Frequency (F) is calculated as:
[tex]F = \frac{1}{T}[/tex]
Where
T = Time to complete a period
From the attachment, the wave complete a cycle or period in 3 seconds..
So:
[tex]F = \frac{1}{3s}[/tex]
[tex]F = 0.333\ Hz[/tex]
[tex]F = 0.3\ Hz[/tex] --- Approximated
Answer:
0.3Hz
Explanation:
I took the test :)
. A falling rubber ball of mass 0.025 kg strikes the ground traveling straight down at 4.0 m/s. Find the magnitude of the impulse that the ground gives to the ball if
a) the ground is soft and the ball stops dead
b) the ground is hard and the ball bounces straight back at 2.0 m/s.
Given that,
Mass of the ball, m = 0.025 kg
Initial speed, u = 4 m/s
To find,
Impulse when (a) the ground is soft and the ball stops dead
(b) the ground is hard and the ball bounces straight back at 2.0 m/s.
Solution,
Impulse = change in momentum
J = m(v-u)
(a) u = 0 (as it stops)
J = 0.025(4-0)
J = 0.1 N-m
(b) v = 2 m/s
J = 0.025(4-2)
= 0.05 N-m
Therefore, this is the required solution.