Answer:
Deflating the tires by pulling the valve stems
Explanation:
When an accident happens, the EMT process is to take off the top of the vehicle of the highly damaged vehicle so that it can easily be assessed to rescue the vehicle occupants. This creates a bigger exit for the rescuers to take out the people. Vehicle extrication means removing the vehicle from around someone involved in a fatal collision, when other other ways of exiting is not advised or possible. to stabilize the vehicle you have to deflate all tires by pulling the valve stems.
(Serious Please) patulong
Answer:
a. Potential energy is highest at Part A
The kinetic energy is highest at Part C and Part D
b. The potential energy is lowest at Part C and Part D
c. The roller coater has equal amount of potential and kinetic energy at Part B, Part D and part F
2) Yes, the mechanical energy is the same from point A to F according to the first law of thermodynamics
Explanation:
The total mechanical energy is constant where the roller coaster moves by only the initial velocity, and the the force of gravity
Total mechanical energy, M.E. = Kinetic energy, K.E. + Potential energy, P.E.
M.E. = K.E. + P.E. = Constant
Therefore, we have;
a. Potential energy is the energy stored in a body, due to its position or elevation, state or arrangement
The higher the elevation, the higher the potential energy, therefore, the highest amount of potential energy is gained when the roller coaster is at the highest point in the motion = Part A
From M.E. = K.E. + P.E. = Constant, the highest kinetic energy is given at the point the roller coaster has the lowest potential energy, which corresponds with the lowest points = Part C and Part D
b. Potential energy, which is the energy of body due to its position or state is lowest at the lowest points = Part C and Part D
c. The value of potential energy, P.E. due to elevation, can be found as follows
P.E. = Mas, m × Gravity, g × Height, h
Therefore, the potential energy will be half the maximum value where the height, h = (Maximum height)/2 and given that M.E. = K.E. + P.E., the kinetic energy, will increase by the same amount, and we have;
K.E. = P.E. at the half the maximum height of the track = Part B, Part D and part F
2) The mechanical energy is the input energy, which according to the first law of thermodynamics cannot be created and destroyed an it is therefore, constant and it is the same from point A to F
The difference between starting and ending
positions is
distance
displacement
Answer:
displacement
Explanation:
Motion can be defined as a change in the location (position) of a physical object or body with respect to a reference point.
This ultimately implies that, motion would occur as a result of a change in location (position) of an object with respect to a reference point or frame of reference i.e where it was standing before the effect of an external force.
A reference point refers to a location or physical object from which the motion (movement) of another physical object or body can be determined.
Mathematically, the motion of an object is described in terms of acceleration, time, distance, speed, velocity, position, displacement, etc.
Displacement can be defined as the change in the position of a body or an object. It is a vector quantity because it has both magnitude and direction.
This ultimately implies that, the difference between the starting and ending positions of a physical object is generally referred to as displacement
convert the following as instructed
67 kg into gram
explain step by step
please reply quickly its urgent
Answer:
1kg = 1000g
67kg therefore would be equal to
67 x 1000 = 67,000g = 6.7 x 10⁴g.
Se lanza una pelota de béisbol desde la azotea de un edificio de 25 m de altura con velocidad inicial de magnitud 10 m/s y dirigida con un ángulo de 63.1° sobre la horizontal. A) ¿Qué rapidez tiene la pelota justo antes de tocar el suelo? Use métodos de energía y desprecie la resistencia del aire.
Answer:
v_f = 24.3 m / s
Explanation:
A) In this exercise there is no friction so energy is conserved.
Starting point. On the roof of the building
Em₀ = K + U = ½ m v₀² + m g y₀
Final point. On the floor
Em_f = K = ½ m v_f²
Emo = Em_g
½ m v₀² + m g y₀ = ½ m v_f²
v_f² = v₀² + 2 g y₀
let's calculate
v_f = √(10² + 2 9.8 25)
v_f = 24.3 m / s
A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop. (a) Assuming a constant angular acceleration, find the time for it to come to rest. (b) What is its angular acceleration
Answer:
The time of motion is 333.3 s
The angular acceleration is -0.0045 rad/s²
Explanation:
Given;
angular distance of the flywheel, θ = 40 rev
initial angular speed, [tex]\omega_i[/tex] = 1.5 rad/s
When the wheel comes to rest, the final angular speed, [tex]\omega_f[/tex] = 0
The angular acceleration is calculated as follows;
[tex]\omega_f^2 = \omega_i^2 + 2\alpha \theta \\\\0 = (1.5 \ rad/s)^2 + 2\alpha (40 \ rev\times \frac{2\pi \ rad}{1 \ rev} )\\\\0 = 2.25 + 160\pi \alpha\\\\160\pi \alpha = - 2.25\\\\\alpha = -\frac{2.25 }{160\pi} \\\\\alpha = -0.0045 \ rad/s^2[/tex]
The time of motion is calculated as;
[tex]\omega_f = \omega _i + \alpha t\\\\0 = 1.5 + (-0.0045t)\\\\0 = 1.5 - 0.0045t\\\\0.0045t = 1.5\\\\t = \frac{1.5}{0.0045} = 333.3 \ s[/tex]
a wave in which particles of the medium vibrate at right angles to the direction that the wave travels is called?
Explanation:
Longitudinal waves are waves in which the vibration of the medium is parallel to the direction the wave travels and displacement of the medium is in the same (or opposite) direction of the wave propagation. Mechanical longitudinal waves are also called compressional or compression waves, because they produce compression and rarefaction when traveling through a medium, and pressure waves, because they produce increases and decreases in pressure.
Answer:
In transverse waves, particles of the medium vibrate to and from in a direction perpendicular to the direction of energy transport.
Explanation:
hope it will help u
A dart is thrown horizontally with an initial speed of 19 m/s toward point P, the bull's-eye on a dart board. It hits at point Q on the rim, vertically below P, 0.19 s later. (a) What is the distance PQ
Answer:
Explanation:
The dart will go forward horizontally with velocity of 19 m/s. It will also fall downwards with initial velocity of zero and gravitational acceleration of 9.8 m/s².
Distance PQ covered by the dart can be calculated using the following formula.
s = ut + 1/2 at²
u is initial velocity , a is acceleration and t is time.
Putting the values
s = 0 + 1/2 x 9.8 x .19²
= .1769 m
= 17.69 cm.
the unit of energy is same as that of work i.e joule give reason
"Energy" is the ability to do work.
"Work" is the process of using energy.
Why Newton's law of gravitation also called universal law?
Answer:
Explanation:
Newton's law of gravitation states that every particle of matter attracts any other particle in the universe with a force directly proportional to the product of there masses and inversely proportional to the square of the distance between them.
This law is also called universal law because it is applicable to all masses at all distances irrespective of the medium.
a trampoline launches a 50kg person 2m into the air. if the springs push with 1960N of force, how much displacement was there in the trampoline
Answer: 0.5 m
Explanation:
Given
Mass of the person is [tex]m=50\ kg[/tex]
Trampoline launches the person into the air up to height of [tex]h=2\ m[/tex]
Force experience by springs is [tex]F=1960\ N[/tex]
Here, the work done on displacing the springs is equivalent to the Potential energy acquired by the person i.e.
[tex]\Rightarrow F\cdot x=mgh\quad [\text{x=displacement of the trampoline}]\\\\\text{Insert the values}\\\\\Rightarrow x=\dfrac{50\times 9.8\times 2}{1960}\\\\\Rightarrow x=\dfrac{980}{1960}\\\\\Rightarrow x=0.5\ m[/tex]
Answer:
0.5
Explanation:
A spherical cell has a radius of 2.3 μm , and the phospholipid bilayer that constitutes its membrane has a thickness of 3.75 nm . At its normal resting state, the outer membrane is at a voltage of 0V, and the inner membrane is at a voltage of -70mV. The dielectric constant of the cell membrane is roughly 9.0. What is the capacitance of the cell? (Note: the surface area of a sphere is 4πr2)
Answer:
Explanation:
The cell is acting like a shell .
Capacity of a shell is given by the following expression.
C= 4πk ε₀ x ab / (b-a )
k is dielectric constant , b and a are outer and inner radius ε₀ is a constant .
Here a = 2.3 x 10⁻⁶ m
b = 2.3 x 10⁻⁶ + .00375 x 10⁻⁶
= 2.30375 x 10⁻⁶ m
b -a = 3.75 x 10⁻⁹ m .
k = 9
4π ε₀ = 1 / 9 x 10⁹ = .111 x 10⁻⁹
C= .111 x 10⁻⁹ x 9 x 2.3 x 10⁻⁶ x 2.30375 x 10⁻⁶ / 3.75 x 10⁻⁹
= 1.41 x 10⁻¹² F
= 1.41 pF .
URGENT
A student runs at 4.5 m/s [27° S of W] for 3.0 minutes and then he turns and runs at 3.5 m/s [35° S of E] for 4.1 minutes.
a. What was his average speed?
b. What was his displacement?
PLEASE SHOW ALL WORK
Answer:
(a) 3.93 m/s
(b) 861.66 m
Explanation:
A = 4.5 m/s [27° S of W] for 3.0 minutes
B = 3.5 m/s [35° S of E] for 4.1 minutes
Distance A = 4.5 x 3 x 60 = 810 m
Distance B = 3.5 x 4.1 x 60 = 861 m
(a) The average speed is defined as the ratio of the total distance to the total time.
Total distance, d = 810 + 861 = 1671 m
total time, t = 3 + 4.1 = 7.1 minutes = 7.1 x 60 = 426 seconds
The average speed is
[tex]v=\frac{1671}{426}=3.93 m/s[/tex]
(b)
[tex]\overrightarrow{A} = 810(- cos 27 \widehat{i} - sin 27 \widehat{j})=- 721.7 \widehat{i} - 367.7 \widehat{j}\\\\\overrightarrow{B} = 861( cos 35 \widehat{i} - sin 35 \widehat{j})= 705.3 \widehat{i} - 493.8 \widehat{j}\\\\\overrightarrow{C} = (- 721.7 + 705.3) \widehat{i} - (367.7 + 493.8) \widehat{j} \\\\\overrightarrow{C}= - 16.4 \widehat{i} - 861.5 \widehat{j}[/tex]
The magnitude is
[tex]C =\sqrt{16.4^2+861.5^2} = 861.66 m[/tex]
a stone of mass 250kg and another stone of mass 400 kg are kept at a distance of 100m what amount of gravitational force develops between them?
Explanation:
Hey there!
Given;
Mass of one object (m1) = 250kg
Mass of another object (m2) = 400 kg
Distance (d) = 100 m
Gravitational constant (g) = 6.67*10^-11
Now;
[tex]f = \frac{g.m1.m2}{ {d}^{2} } [/tex]
Keep all values;
[tex]f = \frac{6.67 \times {10}^{ - 11} \times 250 \times 400}{ {(100)}^{2} } [/tex]
Simplify
[tex]f = \frac{6.67 \times {10}^{ - 11} {10}^{5} }{10000} [/tex]
[tex]f = \frac{6.67 \times {10}^{ - 6} }{10000} [/tex]
Therefore, gravitational force is 6.67*10^-10.
Hope it helps!
An acceleration of 2m/s^2 is produced on a body by applying an effort of 50N. calculate mass of the body
Answer:
25 kg
Explanation:
Given,
Acceleration ( a ) = 2 m/s^2
Force ( F ) = 50 N
To find : Mass ( m ) = ?
Formula : -
F = ma
m = F / a
= 50 / 2
m = 25 kg
So, the mass of the body is 25 kg.
Q011) The Doppler effect a. occurs when the frequency of sound waves received is lower if the wave source is moving toward you than if it's moving away. b. occurs when the pitch of a sound gets lower if the source is receding. c. is the basic explanation for the blue shift of light in our Universe. d. can be applied only to sound waves.
Answer:
Option (c) is correct.
Explanation:
The apparent change in the frequency of light due to the relative motion between the source and the observer is called Doppler's effect.
When the source is moving towards the observer which is at rest, the apparent frequency increases and if the observer is moving away the frequency of sound decreases.
It occurs for both light and sound.
So, to explain the blue shift of light in the universe is due to the Doppler's effect of light.
Please help
Will give the brainliest!
Answer:
both answer is option C
Explanation:
tag me brainliest
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the moon is (6.10x10^11). What is the gravitational force between the two objects?
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the moon is (6.10x10^11). What is the gravitational force between the two objects?
Answer:
Explanation:
This is a simple gravitational force problem using the equation:
[tex]F_g=\frac{Gm_1m_2}{r^2}[/tex] where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.
Filling in:
[tex]F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2}[/tex] I'm going to do the math on the top and then on the bottom and divide at the end.
[tex]F_g=\frac{2.4012*10^{40}}{3.721*10^{23}}[/tex] and now when I divide I will express my answer to the correct number of sig dig's:
[tex]Fg=[/tex] 6.45 × 10¹⁶ N
Un móvil recorre una trayectoria en línea recta de 6000 metros y demora 30 minutos. ¿Cuál es su rapidez expresada en Km/h?
Answer:
La rapidez del móvil es 12 kilómetros por hora.
Explanation:
Asumamos que el móvil experimenta un movimiento rectilíneo uniforme, cuya ecuación cinemática es la siguiente:
[tex]v = \frac{x}{t}[/tex] (1)
Where:
[tex]x[/tex] - Distancia recorrida, en kilómetros.
[tex]t[/tex] - Tiempo, en horas.
[tex]v[/tex] - Rapidez, en kilómetros por hora.
Si tenemos que [tex]x = 6000\,m[/tex] y [tex]t = 30\,min[/tex], entonces la rapidez del móvil es:
[tex]v = \frac{6000\,m\times \frac{1}{1000}\,\frac{km}{m}}{30\,min \times \frac{1}{60}\,\frac{h}{min} }[/tex]
[tex]v = 12\,\frac{km}{h}[/tex]
La rapidez del móvil es 12 kilómetros por hora.
A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predetermined value. Suppose that the material to be used in a fuse melts when the current density rises to 540 A/cm2. What diameter of cylindrical wire should be used to make a fuse that will limit the current to 0.57 A
Answer:
0.0366 m
Explanation:
We are given;
Current density; J = 540 A/cm² = 540 × 10⁴ m
Current; I = 0.57 A
Now, formula for current density is;
J = I/A
Where A is area = πr²
Thus;
J = I/(πr²)
r = √(I/(Jπ))
r = √(0.57/(540π))
r = 0.0183 m
Diameter = 2 × radius
Diameter = 2 × 0.0183
Diameter = 0.0366 m
A delivery boy on a bicycle drags a wagon full of newspapers by pedaling at 0.90 m/s for 45 minutes using a force of 40 N. How much work has the boy done?
Answer:
Explanation:
The equation for work is
W = FΔx
We are looking for work, so that means we have to be able to fill in the Force and the displacement. We have Force, but we don't have displacement. But the thing we need to do first is change the 45 minutes to seconds because the velocity is in m/s, not m/min.
45 minutes is 2700 seconds.
That means that the displacement is
Δx = (.90)(2700) so
Δx = 2430 m
Now we plug that in to find work, along with the given Force:
W = 40(2430) so
W = 97200 J (and that is not the correct number of sig fig's but I have a feeling you're not too into that in class, because if you were, the 40 N would be expressed as 40.0 or 4.0 × 10¹)
Juanita ran one mile around her school track in six minutes. What is
her average speed, and what is the magnitude of her average velocity?
10 mph, 0 mph
6 mph, 0 mph
6 mph, 6 mph
10 mph, 10 mph
Answer:
The correct option is a) 10 mph, 0 mph.
Explanation:
1. The average speed (S) is a magnitude given by:
[tex] S = \frac{D}{T} [/tex]
Where:
D: is the total distance = 1 mi
T: is the total time = 6 min
[tex] S = \frac{D}{T} = \frac{1 mi}{6 min}*\frac{60 min}{1 h} = 10 mph [/tex]
Hence, the average speed is 10 mph.
2. The average velocity is a vector:
[tex] V = \frac{\Delta d}{\Delta t} = \frac{d_{f} - d_{i}}{t_{f} - t_{i}} [/tex]
Where:
[tex]d_{f}[/tex]: is the final distance
[tex]d_{i}[/tex]: is the initial distance
[tex]t_{f}[/tex]: is the final time
[tex]t_{i}[/tex]: is the initial time
Since Juanita ran one mile around her school track, the final position is the same that the initial position, so the magnitude of the average velocity is zero.
Therefore, the correct option is a) 10 mph, 0 mph.
I hope it helps you!
(12 points) Analysis from the point where the block is released to the point where it reaches the maximum height i) Calculate the highest height reached by the block (or the largest distance travelled along the ramp.) ii) Calculate the work done by the gravitational force. iii) Calculate the work done by the normal force. iv) Calculate the work done by the friction force.
Answer:
i) a₁ = -g (sin θ + μ cos θ), x = v₀² / 2a₁
ii) W = mg L sin θ , iii) Wₙ = 0
iv) W = - μ m g L cos θ x
Explanation:
With a drawing this exercise would be clearer, I understand that you have a block on a ramp and it is subjected to some force that makes it rise, for example the tension created by a descending block.
The movement is that when the system is released, the tension forces are greater than the friction and the component of the weight and therefore the block rises up the ramp
At some point the tension must become zero, when the hanging block reaches the ground, as the block has a velocity it rises with a negative acceleration to a point and stops where the friction force and the weight component would be in equilibrium along the way. along the plane
i) Let's use Newton's second law
the reference system is with the x axis parallel to the ramp
Axis y
N - W cos θ = 0
X axis
T - W sin θ - fr = ma
the friction force is
fr = μ N
fr = μ mg cos θ
we substitute
T - m g sin sin θ - μ mg cos θ = m a
a = T / m - g (sin θ + μ cos θ)
With this acceleration we can find the height that the block reaches, this implies that at some point the tension becomes zero, possibly when a hanging block reaches the floor.
T = 0
a₁ = -g (sin θ + μ cos θ)
v² = v₀² - 2a1 x
v = 0 at the highest point
x = v₀² / 2a₁
ii) the work of the gravitational force is
W = F .d
W = mg sin θ L
iii) the work of the normal force
the force has 90º with respect to the displacement so cos 90 = 0
Wₙ = 0
iv) friction force work
friction force always opposes displacement
W = - fr d
W = - μ m g cos θ L
definition of matter . A object which cover the place and have mass is called matter
Answer:
you have written the definition so what are you asking
The designer tested how the number of cells affected the number of steps that could
be taken before the lights stopped working.
Figure 2 shows the results.
Figure 2
1600
1400
1200
1000
Number of
steps in
800
thousands
600
400
200
0
1
5
2 3 4
Number of cells
1
3
Determine how many more steps could be taken when the number of cells was
increased from 3 to 5
[2 marks
Number of steps
=
thousand
Answer:
1000.
Explanation:
Thousand more steps could be taken when the number of cells was increased from 3 to 5 because in the data, the number of cells i.e. 5 was achieved on 1000 steps. According to the data, the number of cells i.e. 1 was achieved on 800 steps, the number of cells i.e. 5 was achieved on 1000 steps, the number of cells i.e. 2 was achieved on 600 steps, the number of cells i.e. 3 was achieved on 400 steps and the number of cells i.e. 4 was achieved on 200 steps. So we can say that 1000 steps are needed for the number of cells increased from 3 to 5.
your initial speed is zero. if you increased your speed by 4 m/s after only 2 seconds and continue to accelerate at the same rate, what will your speed be after 10 seconds?
Answer:
20 m/ s
Explanation:
change in velocity = 4m/ s
[tex]acceleration = \frac{rate \: of \: change \: of \: velocity}{time} [/tex]
[tex]acc = \frac{4}{2} \\ = 2[/tex]
initial velocity = 0m/ s
let velocity after 10s be v[tex]2 = \frac{v - 0}{10} [/tex]
cross multiplying
[tex]20 = v[/tex]
so the velocity after 10s will be 20m/s
sam says that beta particles are not used to irradiate food because they make it radioactive. Jo says that the reason is that beta particles would not be able to penetrate all the way through think packages of food. Who is correct, Sam or Jo?
Answer:
Jo
Explanation:
irradiating food doesn't make it radioactive (that's contamination) and beta particles are stopped by aluminium foil or thin metals, so it may not pass through thick packaging (usually gamma is used to irradiate foods as it can pass through the packaging)
coulomb what is the meaning in physics??
Answer:
Coulomb, unit of electric charge in the metre- kilogram- second- ampere system, the basis of SI system of physics unit. The coulomb is defined as the quantity of electricity transported in one second by a current of one ampere.
In a science museum, a 130 kg brass pendulum bob swings at the end of a 14.4 m -long wire. The pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.7 m to the side and releasing it. Because of its compact shape and smooth surface, the pendulum's damping constant is only 0.010kg/s. You may want to review (Pages 405 - 407) . Part A At exactly 12:00 noon, how many oscillations will the pendulum have completed
Answer:
The time in which the pendulum does a complete revolution is called the period of the pendulum.
Remember that the period of a pendulum is written as:
T = 2*pi*√(L/g)
where:
L = length of the pendulum
pi = 3.14
g = 9.8 m/s^2
Here we know that L = 14.4m
Then the period of the pendulum will be:
T = 2*3.14*√(14.4m/9.8m/s^2) = 7.61s
So one complete oscillation takes 7.61 seconds.
We know that the pendulum starts moving at 8:00 am
We want to know 12:00 noon, which is four hours after the pendulum starts moving.
So, we want to know how many complete oscillations happen in a timelapse of 4 hours.
Each oscillation takes 7.61 seconds.
The total number of oscillations will be the quotient between the total time (4 hours) and the period.
First we need to write both of these in the same units, we know that 1 hour = 3600 seconds
then:
4 hours = 4*(3600 seconds) = 14,400 s
The total number of oscillations in that time frame is:
N = 14,400s/7.61s = 1,892.25
Rounding to the next whole number, we have:
N = 1,892
The pendulum does 1,892 oscillations between 8:00 am and 12:00 noon.
physics class 9 chapter 8 please tell please
Answer:
(a) The motion is uniform
(b) 11.11 m/s
Explanation:
(a)
From the table below, the motion of the bus is uniform.
(b)
Speed(s) = Δd/Δt
s = Δd/Δt............. Equation 1
From the table,
Given: Δd = 10 km = 10000 m, Δt = 15 minutes = (15×60) = 900 seconds
Substitute these values into equation 1
s = 10000/900
s = 11.11 m/s
An object weighs 2.2 pounds on Earth and has a mass of 1 kilogram. What are the weight and mass of the same object in space where there is no gravity acting on it?
Answer:
Heavier than 2.2 pounds
Explanation:
