Answer:
a) W_total = mg (2h + d) , b) E_total = - mg (h + d)
Explanation:
a) We must solve this problem in two parts, the first for the accelerated movement and the second for the movement with constant speed
Let's look for work for the part that is in free fall
y = y₀ + v₀ t - ½ g t²
when he jumps out of a plane his vertical speed is zero
y =y₀ - ½ g t²
dy = 0 - ½ g 2t dt
the work in this first part is
W₁ = ∫ F dy
W₁ = mg ∫ g t dt
W₁ = m g² t² / 2
the time it takes to travel the distance y₀-y = h is
y₀-y = ½ g t²
t =[tex]\sqrt{2h/g}[/tex]
we substitute
W₁ = m g² 2h / g
W₁ = m g 2h
now we look for the work for the part with constant speed
since the velocity is constant let's use the uniform motion ratio
W₂ = F d
W₂ = mg d
the total work is
W_total = W₁ + W₂
W_total = 2mgh + m gd
W_total = mg (2h + d)
b) The change in gravitational potential energy
U = mg Δy
in the part with accelerated movement
U₁ = mg h
in the part with uniform movement
U₂ = mg d
the total potential energy is
E_total = U₁ + U₂
E_total = - mg (h + d)
PLZ help asap :-/
............................
Explanation:
[16][tex]\underline{\boxed{\large{\bf{Option \; A!! }}}} [/tex]
Here,
[tex]\rm { R_1} [/tex] = 2Ω[tex]\rm { R_2} [/tex] = 2Ω[tex]\rm { R_3} [/tex] = 2Ω[tex]\rm { R_4} [/tex] = 2ΩWe have to find the equivalent resistance of the circuit.
Here, [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] are connected in series, so their combined resistance will be given by,
[tex]\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ [/tex]
[tex]\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ [/tex]
[tex]\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ [/tex]
Now, the combined resistance of [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] is connected in parallel combination with [tex]\rm { R_3} [/tex], so their combined resistance will be given by,
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ [/tex]
Reciprocating both sides,
[tex]\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ [/tex]
Now, the combined resistance of [tex]\rm { R_1} [/tex], [tex]\rm { R_2} [/tex] and [tex]\rm { R_3} [/tex] is connected in series combination with [tex]\rm { R_4} [/tex]. So, equivalent resistance will be given by,
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ [/tex]
Henceforth, Option A is correct.
_________________________________[17][tex]\underline{\boxed{\large{\bf{Option \; B!! }}}} [/tex]
Here, we have to find the amount of flow of current in the circuit. By using ohm's law,
[tex] \longrightarrow [/tex] V = IR
[tex] \longrightarrow [/tex] 3 = I × 3.33
[tex] \longrightarrow [/tex] 3 ÷ 3.33 = I
[tex] \longrightarrow [/tex] 0.90 Ampere = I
Henceforth, Option B is correct.
____________________________[tex] \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ [/tex]
A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 60.0-g suspended mass. The suspended mass is given an initial downward speed of 1.60 m/s .
How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)
Express your answer using two significant figures.
Answer:
0.5
Explanation:
because the block is attached to the pulley of the string
Energy from the sun comes to Earth as radiant energy. Which of these is an example of radiant energy being converted to heat energy?
A Turning windmills transform mechanical energy into electrical energy.
B Black shirts feel hotter than light-colored shirts on a sunny day.
C Solar cells convert sunlight into electrical energy.
D Green plants use sunlight in photosynthesis.
Answer:
B
Explanation:
The radiant energy form the sun is absorbed by the black shirt and is converted to heat energy.
Answer:
B Black shirts feel hotter than light-colored shirts on a sunny day.
Explanation:
The energy from the sun also called solar energy is an energy source which reaches the earth as a form of radiant energy, that is it is transmitted without the movement of mass. Solar cells absorbs radiant energy from the sun into electrical energy for powering electrical devices.
During photosynthesis, sunlight absorbed by the chlorophyll of green plants is converted into chemical energy.
In black body, radiant energy abosrde are stored and converted to heat energy, reason dark colored clothes feels hotter than light colored on sunny days.
Which is a mixture?
'a' sodium metal
'b' chlorine gas
'c' sodium metal and chlorine gas
'd' sodium chloride (salt) and water
Answer:
d. Sodium chloride (salt) + water
Explanation:
A mixture is made up of two or more substance combined together (combined chemically).NaCl (salt) can completely dissolve in water and sodium chlorine (aqueous) is a homogeneous mixture.sodium metal when extracted is a soft, silvery white solid.chlorine gas is a pure gas.sodium metal and chlorine gas are at pure state hence they are not mixture.learn more: https://brainly.com/question/2331419
Answer:D. Sodium chloride (salt) and water
Explanation:
I got it right on edge 2023
hope this is helpful!
How much work is required to stretch an ideal spring of spring constant (force constant) 40 N/m from x
Answer:
The work done will be "0.45 J".
Explanation:
Given:
K = 40 N/m
x₁ = 0.20 m
x₂ = 0.25 m
Now,
The required work done will be:
= [tex]\frac{1}{2}k[x_2^2-x_1][/tex]
By putting the values, we get
= [tex]\frac{40}{2}[(0.25)^2-(0.20)^2][/tex]
= [tex]20\times 0.0225[/tex]
= [tex]0.45 \ J[/tex]
A 2090-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that its vertical velocity as a function of time is given by v(t) =At+Bt^2 , where A and B are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of 1.50m/s 2 at the instant of ignition and, 1.00 s later, an upward velocity of 2.00 m/s. (a) Determine A and B , including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket’s weight. (d) What was the initial thrust due to the fuel?
Answer:
a) A = 1.50 m / s², B = 1.33 m/s³, b) a = 12.1667 m / s²,
c) I = M (1.5 t + 1.333 t²) , d) ΔI = M 2.833 N
Explanation:
In this exercise give the expression for the speed of the rocket
v (t) = A t + B t²
and the initial conditions
a = 1.50 m / s² for t = 0 s
v = 2.00 m / s for t = 1.00 s
a) it is asked to determine the constants.
Let's look for acceleration with its definition
a = [tex]\frac{dv}{dt}[/tex]
a = A + 2B t
we apply the first condition t = 0 s
a = A
A = 1.50 m / s²
we apply the second condition t = 1.00 s
v = 1.5 1 + B 1²
2 = 1.5 + B
B = 2 / 1.5
B = 1.33 m/s³
the equation remains
v = 1.50 t + 1.333 t²
b) the acceleration for t = 4.00 s
a = 1.50 + 1.333 2t
a = 1.50 + 2.666 4
a = 12.1667 m / s²
c) The thrust
I = ∫ F dt = p_f - p₀
Newton's second law
F = M a
F = M (1.5 + 2 1.333 t) dt
we replace and integrate
I = M ∫ (1.5 + 2.666 t) dt
I = 1.5 t + 2.666 t²/2
I = M (1.5 t + 1.333 t²) + cte
in general the initial rockets with velocity v = 0 for t = 0, where we can calculate the constant
cte = 0
I = M (1.5 t + 1.333 t²)
d) the initial push
For this we must assume some small time interval, for example between
t = 0 s and t = 1 s
ΔI = I_f - I₀
ΔI = M (1.5 1 + 1.333 1²)
ΔI = M 2.833 N
A particle of mass 1.2 mg is projected vertically upward from the ground with a velocity of 1.62 x 10 cm/h. Use the above information to answer the following four questions: 7. The kinetic energy of the particle at time t = 0 s is A. 1.215 x 10-3 J B. 2.430 J C. 1215 J D. 9.72 x 106 J E. OJ (2)
Answer:
K = 0 J
Explanation:
Given that,
The mass of the particle, m = 1.2 mg
The speed of the particle, [tex]v=1.62\times 10\ cm/h[/tex]
We need to find the kinetic energy of the particle at time t = 0 s.
At t = 0 s, the particle is at rest, v = 0
So,
[tex]K=\dfrac{1}{2}mv^2[/tex]
If v = 0,
[tex]K=0\ J[/tex]
So, the kinetic energy of the particle at time t = 0 s is 0 J.
a vechile having a mass of 500kg is moving with a speed of 10m/s.Sand is dropped into it at the rate of 10kg/min.What force is needed to keep the vechile moving with uniform speed
Answer:
1.67 N
Explanation:
Applying,
F = u(dm/dt)+m(du/dt)................ Equation 1
Where F = force, m = mass of the vehicle, u = speed.
Since u is constant,
Therefore, du/dt = 0
F = u(dm/dt)............... Equation 2
From the question,
Given: u = 10 m/s, dm/dt = 10 kg/min = (10/60) kg/s
Substitute these values into equation 2
F = 10(10/60)
F = 100/60
F = 1.67 N
An object accelerates from rest, and after traveling 145 m it has a speed of 420 m/s. What was the acceleration of the object?
I am not sure how to calculate acceleration without being given the time directly.
Explanation:
Here,we've been given that,
Initial velocity (u) = 0 m/s (as it starts from rest)Distance (s) = 145 mFinal velocity (v) = 420 m/sWe've to find the acceleration of the object. By using the third equation of motion,
→ v² - u² = 2as
→ (420)² - (0)² = 2 × a × 145
→ 176400 - 0 = 290a
→ 176400 = 290a
→ 176400 ÷ 290 = a
→ 608.275862 m/s² = a
If you know initial speed and final speed, you can find the average speed. Then, knowing distance, you can find the time.
KimYurii posted the first answer to this question.
That answer is well organized, well presented, elegant and correct, and it deserves to be awarded "Brainliest" and several merit badges.
My problem is that I can never remember all the different formulas. I guess I had to work with so many uvum in all the Physics, Geometry, and Calculus classes that I took, I filled up all the memory slots with formulas, and over the years they all eventually merged into a big glob of goo. Now, the only formulas I can remember are the ones I had to use as an Electrical Engineer.
When I see this kind of question, I can only remember one or two simple formulas, and I reason it out like this:
Starting speed . . . zero
Ending speed . . . 420 m/s
Formula: Average speed . . . (1/2)·(0 + 420) = 210 m/s
Distance covered . . . 145 m
Formula: Time taken = (distance) / (average speed) = (145/210) second
(Now you have the time.)
Formula: Distance = (1/2)·(acceleration)·(time²)
145 m = (1/2)·(acceleration)·(145/210 sec)²
Acceleration = 290 m / (145/210 s)²
Acceleration = 608.28 m/s²
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. How long will it take to get to the top of its trajectory? A. 3 seconds B. 4 seconds C. 2 seconds D. 6 seconds
Answer:
A (3 seconds)
Explanation:
Well here we have a type of motion called projectile motion and it is pretty similar to an upside down parabola. The top of the trajectory is the vertex of the parabola and is also when v=0.
Lets identify our givens.
Givens:
Horizontal speed= 30m/s
Vertical Speed= 30 m/s
Since the ball is in freefall after being launched ay=-g(take up to be positive) and ax=0
The ball is launched from the ground so y0=0
Final vertical velocity= 0
This problem is now relatively easy because we only need to find the vertical distance so we can ignore horizontal speed and use
vy=vy0+ayt
Plug in our givens
0=30-10t
solve for t
t=3 seconds
Many collisions, like the collision of a bat with a baseball, appear to be instantaneous. Most people also would not imagine the bat and ball as bending or being compressed during the collision. Consider the following possibilities: The collision is instantaneous. The collision takes a finite amount of time, during which the ball and bat retain their shapes and remain in contact. The collision takes a finite amount of time, during which the ball and bat are bending or being compressed. How can two of these be ruled out based on energy or momentum considerations?
The collision is instantaneous.
The collision takes a finite amount of time, during which the ball and bat retain their shapes and remain in contact.
The collision takes a finite amount of time, during which the ball and bat are bending or being compressed.
How can two of these be ruled out based on energy or momentum considerations?
Answer:
The collision takes a finite amount of time, during which the ball and bat are bending or being compressed
Explanation:
These two conditions can be ruled out on the fact that :The collision takes a finite amount of time, during which the ball and bat are bending or being compressed
The rule of energy is been broken here because during the collision of objects energy and momentum is conserved. i.e. the change in shape of the ball when hit by the bat should not be noticed because the compression and returning to normal shape happens instantaneously
A car is moving with a velocity of45m/s. Is brought to rest in 5s.the distance travelled by car before it comes to rest is
Answer:
The car travels the distance of 225m before coming to rest.
Explanation:
Here,
v = 45m/s
t = 5s
d = v × t
Therefore,
d = 45 × 5
= 225m
how did kepler discoveries contribute to astronomy
Answer:
They established the laws of planetary motion. They explained how the Sun rises and sets. They made astronomy accessible to people who spoke Italian.
Explanation:
General Circulation Models (GCM) :_________
a) use data collected exclusively from high-resolution satellites.
b) use spectral models derived from energy released from the earth and clouds.
c) can be run on powerful home computers, allowing citizen scientists to run models.
d) use complicated two-dimensional grid systems that change temporally.
Answer:
b)
Explanation:
GCMs (general circulation models) are useful instruments for gaining a quantitative knowledge of climate processes. Physical processes in the atmosphere, cryosphere, and land surface are represented by them. They are used for modeling the global climate system's reaction to rising greenhouse gas concentrations available at the moment by utilizing spectral models based on the energy emitted by the biosphere and clouds.
Many types of decorative lights are connected in parallel. If a set of lights is connected to a 110 V source and the filament of each bulb has a hot resistance of what is the currentthrough each bulb
Answer:
i₀ = V / R_i
Explanation:
For this exercise we use Ohm's law
V = i R
i = V / R
the equivalent resistance for
[tex]\frac{1}{R_{eq}}[/tex] = ∑ [tex]\frac{1}{R_i}[/tex]
if all the bulbs have the same resistance, there are N bulbs
[tex]\frac{1}{ R_{eq}} = \frac{N}{R_i}[/tex]
R_{eq} = R_i / N
we substitute
i = N V / Ri
where i is the total current that passes through the parallel, the current in a branch is
i₀ = i / N
i₀ = V / R_i
The current in a conductor is 2.5A .explain the meaning of this statement
Answer:
In 1 second the amount of charge flowing through the conductor is 2.5 Q.
Explanation:
In order to keep a leaking ship from sinking, it is necessary to pump 12.0 lb of water each second from below deck up a height of 2.00 m and over the side. What is the minimum horse-
power motor that can be used to save the ship?
Answer:
P = 0.14 hp
Explanation:
The power required by the ship is given as:
[tex]P = \frac{Work}{Time} = \frac{Potential\ Eenrgy}{t}\\\\P = \frac{mgh}{t}[/tex]
where,
P = Power = ?
m = mass to pump = (12 lb)(1 kg/2.20 lb) = 5.44 kg
g = acceleration due to gravity = 9.81 m/s²
h = height = 2 m
t = time = 1 s
Therefore,
[tex]P = \frac{(5.44\ kg)(9.81\ m/s^2)(2\ m)}{1\ s}\\\\P = 106.8\ W[/tex]
Converting to horsepower (hp):
[tex]P = (106.8\ W)(\frac{1\ hp}{746\ W})[/tex]
P = 0.14 hp
Topic: Chapter 19: Some wiggle room
A hummingbird flaps its wings up to 70 times per second, producing a 70 Hz
hum as it flies by. If the speed of sound is 340 m/s, how far does the sound
travel between wing flaps?
= 4.86 m
= 58.9 m
= 0.206 m
= 23,800 m
Answer:
4.86 m
Explanation:
Given that,
The frequency produced by a humming bird, f = 70 Hz
The speed of sound, v = 340 m/s
We need to find how far does the sound travel between wing flaps. Let the distance is equal to its wavelength. So,
[tex]v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{340}{70}\\\\\lambda=4.86\ m[/tex]
So, the sound travel 4.86 m between wings flaps.
del tema de fuerza centripeta
1.- Un chico va en bicicleta a 10m/s por una curva plana de 200m de radio.
a) ¿Cuál es la aceleración?
b) si el chico y la bicicleta tienen una masa total de 70kg, ¿Qué fuerza se necesita para producir esta aceleración?
Answer:
a. C = 0.5 m/s²
b. F = 35 Newton
Explanation:
Given the following data;
Radius, r = 200 m
Velocity, v = 10 m/s
Mass, m = 70 kg
a. To find the centripetal acceleration;
Mathematically, centripetal acceleration is given by the formula;
C = v²/r
Where:
C is the centripetal acceleration
v is the velocity
r is the radius
Substituting into the formula, we have;
C = 10²/200
C = 100/200
C = 0.5 m/s²
b. To find the force;
F = mv²/r
F = (70*10²)/200
F = (70 * 100)/200
F = 7000/200
F = 35 Newton
Why don’t you see tides ( like those of the ocean ) in your swimming pool ?
state the story of archimedes
Answer:
Archimedes was born about 287 BCE in Syracuse on the island of Sicily. He died in that same city when the Romans captured it following a siege that ended in either 212 or 211 BCE. One story told about Archimedes' death is that he was killed by a Roman soldier after he refused to leave his mathematical work.
Question 7 of 10
Which statement best describes diffraction?
A. Waves bend as they pass through an opening.
B. Waves and vibrations are oriented in a single direction.
.
C. Waves bounce off a surface.
D. Waves change direction as they enter a new material.
Answer:
A. Waves bend as they pass through an opening.
Explanation:
important word here is "opening"
diffraction example is a CD reflecting rainbow colors
A. Waves bend as they pass through an opening best describes diffraction.
Diffraction is the spreading out or bending of waves as they pass through an aperture or around an object. If we talk about light waves, diffraction of light occurs when a light wave passes by a corner or through a slit or opening. The slit or opening can be physically approximately the size of, or even smaller than that light's physical wavelength. An example of diffraction is the diffraction of sunlight by the clouds.
To know more, refer to,
brainly.com/question/10709914
The attached picture shows the diffraction of light through a single slit.
#SPJ2
If 5kg Stone and 1kg stone throw the from the building which will land more fa ster and why?
Answer:
Both stones will land at the same time because both stones will fall with the same acceleration through the same height.
Explanation:
We are given that
Mass of stone ,m1=5 Kg
Mass of stone, m2=1 kg
We have to find which stone more faster will land and why.
[tex]h=u+\frac{1}{2}gt^2[/tex]
Initial velocity of both stones=0
[tex]h=\frac{1}{2}gt^2[/tex]
[tex]t^2=\frac{h}{g}[/tex]
[tex]t=\sqrt{\frac{h}{g}}[/tex]
[tex]t_1=t_2=\sqrt{\frac{h}{g}}[/tex]
Because both stones are thrown from the same height.
Both stones will land at the same time because both stones will fall with the same acceleration through the same height and the acceleration does not depend of its mass.
A 15.0-m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60.08 angle with the horizontal. (a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 800-N firefighter has climbed 4.00 m along the ladder from the bottom. (b) If the ladder is just on the verge of slipping when the firefighter is 9.00 m from the bottom, what is the coefficient of static friction between ladder and ground
Answer:
a) fr = 266.92 N, fy = 1300 N, b) μ = 0.36
Explanation:
a) This is a balancing act.
Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive
-w x - W x₂ + R y = 0 (1)
usemso trigonometry to find distances
cos 60.08 = x / 7.5
x = 7.5 cos 60.08
x = 3.74 m
fireman
cos 60.08 = x₂ / 4
x2 = 4 cos 60
x2 = 2 m
wall support
sin 60.08 = y / 15
y = 15 are 60.08
y = 13 m
we substitute in equation 1
R y = w x + W x2
R = (w x + W x2) / y
R = (500 3.74 +800 2) / 13
R = 266.92 N
now let's write the expressions for the translational equilibrium
X axis
R -fr = 0
R = fr
fr = 266.92 N
Y Axis
Fy - w-W = 0
fy = 500 + 800
fy = 1300 N
b) ask the friction coefficient
the firefighter's distance is
cos 60.08 = x₃ / 9.00
x₃ = 9 cos 60
x₃ = 5.28 m
from equation 1
R = (w x + W x₃) / y
R = 500 3.74 + 800 5.28) / 13
R = 468.769 N
we saw that
fr = R = 468.769
The expression for the friction force is
fr = μ N
in this case the normal is the ratio to pesos
N = Fy
N = 1300 N
μ N = fr
μ = fr / N
μ = 468,769 / 1300
μ = 0.36
After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at 19.0 with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits
Answer:
[tex]$\frac{d}{\lambda} = 1.54$[/tex]
Explanation:
Given :
The first dark fringe is for m = 0
[tex]$\theta_1 = \pm 19^\circ$[/tex]
Now we know for a double slit experiments , the position of the dark fringes is give by :
[tex]$d \sin \theta=\left(m+\frac{1}{2}\right) \lambda$[/tex]
The ratio of distance between the two slits, d to the light's wavelength that illuminates the slits, λ :
[tex]$d \sin \theta=\left(\frac{1}{2}\right) \lambda$[/tex] (since, m = 0)
[tex]$d \sin \theta=\frac{\lambda}{2}$[/tex]
[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin \theta}$[/tex]
[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin 19^\circ}$[/tex]
[tex]$\frac{d}{\lambda} = 1.54$[/tex]
Therefore, the ratio is [tex]$\frac{1}{1.54}$[/tex] or 1 : 1.54
A dandelion seed floats to the ground in a mild wind with a resultant velocity of 26.0 cm/s. If the horizontal component velocity due to the wind is 10.0 cm/s, what is the vertical component velocity? Show all work.
Answer:
24 cm/s
Explanation:
Applying
Pythagoras theorem,
a² = b²+c²............. Equation 1
Where a = resultant, b = vertical component, c = horizontal component
From the question,
Given: a = 26 cm/s, c = 10 cm/s
Substitute these values into equation 1
26² = b²+10²
676 = b²+100
b² = 676-100
b² = 576
b = √576
b = 24 cm/s
Is the actual height the puck reached greater or less than your prediction? Offer a possible reason why this might be.
Answer:
Answer to the following question is as follows;
Explanation:
The puck's real altitude is lower than ones projection. That's because the mechanism may not be completely frictionless. Electricity is nevertheless wasted owing to particle interactions such as friction, which might explain why the present the results is lower than predicted.
Stationary waves are
A) transverse waves
B) longitudinal waves
C) mechanical waves
Answer:
stationary waves are transverse waves
Which parts of The Action Potential Are Represented On The ECG?
Answer:
The phases of the cardiac action potential correspond to the surface ECG (ECG) (Figure). The P wave reflects atrial depolarization (phase 0), the PR interval reflects the conduction velocity through the AV node, the QRS complex the ventricular depolarization and QT interval the duration potential ventricular action.
You are driving in such a way that the car is accelerating at a constant rate in the positive direction. When you pass the first sign, you are traveling at 4 m/s. When you pass the second sign 50 m down the road, you note that the seconds indicator of your clock reads 45 seconds. You also note that your velocity is now 9 m/s.
Required:
a. What is your acceleration?
b. What was the clock’s seconds indicator reading when you passed the first sign?
Answer:
Explanation:
a)
v² = u² + 2 a s
v = 9 m/s
u = 4 m/s
s = 50 m
9² = 4² + 2 x a x 50
a = 0.65 m /s²
Acceleration is 0.65 m /s²
b )
time elapsed before velocity changed from 4 m/s to 9 m/s with acceleration of .65 m /s ²
(v - u ) / t = a
(v - u ) / a = t
(9 - 4 ) / .65 = t
t = 7.7
time when passing the first sign will be 7.7 s earlier .
Reading of time indicator = 45 - 7.7
= 37.3 seconds.
Answer:
(a) 0.45 m/s^2
(b) 33.9 s
Explanation:
initial velocity, u = 4 m/s
final velocity, v = 9 m/s
distance, s = 50 m
(a) Let the acceleration is a.
Use third equation of motion
[tex]v^2 = u^2 + 2 as \\\\9^2 = 4^2 + 2\times a\times 50\\\\a = 0.45 m/s^2[/tex]
(b) Let the time is t.
Use first equation of motion
v = u + at
9 = 4 + 0.45 x t
t = 11.1 s
So, the initial time, t' = 45 - 11.1 = 33.9 s