Three different activation stroke types are employed during periodontal instrumentation: the assessment stroke (exploratory stroke), the calculus removal stroke (scaling stroke), and the root planing stroke.
What strokes are used during instrumentation?
It is advised to maintain cerebral perfusion pressure between 61 and 80 mmHg if the systolic blood pressure is greater than 180 mmHg and there is proof or reason to suspect that intracranial pressure is raised. Because a standard curet blade positioned for a vertical or oblique stroke is frequently too large to adapt to root curvatures or for deep insertion in the pocket, especially in narrow pockets or regions with tight tissue tone, horizontal strokes are essential for successfully removing calculus in deep pockets. On the mesial, facial, distal, and lingual surfaces of the front teeth as well as the mesial and distal surfaces of the posterior teeth, vertical strokes are utilized.To clear the periodontal pocket and remove dental calculus, there are three main categories of electronic devices. Head is essentially upright.Head in the least strenuous vertical and horizontal posture.In order to avoid neck and head pain, the eyes are pointed downward. The thumb is positioned on the opposite side of the instrument from the four fingers.Dental instruments are handled using the standard pen grip, which involves holding a pen between the tips of the thumb and forefinger and the lateral side of the distal (final) phalanx of the middle finger. No matter where we position our fulcrum, we should always apply lateral pressure, or pressure to the side of the instrument.With this technique, you may make sufficient strokes without slipping off the tooth and damaging delicate tissue. The blade's shape is the primary distinction between a scaler and a curette in terms of design.A curette's blade is semicircular in cross section, whereas a scaler's blade is triangular.To learn more about instrumentation stroke refer
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a car carrying a 80-kgkg test dummy crashes into a wall at 28 m/sm/s and is brought to rest in 0.10 ss. part a what is the magnitude of the average force exerted by the seat belt on the dummy? express your answer to two significant figures and include the appropriate units
The magnitude of the average force exerted by the seat belt on the dummy is 224N .
What is an average force ?The average force is the force produced by an object moving over a specific period of time at a given rate of speed, or velocity. This velocity is not instantaneous or precisely measured, as the word "average" indicates.
Briefing:mass of the dummy (m) = 80kg
velocity of the dummy (v) = 28 m/s
time (t) = 10 seconds
Average force exerted (F)
To calculate the average force;
According to the formula;
F = (m × v) ÷ t
Where;
F represents the force exerted
m represents the mass of the dummy
v represents the velocity
t represents the time
F = m * v/t
F = 80 *28/10
F = 224
F = 224 N
The force exerted by the seat belt on the dummy is 224N .
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To find the density of an object you would use a graduated cylinder to find what characteristic of the obiect? A)mass b)length c)volume d)weight
Answer:
C Volume
Explanation:
You would use a graduated cylinder with liquid in it to see how much is displaced when the object is placed in it to determine the volume of the object
In 2 - 3 sentences, explain the difference in Kinetic energy and Potential energy.
The main difference between Kinetic energy and Potential energy is that kinetic energy refers to movement while potential energy refers to storage.
What are Kinetic energy and Potential energy?Kinetic energy is a type of energy in motion or energy in movement such as a turbine (mechanical energy), while Potential energy refers to the energy that is stored to be used when required (e.g. chemical bonds of foods).
Therefore, with this data, we can see that Kinetic energy is used as movement, while Potential energy is stored to be used in the future.
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Physics Question A ball of Mass 4. 0kg Moving at 6m5' collided with a vertical wall, and retraces its path in the opposite direction at 3 ms: If the time of impact is 0. 20g, Calculate:
1. Magnitude of the change in the linear momentum of the bofy.
ii magnitude of the loss in the Kinetic energy. Iii magnitude of the force experienced by the body due to collision with the ball.
The ball experiences an equal and opposing force from the wall as a result of Newton's third law .Therefore, in a perfect world, the ball would return at the same speed it originally did, preserving momentum.
explain magnitude of the loss in the Kinetic energy?
An object's mass times its velocity change equals the change in momentum of that object.Δp=m⋅(Δv)=m⋅(vf−vi) Pi=m1v1 will be the system's initial momentum.Therefore, the difference between the initial and final kinetic energies is non-zero, demonstrating that kinetic energy is always lost in inelastic collisions. The magnitude of this loss is Loss=12m1v12[m2m1+m2]. The ball is severely deformed as it is struck due to the bat's tremendous force.The average force acting during the bat-ball collision is therefore about two tons, with a peak force of nearly four tons, during the 0.7 millisecond contact time. By multiplying the magnitudes of two forces acting in the same direction along a straight line, you can quickly calculate the combined force.To learn more about magnitude refer
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rocky the flying squirrel is carrying a nut of mass 0.5 kg while flying horizontally at a height of 15 m above the ground at a speed of 12 m/s. bullwinkle is eagerly awaiting the delivery of the nut on the ground. rocky releases the nut as he is directly above bullwinkle. how far from bullwinkle will the nut land if bullwinkle does not move
The nut distance from the Bullwinkle after uniform motion is 21 m.
We need to know about the uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule
vt = vo + a . t
vt² = vo² + 2a . s
s = vo . t + 1/2 . a . t²
where vt is final velocity, vo is initial velocity, a is acceleration, t is time and s is displacement.
From the question above, the parameters given are
m = 0.5 kg
s = 15 m
vx = 12 m/s
vo = 0 m/s
a = g = 9.8 m/s²
Find the time taken of the nut for landing
s = vo . t + 1/2 . a . t²
15 = 0 + 1/2 . 9.8 . t²
t² = 3.06
t = 1.75 s
Find the distance of nut in horizontal direction
vx = x / t
12 = x / 1.75
x = 12 . 1.75
x = 21 m
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two blocks are released from the top of a building. one falls straight down while the other slides down a smooth ramp. if all friction is ignored, which one is moving faster when it reaches the bottom?
The first and second blocks move at the same speed.
What is the law of conservation of mechanical energy ?
The law of conservation of mechanical energy states that the total mechanical energy of a system is constant if the only forces acting on the system are conservative.
The total mechanical energy of a system is the sum of kinetic and potential energy.
E = KE + PE
This is the total energy, kinetic energy and potential energy of the system.
The kinetic energy of the
system is:
KE = 1/2 x m x v^{2}
This is the mass and velocity of the object.
The gravitational potential energy is:
PE = mgh
Here is the acceleration of gravity and altitude.
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according to modern ideas and observations, what can be said about the location of the center of our expanding universe? according to modern ideas and observations, what can be said about the location of the center of our expanding universe? the milky way galaxy is at the center. the universe does not have a center. earth is at the center. the sun is at the center.
The correct answer is option C:
According to modern ideas and observations, the location of the center of our expanding universe explains that the universe does not have a center.
Our universe is composed of three kinds of substances that are dark matter, normal matter, and dark energy.
Everything in the universe is made of normal matter i.e. planets, stars, human beings, and all other visible objects.
Astronomers have theorized that the universe is expanding at a faster rate. Between any two given points, the distance of the universe is increasing and so we can predict that the universe will be 2x in size after ten billion years.
In general, the universe has no central point and it is expanding continuously because of a dark, mysterious force that is pulling the universe apart.
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Calculate the area of the plates of a 1 pF parallel plate capacitor in a vacuum if the separation of the plates is 0.1 mm. [ε0=8.85×10-12C2N-1m-2 ]Select one:6.2 m214.4 m211.3 m218.4 m2
Apply:
C = A e0 / d
Where:
C = capacitance = 1pf = 1 x10^-12 F
A = area
ε0 = 8.85×10^-12
1 pf = 1 x 10^-12 F
A = Cd / ε0 = 1 x 10^-12 ( 1 x 10^-4 ) / 8.85×10^-12 = 1.13 x10 ^-5 m^2
What stage ends when the piglets
are weaned at about 2-4 weeks?
A. Farrowing
B. Loading
C. Breeding
D. Finishing
the answer to this is finishing
If a magnifying glass uses a convex lens of focus length 6.25 cm when it is held 5.20 cm in front of an object what is the image distanceMind your minus signs Unit cm
Answer:
30.95 cm
Explanation:
The focal length f, the object distance do, and the image distance di are related with the following equation:
[tex]\frac{1}{f}=\frac{1}{d_0}+\frac{1}{d_i}[/tex]So, replacing the values, we get:
[tex]\frac{1}{6.25}=\frac{1}{5.20}+\frac{1}{d_i}[/tex]Then, solving for di, we get:
[tex]\begin{gathered} \frac{1}{6.25}-\frac{1}{5.2}=\frac{1}{d_i} \\ -\frac{21}{650}=\frac{1}{d_i}_{} \\ -\frac{650}{21}=d_i \\ -30.95\operatorname{cm}=d_i \end{gathered}[/tex]Therefore, the image distance is 30.95 cm
An 85.0 kg man stands on a spring scale in an elevator when it begins to accelerate upwards. If the scale at that moment indicates that his "weight" is 1020 N, what is the acceleration of the elevator?
2.20 m/s^2 is the acceleration of the elevator.
Option c is correct answered .
What does the scale read when the elevator accelerates upward?If you stand on a scale in an elevator accelerating upward, you feel heavier because the elevator's floor presses harder on your feet, and the scale will show a higher reading than when the elevator is at rest. On the other hand, when the elevator accelerates downward, you feel lighter.
How do you find the acceleration of an elevator with weight?support force F = mass x acceleration + weight
For a mass m= kg, the elevator must support its weight = mg = Newtons to hold it up at rest. If the acceleration is a= m/s² then a net force= Newtons is required to accelerate the mass.
Fnet = m a
Fn + (-Fg) = m a
a = 1020 - (85)(9.8) /85
a = 2.20 m/s^2
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an ice hockey forward with mass 70.0 kg is skating due north with a speed of 5.5 m/s. as the forward approaches the net for a slap shot, a defensive player (mass 110 kg) skates toward him in order to apply a body-check. the defensive player is traveling south at 4.0 m/s just before they collide. if the two players become intertwined and move together after they collide, in what direction and at what speed do they move after the collision? friction between the two players and the ice can be neglected.
The speed of the two players after the collision is 0.31 m/s in a direction of due south.
What is the final speed of the two players?The final speed of the two players after the collision is calculated as follows.
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
m₁ is the mass of the first playerm₂ is the mass of the second playeru₁ is the initial velocity of the first playeru₂ is the initial velocity of the second playerv is the common speed of both players after the collision.(70 x 5.5) + (-110 x 4) = v(70 + 110)
385 - 440 = 180v
-55 = 180v
v = -55/180
v = -0.31 m/s
Thus, the direction of the speed of the two players after the collision is due south.
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while riding a roller coaster, a girl drops an object. the roller coaster was rising vertically at a velocity of 11.0m/s and was 5.00m above the ground when the object was dropped. how long does it take to reach the ground
Answer:
Approximately [tex]0.388\; {\rm s}[/tex], assuming that air resistance is negligible and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
Initial vertical velocity of the object: [tex]u = 11.0\; {\rm m\cdot s^{-1}}[/tex], upwards (same as that of the rollercoaster.)
Initial height of the object: [tex]h_{0} = 5.00\; {\rm m}[/tex].
If air resistance is negligible, this object will accelerate downwards at a constant [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex]. Note that [tex]a[/tex] is negative since the object is accelerating downwards.
The SUVAT equation [tex]h = (1/2)\, a\, t^{2} + u\, t + h_{0}[/tex] gives the height [tex]h[/tex] of this object at time [tex]t[/tex]. Note that while the initial height is [tex]5.00\; {\rm m}[/tex], [tex]h = 0[/tex] when the object reaches the ground.
Since acceleration [tex]a[/tex], initial velocity [tex]u[/tex], and initial height [tex]h_{0}[/tex] are all given, setting [tex]h\![/tex] to [tex]0[/tex] and solving for [tex]t[/tex] will give the time it takes for this object to reach the ground:
[tex](1/2)\, a\, t^{2} + u\, t + h_{0} = 0[/tex].
[tex](1/2)\, (-9.81) \, t^{2} + 11.0\, t + 5.00 = 0[/tex].
[tex](-4.905)\, t^{2} + 11.0\, t + 5.00 = 0[/tex].
Solve this equation for [tex]t\![/tex] using the quadratic formula. Note that [tex]t > 0[/tex] since [tex]t[/tex] denotes the amount of time required for the object to reach the ground.
[tex]\begin{aligned} t &= \frac{-11.0 + \sqrt{11.0^{2} - 4 \times (-4.905) \times 5.00}}{2\times (-4.905)} \\ &\approx 0.388\; {\rm s}\end{aligned}[/tex].
The other root of this quadratic equation is negative and isn't a valid solution to the question.
In other words, it will take approximately [tex]0.388\; {\rm s}[/tex] for this object to reach the ground.
a very thin horizontal, 2.00-m long, 5.00-kg uniform beam that lies along the east-west direction is acted on by two forces. at the east end of the beam, a 200-n force pushes downward. at the west end of the beam, a 200-n force pushes upward. what is the angular acceleration of the beam?
The angular acceleration of the beam of length 2 meters and mass 5 kg is 240 radian/s².
The length of the rod is 2 meters, the mass of the rod is 5kg, a force of 200N is applied downward in the east direction while a force of 200N applied upward in west direction.
We know,
That the torque on a body is,
T = Fr
T is torque,
F is the force,
r is the perpendicular distance from the axis of rotation.
So, assuming the centre of the beam as the axis of rotation, the torque on the body is,
T = 200(1)+200(1)
T = 400N-m
We also know,
That torque on the body is,
T = IA
Where I is moment of inertia of the beam,
A is the angular acceleration of the body,
We know, moment of inertia of the beam assuming the centre of the rod as axis of rotation is,
I = MR²/12
M is the mass of the body,
R is the length of the rod,
So,
I = 5(2)²/12
I = 20/12 kg-m²
So, now we can write,
T = 20/12A
As both are value of torque, so,
20/12A = 400
A = 240 radian/s²
The angular acceleration is 240 radian/s².
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suppose you were hired to build a dam. what features would you look for in a site? be sure to consider the impact on living things as well as the physical characteristics of the site.
I need to get it done right now
Answer: The wall and base
Explanation:
what is an elliptical galaxy? what is an elliptical galaxy? any galaxy with an elliptical halo when observed at radio wavelengths. a spiral galaxy seen from an angle, giving it an elliptical profile. a galaxy with an elliptical outline and a smooth distribution of brightness (no spiral arms). a spiral galaxy with an elliptically shaped nuclear bulge and the spiral arms starting from the ends of the ellipse.
An elliptical galaxy is a galaxy with an elliptical outline and a smooth distribution of brightness (no spiral arms).
In the field of astronomical physics, we can describe elliptical galaxies as stretched galaxies that have an elliptical outline. This type of galaxy is usually made through old stars and has an even, smooth distribution of light.
Although the elliptical galaxies are not a major part of the universe they make up 10–15% in the Virgo Supercluster.
As the elliptical galaxy is made from broken stars, their size may vary from being a dwarf elliptical to a supergiant.
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Find the coefficient of static friction between the ramp and object with the following
Horizontal length of ramp: 75cm
Vertical length of ramp: 19cm
mass of object: 128g
The coefficient of static friction between the ramp of 75 cm horizontal length and a vertical length of 19 cm and object with a mass of 128 g is 0.25
Since the ramp forms a right angled triangle,
tan θ = Opposite side / Adjacent side
Opposite side = Vertical length of ramp = 19 cm
Adjacent side = Horizontal length of ramp = 75 cm
tan θ = 19 / 75
θ = [tex]tan^{-1}[/tex] ( 0.25 )
θ = 14°
m = 128 g = 0.128 kg
g = 9.8 m / s²
W = m g
W = 0.128 * 9.8
W = 1.25 N
Since the object is in static condition,
∑ [tex]F_{y}[/tex] = 0
N - [tex]W_{y}[/tex] = 0
N - W cos θ = 0
N = 1.25 * cos 14°
N = 1.25 * 0.97
N = 1.21 N
F[tex]_{s}[/tex] = μ[tex]_{s}[/tex] N
F[tex]_{s}[/tex] = Static frictional force
μ[tex]_{s}[/tex] = Co-efficient of static friction
N = Normal force
Since the object is in static condition,
∑ [tex]F_{x}[/tex] = 0
F[tex]_{s}[/tex] - [tex]W_{x}[/tex] = 0
F[tex]_{s}[/tex] = W sin θ
F[tex]_{s}[/tex] = 1.25 * sin 14°
F[tex]_{s}[/tex] = 1.25 * 0.24
F[tex]_{s}[/tex] = 0.3 N
μ[tex]_{s}[/tex] = F[tex]_{s}[/tex] / N
μ[tex]_{s}[/tex] = 0.3 / 1.21
μ[tex]_{s}[/tex] = 0.25
Therefore, the coefficient of static friction between the ramp and object is 0.25
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When a red light with a wavelength of 670 nm shines on a piece of metal, an electron is ejected. What is the energy of the photon absorbed by the electron to overcome the attractive forces?.
The energy of the photon of light as we can see here is 2.9 * 10-19 J.
What is the energy of the light?We know that light has to do with any of the components of the electromagnetic spectrum. These are the kinds of radiation that are able to pass through vacuum because they do not need a medium for the propagation of the wave. We are now asked to obtain the energy of the photon based in the wavelength that we have been supplied in the question here.
We have that the wavelength of the photon from what we can see in the question have been given as 670 nm. Then;
E = hc/λ
E = energy of the light
h = Plank's constant
c = speed of light
λ = wavelength of the light
It then follows that;
E = 6.6 * 10^-34 * 3 * 10^8/670 * 10^-9
E = 2.9 * 10-19 J
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if this wire is used as a heating element, how much time is required for this wire to raise the temperature of 25.0 g of water from 27.0 o c to 60.0 o c? the wire is completely submerged in the water and all the energy dissipated in the wire is absorbed by the water.
The time required to gain the thermal energy is 3465/P second.
We need to know about thermal energy to solve this problem. The total heat received by the system will equal to total heat released by objects. It should follow
Q released = Q received
The heat can be defined by
Q = m . c . ΔT
where Q is thermal energy, m is mass, c is the specific heat constant and ΔT is the change in temperature.
The given parameters are
m = 25 g = 0.025 kg
c = 4200 J/kgK
ΔT = 60 ⁰C - 27 ⁰C = 33 ⁰C
The thermal energy released by the heating element per second is equal to the power. Let's assume that the heating element has the power of P
Find the thermal energy received by the water
Q = m . c . ΔT
Q = 0.025 . 4200 . 33
Q = 3465 joule
Find the time required to raise the temperature
P = Q / t
P = 3465 / t
t = 3465 / P second
where P is power in the watt unit.
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HELP!
Noble gases are the ____________________.
A. most reactive of all elements
B. least consistent of all elements
C. most consistent of all elements
D. least reactive of all elements
Answer: Most consistent of all elements (if consistency here denotes stability)
if not, then the answer will be (D)
Explanation:
Nobel gasses have full outermost electron shells which means that they are very stable and do not need to lose, gain, or share electrons to gain stability. So they do not react with other elements to form ionic or covalent bonds
A car is traveling on a highway. the distance (in miles) from its destination and the time (in hours) is given by the equation d = 420 minus 65 t. what is the slope of the line represented by the equation? what is the practical meaning of the slope? a. m = 420; the total distance is 420 miles. b. m = 65; the car is traveling 65 miles per hour. c. m = negative 65; the car is traveling 65 miles per hour. d. m = negative 420; the total distance is 355 miles.
True or false. Objects tend to stay moving because of a force called inertia
Answer:
true
Explanation:
suppose a spring with spring constant 6 n/m is horizontal and has one end attached to a wall and the other end attached to a mass. you want to use the spring to weigh items. you put the spring into motion and find the frequency to be 1.4 hz (cycles per second). what is the mass? assume there is no friction.
When a mass attached to a spring, it will make the spring to produce a harmonic motion with frequency 1.4 Hz. Then the mass is 77.48 gram.
The frequency of a harmonic motion caused by a displaced spring is:
f = 1/2π . √(k/m)
Where:
k = spring constant
m = mass
Parameters given from the problem:
f = 1.4 Hz
k = 6 N/m
Plug those parameters into the formula:
1.4 = 1/2π . √(6/m)
√(6/m) = 8.8
m = 0.07748 kg = 77.48 gram
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27. Beyond "c", the speed of the rocket is;
(A) constant.
(B) continuously increasing.
(C) continuously decreasing.
(D) increasing for a while and constant thereafter.
(E) constant for a while and decreasing thereafter.
Beyond "c", the speed of the rocket is; (D) increasing for a while and constant thereafter.
How quickly does a rocket move?The space shuttle must accelerate from zero to 8,000 metres per second (nearly 18,000 miles per hour) in eight and a half minutes to reach the minimum height necessary to circle the Earth.
The orbital velocity is 7.9 kilometres per second, which translates to more than 20 times the speed of sound. A rocket attaining orbital velocity (1st cosmic velocity) will enter orbit around the Earth (C), whereas a faster rocket would follow an elliptical trajectory (D).
On Earth, air tends to prevent exhaust gases from exiting the engine. This lessens the thrust. However, because there is no atmosphere in space, the exhaust gases may depart considerably more easily and quickly.
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what is the magnitude of the acceleration aaa of the chair? what is the magnitude of the normal force fnfnf n acting on the chair? express your answers, separated by a comma, in meters per second squared and newtons to three significant figures.
535.1 Newton Force is the magnitude of the acceleration of the chair.
Fnet=Fp×cos35-F
=164×cos35-91
=43.34 N
a=Fnet/m
=43.34/45
=0.963 m/s^2
FN = mg + Fp×sin35
=45×9.8+164×sin35
=535.1 N
Newton is the abbreviation for the SI unit of absolute force known as newton (N). It is described as the amount of force required to accelerate a kilogram of mass by one meter per second. In the foot pound second system, one newton is equivalent to around 0.2248 pounds of force or 100,000 dyne in the centimeter gram second (CGS) system. The newton was given its name in of Sir Isaac Newton, whose second rule of motion outlines the modifications that a force may cause in a body motion.
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2200 J of work was done in 10 s. What is the amount of power?
We will have that the power will be:
[tex]\begin{gathered} P=\frac{W}{\Delta t}\Rightarrow P=\frac{2200J}{10s} \\ \\ \Rightarrow P=220J/s\Rightarrow P=220W \end{gathered}[/tex]So, the power was 220 Watts.
I need help with question 7, I just need the answer you don’t have to explain.
Explanation
Step 1
free body diagram
so, for m1
m1= 10 kg
so,
[tex]\begin{gathered} \sum ^{forces}_{\text{ y}}=T_1-mg=ma \\ T_1=m_1(a+g)\Rightarrow equation(1) \\ T_1=10(a+g) \end{gathered}[/tex]and for m2
[tex]\begin{gathered} \sum ^{forces}_{\text{ x}}=m_2g\sin (37)-T_1=m_2a \\ \text{solve for T}_1 \\ T_1=m_2g\sin (37)-m_2a \\ \text{replace} \\ T_1=(3.6\text{ kg)(9.8 }\frac{m}{s^2})\sin 37-3.6a \\ T_1=21.23\text{ -}3.6a\Rightarrow eq(2) \end{gathered}[/tex]Step 2
now, replace in equaiotn (1) and solve for a
[tex]\begin{gathered} T_1=m_1(a+g)\Rightarrow equation(1) \\ 21.23\text{ -}3.6a=10(a+g) \\ 21.23\text{ -}3.6a=10a+10g \\ -7.2a=10a-98.1 \\ -17.2a=-98.1 \\ a=-\frac{98.1}{-17.2} \\ a=5.703\text{ }\frac{m}{s^2} \end{gathered}[/tex]finally, replace in equation (2) to find Tension
[tex]\begin{gathered} T_1=21.23\text{ -}3.6a \\ T_1=21.23\text{ -}3.6(5.703) \\ T_2=3.891\text{N} \end{gathered}[/tex]I hope this helps you
a rocket started at rest.after initiating the launch,it had an acceleration of 90m/s2.how long did it take to reach a speed of 2430 m/s2
The time taken for the rocket to reach a speed of 2430m/s from rest is 27.0 seconds.
What is the time taken for the rocket to reach a speed of 2430m/s?Motion is simply the change in position of an object over time.
From the First Equation of Motion;
v = u + at
Where v is final velocity, u is initial velocity, a is acceleration and t is time elapsed.
Given the data in the question;
Since it starts from rest, Initial velocity u = 0m/sFinal velocity v = 2430m/sAcceleration a = 90m/s²Elapsed time t = ?To determine the time, substitute the given values into the formula above and solve for t.
v = u + at
2430m/s = 0 + ( 90m/s² × t )
2430m/s = 90m/s² × t
t = 2430m/s / 90m/s²
t = 27.0s
Therefore, the elapsed time is 27.0 seconds.
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A rocket launched with an acceleration of 90 m/s², and a final speed of 2430 m/s. So the time required to reach the final speed is 27 seconds.
What is acceleration?The ratio of change in velocity to time is referred to as acceleration. It has a SI unit m/s² with dimension formula MLT⁻².
According to the question, the given values are,
Final velocity, v = 2430 m/s
Initial velocity, u = 0 m/s
The acceleration is given as, a = 90 m/s².
Use the equation of motion to find the time required to reach the final speed,
v = u + at
2430 = 0 + ( 90 × t )
2430 = 90 × t
t = 2430 / 90
t = 27.0 seconds.
Hence, the time taken to reach the final speed is 27 seconds.
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What is the force of gravity pulling on a 12kg block accelerating at 10m/s²?
120N is the force of gravity pulling on a 12kg block accelerating at 10m/s²
F = m × a
F = 10 ×12
F= 120N
How is the potential energy of gravity stored?The mass (m) and height (h) of the item determine how much gravitational potential energy is there. Based on an object's high position relative to a lower location, gravitational energy is potential energy that is stored in the object.
This energy can be utilised later to move an item since it can be stored and utilised at a later time. Three variables—mass, gravity, and height—determine the gravitational potential energy. Energy is directly inversely proportional to all three variables.
Because the highest amount of energy that may be converted into other forms is called potential energy. You need more energy to move the thing at a greater height.
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A 0. 20-kg baseball is struck with a force of 100 n from a 0. 94-kg baseball bat. What will be the acceleration of the ball and the bat, in that order, while the bat and ball are in contact?.
The acceleration of the baseball is 500 m/s².
A force is an influence that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull.
The mass of the baseball, m = 0.2 kg
The force acting on the baseball is 100 N.
By Newton's second law of motion,
F = ma
Substituting the values in the above equation,
100 N = 0.2 kg × a
a = 100 / 0.2
a = 500 m/s²
The total mass when the baseball and baseball bat are in contact is:
m = 0.2 + 0.94 = 1.14 kg
Then the acceleration will be:
F = ma
100 = 1.14 × a
a = 87.7 m/s²
The acceleration of the ball after the contact is 500 m/s².
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