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.
In 2009 Usain Bolt set the world record time by running 100 meters in 9.58 s. Assume that during this race he ran in a straight line with constant acceleration a. What would be the required constant acceleration a
In 2009 Usain Bolt set the world record time by running 100 meters in 9.58 seconds, assuming that he ran this race with constant acceleration, then the required constant acceleration would have been
What are the three equations of motion?There are three equations of motion given by Newton
v = u + at
S = ut + 1/2×a×t²
v² - u² = 2×a×s
By using the second equation of motion given by Newton,
S = ut + 1/2at²
100= 0 + 0.5*a*9.58²
a = 2.17 meters / second²
Thus,the required constant acceleration of Usain Bolt would have been 2.17 meters / second².
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In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its unstretched length
Answer: hello below is the missing part of your question
A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5029 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.4 m/s.
answer
x = 0.0962 m
Explanation:
First step :
Determine the length of the rough patch/spot
F = Uₓ (mg)
and w = F.d = Uₓ (mg) * d
hence;
d( length of rough patch) = w / Uₓ (mg) = 46.55 / (0.49 * 10 * 9.8) = 0.9694 m
next :
work done on unstretched spring length
Given that block travels halfway i.e. d = 0.9694 / 2 = 0.4847 m
w' = Uₓ (mg) * d
= 0.49 * 10 * 9.81 * 0.4847 = 23.27 J
also given that the Elastic energy of spring = work done ( w')
1/2 * kx^2 = 23.27 J
x = [tex]\sqrt{\frac{2*23.27}{5029} }[/tex] = 0.0962 m
Which best describes the relationship between heat,intemal energy, and thermal energy?
Internal energy is heat that flows and heat is the part of thermal energy that can be transferred
Internal energy is thermal energy that flows, and thermal energy is the part of heat that can be transferred,
Thermal energy is heat that flows, and heat is the part of intemal energy that can be transferred
Heat is thermal energy that flows, and hennal energy is the part of internal energy that can be transferred.
Answer:
It is all a thermodynamic system that is highly related to each other.
Explanation:
Because they are in the physics of thermodynamics it is not wrong to say they follow the same thermodynamic rules and has highly the same properties of energy.
There are two beakers of water on the table. We can compare the average kinetic energy of the water molecules in the two beakers by measuring their
A temperatures.
B volumes.
C densities.
D masses.
Answer: masses
Explanation:
Trust me
When the insulation resistance between a motor winding and the motor frame is tested, the value obtained is 1.0 megohm (106 Ω). How much current passes through the insulation of the motor if the test voltage is 1000 V?
Answer:
0.001 A
Explanation:
Applying,
V = IR.............. Equation 1
Where V = Voltage of the motor, I = current, R = resistance
make I the subject of the equation
I = V/R.............. Equation 2
From the question,
Given: V = 1000 V, R = 1 MΩ = 10⁶ Ω
Substitute these values into equation 2
I = 1000/10⁶
I = 10⁻³ A
I = 0.001 A
High-speed stroboscopic photographs show that the head of a -g golf club is traveling at m/s just before it strikes a -g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at m/s. Find the speed of the golf ball just after impact.
The question is incomplete. The complete question is :
High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 60 m/s just before it strikes a 50 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball just after impact.
Solution :
We know that momentum = mass x velocity
The momentum of the golf club before impact = 0.200 x 60
= 12 kg m/s
The momentum of the ball before impact is zero. So the total momentum before he impact is 12 kg m/s. Therefore, due to the conservation of momentum of the two bodies after the impact is 12 kg m/s.
Now the momentum of the club after the impact is = 0.2 x 40
= 8 kg m/s
Therefore the momentum of the ball is = 12 - 8
= 4 kg m/s
We know momentum of the ball, p = mass x velocity
4 = 0.050 x velocity
∴ Velocity = [tex]$\frac{4}{0.050}$[/tex]
= 80 m/s
Hence the speed of the golf ball after the impact is 80 m/s.
When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not noticeable?
Answer:
Yes, when an apple falls towards the earth, the apple gets accelerated and comes down due to the gravitational force of attraction used by the earth. The apple also exerts an equal and opposite force on the earth but the earth does not move because the mass of the apple is very small, due to which the gravitational force produces a large acceleration in it (a = F/m) but the mass of the earth is very large, the same gravitational force produces very small acceleration in the earth and we don't see the earth rising towards the apple.
12) If, after viewing a specimen at low power, you switch to high-dry power and, after using fine focus, cannot find the specimen, what things could you do to help yourself (before calling me over to assist you?)
Answer:
See the answer below
Explanation:
After seeing an object on a slide at the low-power objective of the microscope and it disappears on changing to high power, the following can be done to resolve the problem
1. Drop a few drops of immersion oil on the slide and view again under high the power objective.
2. If the object is still not visible after the action above, return the microscope to the low-power objective and make sure the object is refocused and centered. Then carefully change back to the high power objective and use the fine adjustment to bring it into focus.
IS ANYONE THERE..??!
Answer:
hmmmmmmmmmmmmmmmmmmmmmmmm y
A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg. How far will block m drop in the first seconds after the system is released?
How long will block M move during above time?
At the time, calculate the velocity of block M
Find out the deceleration of the block M, if the connected string is
removal by cutting after the first second. Then, calculate the time
taken to contact block M and pulley.
Answer:
a) y = 0.98 t², t=1s y= 0.98 m,
b) he two blocks must move the same distance
c) v = 1.96 m / s, d) a = -1.96 m / s², e) x = 0.98 m
Explanation:
For this exercise we can use Newton's second law
Big Block
Y axis
N-W = 0
N = M g
X axis
T- fr = Ma
the friction force has the expression
fr = μ N
fr = μ Mg
small block
w- T = m a
we write the system of equations
T - fr = M a
mg - T = m a
we add and resolved
mg- μ Mg = (M + m) a
a = [tex]g \ \frac{m - \mu M}{m+M}[/tex]
a = [tex]9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}[/tex]
a = 9.8 (6/30)
a = 1.96 m / s²
a) now we can use the kinematic relations
y = v₀ t + ½ a t²
the blocks come out of rest so their initial velocity is zero
y = ½ a t²
y = ½ 1.96 t²
y = 0.98 t²
for t = 1s y = 0.98 m
t = 2s y = 1.96 m
b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.
As the curda is in tension the two blocks must move the same distance
c) the velocity of the block M
v = vo + a t
v = 0 + 1.96 t
for t = 1 s v = 1.96 m / s
t = 2 s v = 3.92 m / s
d) the deceleration if the chain is cut
when removing the chain the tension becomes zero
-fr = M a
- μ M g = M a
a = - μ g
a = - 0.2 9.8
a = -1.96 m / s²
e) the distance to stop the block is
v² = vo² - 2 a x
0 = vo² - 2a x
x = vo² / 2a
x = 1.96² / 2 1.96
x = 0.98 m
the time to travel this distance is
v = vo - a t
t = vo / a
t = 1.96 /1.96
t = 1 s
A 12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall (above the fire truck) if the ladder makes an angle of with the horizontal
Complete Question
A 12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall (above the fire truck) if the ladder makes an angle of
40° 16' with the horizontal.
Answer:
[tex]d=8.01m[/tex]
Explanation:
From the question we are told that:
Length of ladder [tex]l=12.5m[/tex]
Angle [tex]\theta=40° 16'=20.26 \textdegree[/tex]
Generally the Trigonometric equation for distance d it goes up the wall is mathematically given by
[tex]d=l sin \theta[/tex]
[tex]d=12.5 sin 40.26[/tex]
[tex]d=8.01m[/tex]
At which point is the kinetic energy of the pendulum the greatest?
C
A
D
B
Answer:
Point C
Explanation:
Greatest Kinetic Energy means lowest potential energy since energy is conserved. Lowest potential energy means lowest height which is at Point C.
A uniform metre rule of mass 10g is balanced on a knife edge placed at 45cm mark. Calculate the distance of a mass 25g from the pivot
Answer:
The distance of a mass 25g from the pivot is 18cm
Explanation:
Given
[tex]m_1 = 10g[/tex]
[tex]d_1 = 45cm[/tex]
[tex]m_2 = 25g[/tex]
Required
Distance of m2 from the pivot
To do this, we make use of:
[tex]m_1 * d_1 = m_2 * d_2[/tex] --- moments of the masses
So, we have:
[tex]10 * 45= 25* d_2[/tex]
[tex]450= 25* d_2[/tex]
Divide both sides by 25
[tex]18= d_2[/tex]
Hence:
[tex]d_2 = 18[/tex]
Two charged particles exert an electric force of 27 N on each other. What will the magnitude of the force be if the distance between the particles is reduced to one-third of the original separation
Answer:
243 N
Explanation:
The formula for electromagnetic force is F= Kq1q2/r^2
where r is the distance between the charges, if the distance between the charges is reduced by 1/3 then F will increase by 9 [(1/3r)^2 becomes 1/9r which is 9F] so 27*9 is 243N
Every object around you is attracted to you. In fact, every object in the galaxy is attracted to every other object in the galaxy.
a. True
b. False
Answer:
True
Explanation:
With the gravitational pull that our planets have, we are able to remain in orbit. This demonstrates how every object in the galaxy is attracted to every other object. Every object in the universe that has mass exerts a gravitational pull on every other mass. We as humans do it too, but since our force isn't strong, we don't have much of an effect. I hope this helped and please don't hesitate to reach out with more questions!
Three 30 g metal balls, one of aluminum, copper and lead, are placed in a large beaker of hot water for a few minutes. [The specific heats of aluminum, copper, and lead are 903, 385, and 130 J / (kg ° C), respectively].
to. Which of the balls, if any, will reach the highest temperature? Explain.
b. Which of the balls, if any, will have the most heat energy? Explain.
Answer:
The answer is below
Explanation:
Specific heat capacity is an intensive property of a material. The specific heat of a material is the amount of energy required to raise the temperature of one unit mass m of material by one unit of temperature.
a) Temperature is inversely proportional to specific heat capacity. If the same amount of heat is applied to all three balls, the ball that will reach the highest temperature is the ball with the least specific heat capacity.
Hence lead will have the highest temperature since it has the least specific heat capacity.
b) The quantity of heat is directly proportional to the specific heat capacity. Hence if all balls experience the same temperature change, the ball that have the most energy will be that with the highest specific heat capacity.
Hence aluminum will have the most heat since it has the highest specific heat capacity.
write down the following units in the ascending of their value A) mm nm cm um B) 1m 1cm 1km 1mm. convert the following units into SI without changing their values? A)3500g B)2.5km C)2h
Answer:
A) nm, um, mm, cm
B) 1mm, 1cm, 1m, 1km
A) 3500g, B) 2500m, C) 7200 seconds
collisions may take place between
Answer:
Collisions may take place between the reactants.
Explanation:
The collision frequency must be greater than the frequency factor for the reaction. A collision between the reactants must occur.
A hot air balloon is a sphere of volume 2210 m3. The density of the hot air inside is 1.13 kg/m3, while the air outside has a density of 1.29 kg/m3. The balloon itself has a mass of 240 kg. What is the TOTAL NET force acting on the balloon?
[?]N
The total net force acting on the balloon will be 24498 Newtons
Given that
Volume of the balloon = 2210 cubic meter
Density of the air inside the balloon = 1.13 kg/m3
What will be the net force exerted on the balloon ?Here force on the balloon will be equal to the weight of the air displaced by balloon
[tex]F= mass of air displaced\times gravity[/tex]
[tex]F= Density \times volume \times gravity[/tex]
[tex]F=1.13 \times 2210 \times 9.81[/tex]
[tex]F=24498 N[/tex]
The total net force acting on the balloon will be 24498 Newtons
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Explain what a circuit breaker is and how it helps protect your house?
Explanation:
A circuit breaker is an electrical switch designed to protect an electrical circuit from damage caused by overcurrent/overload or short circuit. Its basic function is to interrupt current flow after protective relays detect a fault.
Circuit breakers have been designed to detect when there is a fault in the electricity, so it will “trip” and shut down electrical flow. ... This detection is key to preventing surges of electricity that travel to appliances or other outlets, which can cause them to break down
recognizing forms of energy
Answer:
hi the question isn't obvious and need a photo I guess
Question 8 of 10
What was the name of the book that Ibn al-Haytham wrote?
A. Weather and Air Flow
B. Book of Optics
C. Light and Vision
D. Book of Sound
Answer:b
Explanation:
what is the dimensional formula of force and torque
Answer:
Units. Torque has the dimension of force times distance, symbolically T−2L2M. Although those fundamental dimensions are the same as that for energy or work, official SI literature suggests using the unit newton metre (N⋅m) and never the joule. The unit newton metre is properly denoted N⋅m.
Dimension: M L2T−2
In SI base units: kg⋅m2⋅s−2
Other units: pound-force-feet, lbf⋅inch, ozf⋅in
Answer:
hope it is helpful to you
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A double-slit experiment is performed with light of wavelength 550 nm. The bright interference fringes are spaced 2.3 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 360 nm?
Answer:
[tex]d_2=1.5*10^-3m[/tex]
Explanation:
From the question we are told that:
Initial Wavelength [tex]\lambda_1=550nm=550*10^{-9}[/tex]
Space 1 [tex]d_1=2.3*10^{-3}[/tex]
Final wavelength [tex]\lambda_2=360*10^{-9}[/tex]
Generally the equation for Fringe space at [tex]\lambda _2[/tex] is mathematically given by
[tex]d_2=\frac{d_1}{\lambdaI_1}*\lambda_2[/tex]
[tex]d_2=\frac{2.3*10^{-3}}{550*10^{-9}}*360*10^{-9}[/tex]
[tex]d_2=1.5*10^-3m[/tex]
What is (a) the x component and (b) the y component of the net electric field at the square's center
Answer:
What is (a) the x component and (b) the y component of the net electric field at the square's center
What word chemical equation describes this chemical reaction?
Answer : sodium + chlorine → sodium chloride
Betelgeuse (in Orion) has a parallax of 0.00451 + 0.00080 arcsec,as measured by the Hipparcos satellite. What is the distance to Betelgeuse, and what is the uncertainty in that measurement?
We have that the distance to Betelgeuse, and the uncertainty in that measurement is
[tex]d=(221.7\pm39.33)pc[/tex]Uncertainty U = 0.00080
From the Question we are told that
Betelgeuse (in Orion) has a parallax of 0.00451 + 0.00080
Generally
[tex]Distance\ in\ parsecs =\frac{ 1}{(parallax\ measured\ in\ arcseconds}[/tex]
Where
Parallax [tex]P =0.00451[/tex]
Uncertainty [tex]U = 0.00080[/tex]
Generally the equation for the distance is mathematically given as
[tex]d=(\frac{1}{P}pc\pm(\frac{U}{P}*100\%))[/tex]
Therefore
[tex]d=(\frac{1}{0.00451}pc\pm(\frac{0.00080}{0.00451}*100\%))[/tex]
[tex]d=(221.7\pm39.33)pc[/tex]
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A 0.495-kg hockey puck, moving east with a speed of 4.50 m/s , has a head-on collision with a 0.720-kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed (magnitude of the velocity) of each object after the collision?
Answer:
a) [tex]v_1=-0.833m/s[/tex]
b) [tex]V_2=12.5m/s[/tex]
Explanation:
From the question we are told that:
Hockey puck Mass [tex]m_1=0.495kg[/tex]
Hockey puck Speed [tex]u_1=4.50m/s[/tex]
Puck Mass [tex]m_2=0.720kg[/tex]
Assuming
Initial speed of Puck [tex]u_2=0[/tex]
Generally the equation for Speed of First Puck is mathematically given by
[tex]v_1=(\frac{m_1-m_2}{m_1+m_2})*v_1+(\frac{2m_2}{m_1+m_2})u_2[/tex]
[tex]v_1=(\frac{0.495-0.720}{0.495+0.720})*4.50+0[/tex]
[tex]v_1=-0.833m/s[/tex]
Generally the equation for Speed of Second Puck is mathematically given by
[tex]V_2=(\frac{2m_1}{m_1+m_2})u_2-(\frac{m_1-m_2}{m_1+m_2})v[/tex]
[tex]V_2=(\frac{2*0.495}{0.495*0.720})*4.50-0[/tex]
[tex]V_2=12.5m/s[/tex]
Typhoon signal number 2 is raised. What is the speed of the expected typhoon?
the simple answer is from 61kmph to 120kmph
Explanation:
no explanation is needed
two point charges with charge q are initially separated by a distance d. if you double the charge on both charges, how far should the charges be separated for the potential energy between them to remain the same
Answer:
r ’= 4 r
Explanation:
Electric potential energy is
U = [tex]k \frac{q_1q_2}{r_{12}}[/tex]k q1q2 / r12
in this exercise
q₁ = q₂ = q
U = k q² / r
for when the charge change
U ’= k q’² / r’
indicate that
q ’= 2q
U ’= U
we substitute
U = k (2q) ² / r ’
U = 4 k q² / r ’
we substitute
[tex]k \ \frac{q^2}{r} = 4 k \ \frac{q^2}{r'}[/tex]k q² / r = 4 k q² / r ’
r ’= 4 r
