You take your pulse and observe 80 heartbeats in a minute. What is the period of your heartbeat? What is the frequency of your heartbeat?
Answer:
120 beats per minute.
Explanation:
If I take your pulse and observe 80 heartbeats in a minute. Then the period of your heartbeat is 0.8 s and frequency is 1.3Hz.
What is Heartbeat ?A pulse is the term used in medicine to describe the tactile arterial palpation of the cardiac cycle (heartbeat) by skilled fingertips. Any location where an artery can be compressed close to the surface of the body, such as the carotid artery in the neck, the radial artery in the wrist, the femoral artery in the groyne, the popliteal artery behind the knee, the posterior tibial artery near the ankle joint, and on the foot, can be used to palpate the pulse (dorsalis pedis artery). Heart rate may be determined by monitoring pulse, or the number of arterial pulses per minute. Auscultation, which is the process of counting the heartbeats while listening to the heart using a stethoscope, is another way to determine the heart rate. Typically, three fingers are used to gauge the radial pulse.
Given,
heart beat = 80 beats/min = 1.3 beats/s
Frequency is nothing but how much beats is heart having in one second and that is 1.3 beats/s. Hence frequency of heart is 1.3Hz.
The Period is reciprocal of frequency,
T = 1/f = 0.8 s
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For the following questions, assume the wavelengths of visible light range from 380 nm to 760 nm in a vacuum. (a) What is the smallest separation (in nm) between two slits that will produce a ninth-order maximum for any visible light
Answer:
Explanation:
This is an interference exercise, which the case of constructive interference is described by the expression
d sin θ = m λ
in this case they indicate that we are in the ninth order (m = 9).
To be able to observe the pattern, the dispersion angle must be less than 90º
we substitute
sin 90 = 1
d = m lang
let's calculate
d = 9 λ
d = 9 380 10⁻⁰
d = 3.42 10⁻⁶
d2 = 9 760 10⁻⁹
d2 = 6.84 10₋⁶
A(n) 60.9 kg astronaut becomes separated
from the shuttle, while on a space walk. She
finds herself 36.5 m away from the shuttle
and moving with zero speed relative to the
shuttle. She has a(n) 0.928 kg camera in her
hand and decides to get back to the shuttle
by throwing the camera at a speed of 12 m/s
in the direction away from the shuttle.
How long will it take for her to reach the
shuttle? Answer in minutes.
Answer in units of min.
Answer:
m v = M V conservation of momentum after throwing camera
V = m * v / M = .928 * 12 / 60.9 = .183 m/s
t = S / V = 36.5 / .183 = 199 sec = 3.32 min
Identifying the factors contributing to and acting as determinant factors of health disparities during the program theory and development process is a means of culturally tailoring the program.
a) true
b) false
Answer:
a) True
Explanation:
A program-specific message provided to an individual or group with the intention of raising awareness of a health condition, motivating behavior change, removing perceived barriers to participating in a health habit, or something else relating to the program's aims and objectives. The most effective intervention messages are usually theory-based and culturally adapted.
A person rolls a 7 kg bowling ball down a lane in a bowling alley. The lane is
18 m long. The ball is traveling at 7 m/s when it leaves the person's hand.
What is the ball's kinetic energy at this point?
Answer:
171.5J
Explanation:
K=1/2 *m *U²
K=1/2 *7 *7²
K=171.5 J
Assignment: 06.05 Infections and Health
What is a reasonable measurement for the distance to Neptune?
30 light years
30 kilometers
30 parsecs
30 Astronomical Units
Answer:
30 kilometers is a reasonable measurement
The spring in the figure has a spring constant of 1400 N/m . It is compressed 17.0 cm , then launches a 200 g block. The horizontal surface is frictionless, but the block's coefficient of kinetic friction on the incline is 0.210.
What distance d does the block sail through the air?
Use the work-energy theorem to find the velocity of the block when it's released by the spring. The work done by the spring on the block as it's restored to equilibrium is
W = 1/2 kx ²
where k is the spring constant and x is the compression of the spring. So
W = 1/2 (1400 N/m) (0.170 m)² = 20.23 J
This is equal to the block's change in kinetic energy ∆K,
W = ∆K
and since it starts from rest, the initial K is zero, leaving us with
W = 1/2 mv ²
where m is the mass of the block and v is its speed, so that
20.23 J = 1/2 (0.200 kg) v ²
==> v ≈ 14.2 m/s
The block slides at this speed across the frictionless surface until it hits the incline which introduces friction.
First, you need to find the length of the incline. It forms a 45° angle, and the underlying 45°-45°-90° triangle has a hypotenuse of length √2 (2.0 m) ≈ 2.83 m.
Next, you need to find the total work done on the block as it slides up the incline. Use Newton's second law to examine the forces acting on the block during this phase:
• the net force acting on the block in the direction perpendicular to the incline is
∑ F = n - mg cos(45°) = 0
where n = mg cos(45°) ≈ 1.39 N is the magnitude of the normal force and mg cos(45°) ≈ 1.39 N is the perpendicular component of the block's weight;
• the net force acting on the block parallel to the surface is
∑ F = -f - mg sin(45°) = ma
where f = µn = 0.210n ≈ 0.291 N is the magnitude of kinetic friction, mg sin(45°) ≈ 1.39 N is the parallel component of the weight, and a is the acceleration of the block.
Only the parallel forces do work on the block, and this work is negative because friction and weight oppose the block's sliding up the incline. The total work done on the block is then
W = (-0.291 N - 1.39 N) (2.83 m) ≈ -4.74 J
Use the work-energy theorem again to find the block's new speed v at the top of the incline:
W = ∆K
==> -4.74 J = 1/2 (0.200 kg) v ² - 1/2 (0.200 kg) (14.2 m/s)²
==> v ≈ 12.4 m/s
And now this becomes a projectile problem. The block travels a horizontal distance x after being launched at an angle of 45° with initial speed 12.4 m/s after time t according to
x = (12.4 m/s) cos(45°) t
Its height y from the 2.0 m-high surface at time t is given by
y = (12.4 m/s) sin(45°) t - 1/2 gt ²
The block lands on the surface when y = 0, which occurs after t ≈ 1.79 s, at which point the block has covered a distance d ≈ 15.7 m.
The block sail through the air at the distance of "15.8 m"
Given:
Spring constant,
1400 N/mMass,
200 gBlock's coefficient,
0.210By using Work energy theorem, we get
→ [tex]W_{spring}+W_g+W_f = KE_f-KE_i[/tex]
By substituting the values, we get
→ [tex]\frac{1}{2}\times 1400\times (0.17)^2- (0.2\times 9.8\times 2)-(0.21\times 0.2\times \frac{9.8}{\sqrt{2} }\times \sqrt{2}\times 2 )= \frac{1}{2}\times 0.2\times V_f^2[/tex]
here,
[tex]V_f = 12.44 \ m/s[/tex]
→ [tex]d = \frac{V_f^2 Sin 2 \Theta}{g}[/tex]
[tex]= \frac{(12.44)^2 Sin 90^{\circ}}{9.8}[/tex]
[tex]= 15.8 \ m[/tex]
Thus the answer above is right.
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The red light from a helium-neon laser has a wavelength of 644.6 nm in air. Find the speed, wavelength, and frequency of helium-neon laser light in air, water, and glass. (The glass has an index of refraction equal to 1.50.)speed (m/s)wavelength (nm)frequency (Hz)airwaterglass
Answer:
air f = 4.6527 10¹⁴ Hz
water f = 3.4914 10¹⁴ Hz
glass f = 3.1027 10¹⁴ Hz
Explanation:
The refractive index of a material is given by
n = c / v
where c is the speed of light in a vacuum c = 3 108 m / s and v is the speed of light in the material medium.
the speed of the wave is
v = λ f
we substitute
c / n = λ f
f = [tex]\frac{c}{n \ \lambda}[/tex]
The refractive indices are
air 1,00029
water 1.3330
glass 1.5
let's calculate the frequencies
vaccum
f = 3 10⁸ / 1 644.6 10⁻⁹
f = 4.6540 10¹⁴ Hz
air
f = 3 10⁸ / 1,00029 644.6 10⁻⁹
f = 4.6527 10¹⁴ Hz
Water
f = 3 10⁸ / 1.333 644.6 10⁻⁹
f = 3.4914 10¹⁴ Hz
glass
f = 3 10 ^ 8 / 1.5 644.6 10⁻⁹
f = 3.1027 10¹⁴ Hz
An object produces a sound wave with a wavelength 75.0cm. if the speed of sound is 350.0m/s, the frequency of the sound is
Answer:
wavelength=75.0
speed of sound(v)=350 .0m/s
frequency(f)=?
we know,
v=f*wavelengh
350.0 =f*750
f. =350/75
=4.667
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Consider a point on a bicycle wheel as the wheel turns about a fixed axis, neither speeding up nor slowing down. Compare the linear and angular accelerations of the point.
a. Both are zero.
b. Only the angular acceleration is zero.
c. Only the linear acceleration is zero.
d. Neither is zero
Explain your choice
Answer:
c. Only the linear acceleration is zero.
Explanation:
The linear acceleration is defined as the rate of change of linear velocity. Since the bicycle is moving in the same direction, with the same speed, without speeding up or slowing down. Therefore, there will be no change in linear velocity and as a result, linear acceleration will be zero.
The angular acceleration is the rate of change of angular velocity. Since the angular velocity is changing its direction constantly. Therefore, it has a certain component of acceleration at all times called centripetal acceleration.
Therefore, the correct option is:
c. Only the linear acceleration is zero.
Suppose 3 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The gas is then allowed to expand quickly and adiabatically back to its original volume.
Required:
a. Find the highest temperature attained by the gas.
b. Find the lowest temperature attained by the gas.
c. Find the highest pressure attained by the gas.
d. Find the lowest pressure attained by the gas.
Answer:
a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm
Explanation:
For isothermal expansion PV = constant
So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,
So, P₁V₁ = P₂V₂
P₂ = P₁V₁/V₂
Since V₂/V₁ = 0.19,
P₂ = P₁V₁/V₂
P₂ = 1 atm (1/0.19)
P₂ = 5.26 atm
For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas
So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,
So, P₂V₂ⁿ = P₃V₃ⁿ
P₃ = P₂V₂ⁿ/V₃ⁿ
P₃ = P₂(V₂/V₃)ⁿ
Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19
1/0.19,
P₃ = P₂(V₂/V₃)ⁿ
P₃ = 5.26 atm (0.19)⁽⁵/³⁾
P₃ = 5.26 atm × 0.0628
P₃ = 0.33 atm
Using the ideal gas equation
P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)
P₃V₃/T₃ = P₄V₄/T₄
T₃ = P₃V₃T₄/P₄V₄
T₃ = (P₃/P₄)(V₃/V₄)T₂
Since V₃ = V₄ = V₁ and P₄ = P₁
V₃/V₄ = 1 and P₃/P₄ = P₃/P₁
T₃ = (P₃/P₁)(V₃/V₄)T₂
T₃ = (0.33 atm/1 atm)(1)273 K
T₃ = 90.1 K
So,
a. The highest temperature attained by the gas is T₁ = 273 K
b. The lowest temperature attained by the gas = T₃ = 90.1 K
c. The highest pressure attained by the gas is P₂ = 5.26 atm
d. The lowest pressure attained by the gas is P₃ = 0.33 atm
A man standing on a frictionless ice throws a 1.00kg mass at 20m/s at an angle elevation of 40.0 degrees. What was the magnitude of the mans momentum immediately after the the throl
Answer:
Explanation:
1.00kg×20m/s×cos40=15.3
The engine in the car from question 1 uses a force of 1200 N to cause the car to accelerate at 3.5 m/s2. What is the car's mass?
Answer:
Mass of car's engine = 342.85 kg (Approx.)
Explanation:
Given values:
Force applied by car = 1,200 N
Acceleration of car' engine = 3.5 m/s²
Find:
Mass of car's engine
Computation:
⇒ Mass = Force / Acceleration
⇒ Mass of car's engine = Force applied by car / Acceleration of car
⇒ Mass of car's engine = 1,200 / 3.5
⇒ Mass of car's engine = 342.85 kg (Approx.)
State two (2) examples of osmosis occurring in everyday life
What is the potential energy of a 7kg object 4m off the ground ?
please show your work
Answer:
Gravitational potential energy is mass of the object times the gravitational constant times the height of the object:
U = mgh (I will use 10 for the gravitational constant but you can use 9.8 or 9.81 or something even more accurate)
U = 280
The gravitational potential of the object is 280 joules
The first charged object is exerting a force on the second charged object. Is the second charged object necessarily exerting a forcer on the first?
Answer:
Explanation:
Of course because it's Newton's Law that if body A exerts force on body B, then body B will exert equal but opposite force on body A.
HAPPY LEARNING:)
The first charged object is exerting a force on the second charged object. The second charged object necessarily exerting a force on the first is correct.
What is force?A force is an effect that can alter an object's motion according to physics. 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. A force is a vector quantity since it has both magnitude and direction.
Newton's third law states that for every action (force) in nature there is an equal and opposite reaction, from that we can understand when force is exerted on first charged object is exerting a force on the second charged object. The second charged object necessarily exerting a force on the first.
The first charged object is exerting a force on the second charged object. The second charged object necessarily exerting a force on the first is correct.
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Short-term memory is active, while long-term memory is:
A dynamic
B
reflective.
c) passive
D
recessive.
Answer:
the answer is b. reflective
Now imagine that you are a Haitian taptap driver and want a more comfortable ride. You decide to replace the springs with new springs that can handle the typical heavy load on your vehicle. What spring constant do you want your new spring system to have?
The new springs should have a spring constant that is (slighty larger, substantially larger, slightly smaller, substantially smaller) substantially larger slightly larger slightly smaller substantially smaller than the spring constant of the old springs.
Answer:
We use a spring of large spring constant.
Explanation:
The spring constant is defined as the force applied on the spring per unit extension or compression in length.
F = k x
where, F is the force, x is the extension, k is the spring constant.
Its unit is N/m.
To get the comfortable ride, we use the spring of large spring constant, so that the spring gets stiffer and we get comfort.
A car has a mass of 900 kg is accelerated from rest at a rate of 1.2 m/s calculate the time taken to reach 30/s
Answer:
12+2=24+30+2=66
Explanation:
Determine the minimum horizontal force P required to hold the crate from sliding down the plane. The crate has a mass of 50 kg and the coefficient of static friction between the crate and the plane is . ms
Answer: hello some data related to your question is missing attached below is the missing data and diagram related to the solution
answer:
P = 141.21 N
Explanation:
Given data:
Mass of crate = 50 kg
coefficient of static friction ( μ ) = 0.25
Calculate minimum horizontal force ( P ) that holds the crate from sliding
∑fx = 0
= P + Fcos θ - N*sinθ = 0
= P + 0.25N cos 30° - Nsin30° = 0
∴ P = 0.2835 N = 0
P - 0.2853 N = 0 ------- ( 1 )
∑fy = 0
- 50g + Ncosθ + Fsinθ
- 50*9.81 + Ncos30° + 0.25Nsin30°
∴ N = 494.942 N ----- ( 2 )
input 2 into 1
P - 0.2853 ( 494.942 ) = 0
P = 141.21 N
A kettle operates from a 120 V outlet. It has a heating element with a resistance of 8.0 Ω . Calculate the current going through the element.
Answer:
I = 15A
Explanation:
V = I*R
120V = I*8.0ohms
I = 120V/8.0ohms
I = 15A
Answer:
I=15A
Explanation:
you know what to do.
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Give them the brain list.
concave mirror daily application
Answer:
concave mirror use in headlights and torches
Explanation:
Concave mirrors are used in headlights and torches. The shaving mirrors are also concave in nature since these mirrors can produce enlarged clear images. Doctors use concave mirrors as head mirrors to have a clearer view of eyes, noses, and ears. The dental mirrors used by dentists are also concave.
The extension produced on the wire by the application of 300 N force is 2 mm. The elastic energy stored on the wire is
Answer:
150000 is the answer to the question
Which of these hazmat products are allowed in your FC?
Please choose all that apply.
A GPS unit (lithium batteries)
A subwoofer (magnetized materials)
A can of hairspray (flammable/aerosols)
Fireworks (explosives)
Answer: Hazmat products are allowed in your FC are:
A GPS unit (lithium batteries) A subwoofer (magnetized materials)Explanation:
Hazmat products consist of flammable, corrosive and harmful substances which are actually very hazardous to human health and environment.
Hazardous material allowed in FC are as follows.
Magnetized material products like as speakers.Non-spillable battery products like toy cars.Lithium-ion battery containing products like laptops, mobile phones etc.Non-flammable aerosol.So, hazmat allowed products are GPS unit (lithium batteries) and subwoofer (magnetized materials).
Thus, we can conclude that hazmat products are allowed in your FC are:
A GPS unit (lithium batteries) A subwoofer (magnetized materials)a system absorbs 500J of heat and at the same time 400J of work is done on the system what is change in internal energy
This is the solution of the problem. The problem is calculated from a closed system energy balance equation
Use a projectile motion kinematic equation for the vertical motion to find the time t that the ball is in the air, from when it leaves the track until it strikes the floor.
Answer:
The time of flight is [tex]T = \frac{2 u sin A}{g}[/tex].
Explanation:
Let the initial velocity is u and the angle of projection is A.
Use first equation of motion for vertical motion
Let the time to reach the maximum height is t.
[tex]v = u - gt\\\\0 = u sin A - gt \\\\t = \frac{ u sin A}{g}[/tex]
Total time of flight is
T =2 t
[tex]T = \frac{2 u sin A}{g}[/tex]
What happens to the water when you throw rock into a pond
Answer:
The water usually rushes back too enthusiastically, causing a splash – and the bigger the rock, the bigger the splash. The splash then creates even more ripples that tend to move away from where the rock went into the water.
The position of an object of
mass 5 kg as a function of
time is giving by r = (20m/s4)t4
i + (12 m/s3)t3 j. Find the
force acting on the object as a
function of time. Express the
force in unit vectors. Hint:
Remember that Newton's
second Law relates the force
to the acceleration
Answer:
[tex]F=5(240t^2i+72tj)\ N[/tex]
Explanation:
Given that,
The mass of the object, m = 5 kg
The position vector is, [tex]r=20t^4i+12t^3j[/tex]
Velocity, [tex]v=\dfrac{dr}{dt}=80t^3i+36t^2j[/tex]
Acceleration, [tex]a=\dfrac{dv}{dt}=240t^2i+72tj[/tex]
Newton's second law of motion is given as follows:
F = ma
Put all the values,
[tex]F=5(240t^2i+72tj)\ N[/tex]
Hence, this is the required solution.
It is said that a gas fills all the space available to it. Why then doesn't the atmosphere go off into space?