Answer:
I believe B. The picture is not completely clear
Explanation:
Professor Stauffer conducts basic research on the psychological, biological, and behavioral
factors that contribute to heart disease. She is most likely a
High blood pressure, high LDL cholesterol, diabetes, smoking, exposure to secondhand smoke, obesity, a poor diet, and inactivity are the main risk factors for heart disease and stroke.
Explain about the contribute to heart disease?High blood pressure, high cholesterol, and smoking are the three main risk factors for heart disease that at least half of all Americans (47%) have. Age and family history are two uncontrollable risk factors for heart disease.
The most frequent cause of coronary artery disease is atherosclerosis, which is a buildup of fatty plaques in the arteries. Poor diet, lack of exercise, obesity, and smoking are risk factors. Choosing a healthy lifestyle can help reduce the risk of atherosclerosis.
The most prevalent type of heart disease is coronary heart disease (CHD). It happens when plaque deposits cause the arteries that carry blood to the heart to constrict or stiffen. Fat, cholesterol, and other components found in blood make up plaque. The term "atherosclerosis" also refers to this plaque accumulation.
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Which of these particles are equal in number when an atom is neutral?
A. protons, neutrons
B. protons, electrons
C. neutrons, electrons
D. protons, neutrons, electrons
In a class experiment to determine information about free-fall acceleration, a watermelon and a pumpkin are each set to fall from the back of the stands at your football stadium.
a. If the watermelon and the pumpkin are both dropped at the same time, which one will hit the ground first?
b. If the watermelon is thrown downward with an initial speed of 10 m/s and the pumpkin is dropped, which one will hit the ground first? Show calculation to support your answer.
c. If it takes the watermelon 1 second to reach the ground when it is thrown downward at 10 m/s, how tall are the stands?
d. How long does it take the pumpkin to reach the ground if it is dropped from this height you calculated in part c?
Answer:
a) They both hit the ground at the same time
b) Watermelon will hit first, since its speed is faster than the pumpkin speed
c) h = 29,6 m
d)t = 3,02 sec
Explanation:
Equations for fall free movement are:
vf = v₀ + g*t when v₀ = 0 (dropped case) vf = g*t
h = v₀*t + 1/2*g*t²
a) For both ( watermelon and pumpkin) the equation of speed is the same:
vf = g*t² Both will have the same speed second through second
They both hit the ground at the same time
b) Now is watermelon is thrown with v₀ = 10 m/s
Watermelon will hit first since its speed is faster than the pumpkin speed
vf(watermelon) = 10 + g*t
vf₂ (pumpkin) = g*t
c) h = v₀*t + (1/2)*g*t²
h = (10)*1 + (1/2)*9,8*1
h = 10 + 19,6
h = 29,6 m
d) h = g*t
t = 29,6/9,8
t = 3,02 sec
Is it possible for an object to be accelerating and at rest at the same time? Explain.
Answer:
Yes
Explanation:
acceleration is a change in velocity, so an object might be momentarily at rest but a split-second later have some speed, i.e. it can be changing its velocity even if it is momentarily at rest.
The magnetic field at a distance of 2 cm from a current-carrying wire is 4 µT. Find the magnetic field at a distance of 1 cm from the wire.Group of answer choices2 µT4 µT6 µT8 µT
Given:
The magnetic field is
\(B=\text{ 4 }\mu T\)when the distance is d = 2 cm.
To find the magnetic field when the distance is d' = 1 cm
Explanation:
The relation between magnetic field and distance is
\(\begin{gathered} B=\frac{\mu_0I}{2\pi d} \\ B=\frac{k}{d} \end{gathered}\)Let the k be any arbitrary constant.
\(\begin{gathered} k=\text{ Bd} \\ =4\text{ }\times2 \\ =8\text{ cm }\mu T \end{gathered}\)On substituting the values, the magnetic field will be
\(\begin{gathered} B^{\prime}=\frac{k}{d} \\ =\frac{8\text{ cm }\mu T}{1\text{ cm}} \\ =\text{ 8 }\mu T \end{gathered}\)Final Answer: The magnetic field is 8 micro Tesla when the distance is 1 cm.
Liquid water is nearly 1,000 times denser than air. Thus, for every 32.0 feet (9.75 m) a scuba diver descends below the water's surface, the pressure increases by 1.00 atm. Human lungs have a volume of approximately 3.50 L. If a scuba diver descends to a depth of 80.0 feet where the pressure is 3.50 atm (2.50 atm from the water and 1.00 atm from the air pressure), then by how much does the volume of a 3.50 L surface sample of air decrease
Answer:
ΔV = -2.1 L
Explanation:
To solve this exercise we can use the ideal gas equation for two points
PV = nRT
P₁V₁ = P₂ V₂
where point 1 is on the surface and point 2 is at the desired depth,
V₂ = \(\frac{P_1}{P_2} \ V_1\)
let's calculate
V₂ = ( \(\frac{1 atm}{2.5 atm}\) ) 3.5 L
V₂ = 1.4 L
this is the new volume, the change in volume is
ΔV = V₂ -V₁
ΔV = 1.4-3.5
ΔV = -2.1 L
If the mass = 1.50 kg, F1 = 1.50 N, 0, = 35.0°, 02 = 45.00 and the acceleration of the object in the + x direction is 1.0 m/s?, what must be the force F2? (Shown in image)
The magnitude of force F₂ is determined as 0.38 N.
What is the magnitude of F₂?
The magnitude of force (F₂) is calculated by applying Newton's second law of motion as follows;
F = ma
where;
F is the resultant of the two forcem is the mass of the objecta is the acceleration of the objectThe resultant forces in the x-direction is given as
F₁cosθ₁ + F₂cosθ₂ = maₓ
(1.5 cos35) + F₂(cos45) = 1.5 x 1
1.23 + 0.71F₂ = 1.5
0.71F₂ = 1.5 - 1.23
0.71F₂ = 0.27
F₂ = 0.27/0.71
F₂ = 0.38 N
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Three points A, B and C are located along a horizontal line. A positive test charge is released from rest at C and accelerates towards B. Upon reaching B, the test charge continues to accelerate toward A. Assuming that only motion along the line is possible,
a) Which point has the highest potential; A, B or C?
b) What will a negative test charge do when it is released from rest at B?
Ramp 1 Ramp 2 Ramp 3
Trial 1 0.95 s 0.78 s 1.31 s
Trial 2 0.87 s 0.75 s 1.27 s
Trial 3 0.92 s 0.80 s 1.44 s
Avg. Time 0.91 s 0.78 s 1.34 s
Which of the following conclusions can be made from the above data?
From this data it is clear that the time require to complete ramp 3 is more than any other, hence it can be large ramp or there are more objectless or turns. Ramp 2 requires least time to complete, it can be shorter or having less number of obstacles.
An inclined plane, also known as a ramp, is a flat supporting surface that is slanted at an angle from the vertical direction, with one end higher than the other, and is used to help raise or reduce a weight. The inclined plane is one of the six traditional basic devices established by Renaissance scientists. Inclined aircraft are used to transport big cargoes over vertical obstructions. Examples include a ramp used to load items into a truck, a person going up a pedestrian ramp, and an automobile or railroad train climbing a gradient. it can have obstacles in the path
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If the normal force of the bow on the string is 0.50 N , how far can the string be pulled before it slips if the string is bowed at its center
Due to the high tension in the string, the distance the string is pulled
before it slips is a small fraction of the length of the string.
Response:
The distance the string can be pulled before slipping is 0.55 mmHow does friction force determines the distance the string can be pulled?The given parameters are;
Normal force on the string = 0.5 N
From a similar question posted online, we have;
Length of the violin string, L = 0.33 m
Tension in the string, T = 60 N
Static friction, \(\mu_s\) = 0.8
Required:
The distance the string can be pulled before it slips.
Solution:
Friction force, \(F_f\) = 0.5 N × 0.8 = 0.4 N
According to Newton's law of motion, we have;
\(F_f\) = 2·T × sin(θ)
Which gives;
0.4 N = 2 × 60 N × sin(θ)
Therefore;
\(sin(\theta) = \dfrac{0.4}{2 \times 60} = \mathbf{ \dfrac{1}{300}}\)
The triangle formed by the half length of the string and the displacement gives;
\(sin(\theta) = \dfrac{x}{\frac{L}{2} } = \mathbf{\dfrac{2 \cdot x}{L}}\)Where;
x = The distance the string is pulled before slipping
Which gives;
\(\dfrac{1}{300} = \mathbf{\dfrac{2 \cdot x}{0.33 \, m}}\)
0.33 m = 300 × 2 × x = 600·x
\(x = \dfrac{0.33 \, m}{600} = 0.00055 \, m = \mathbf{0.55 \, mm}\)
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Robert, my neighbor, a forty-five-year-old blacksmith is seven feet tall, and eats all day long. What does he weigh?
The neighbour weighs Iron. As he is a blacksmith, iron is the only thing he weighs the whole day.
What is a riddle?A riddle is a statement, question or phrase having a double or veiled meaning, put forth as a puzzle to be solved.
If Robert, your neighbor, a forty-five-year-old blacksmith is seven feet tall, and eats all day long, based on these statements, we can explain the riddle as Robert weighs iron for the fact that he is a blacksmith, iron is the only thing he weighs the whole day.
Thus, the correct explanation for the riddle is " The neighbour weighs Iron. As he is a blacksmith, iron is the only thing he weighs the whole day.
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What part of the sweating process promotes cooling?
wiping sweat from the surface of the skin
condensation of liquid sweat on the skin
liquid (sweat) sitting on the skin’s surface
evaporation of liquid sweat from the skin
Answer: Condensation of liquid sweat on the skin
Explanation:
H H H O H O N C C N C C H O H H O H H C H H H ALANINE GLYCINE
What is the difference between these two amino acids?
in a typical cop movie we see the hero pulling a gun firing that gun straight up into the air and shouting
It is not recommended to fire a gun straight up into the air.
When a bullet is fired into the air, it will eventually come down and can pose a danger to people and property below. The bullet can still be lethal when it reaches the ground, especially if it lands on a hard surface or hits someone directly.
Additionally, firing a gun in a residential area can be illegal and can result in legal consequences. In general, guns should only be fired in designated shooting ranges or in self-defense situations where there is an immediate threat to life. It is important to handle firearms responsibly and follow all safety guidelines to prevent accidents and injuries.
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A 375 N unbalanced force was applied to a 45 kg object at rest. How far would this object travel after 7 seconds?
Answer:204.2
Explanation:
F = 375 N
M = 45 KG
T = 7 SECONDS
U = 0
S = ?
F = ma
a = F/M
= 8.33m/s
from second equation of motion
s = ut + 1/2at^2
S = 204.166....
a man weights 800n at the earth's surface. What will be the weight of the same person when he is orbiting the earth at double the high of the earth's center
Answer:
200 N
Explanation:
F = G m1 m2 / r^2 G m1 and m2 are constant for this scenario
so this becomes
F is proprtional to 1/r^2
f = 1/r^2
800 = 1/r^2 now change r to 2r
F = 1/(2r)^2 = 1/4 * 1/r^2 <======shows force will be 1/4 of original , or 200N
an object has the mass of 5.7kg what is the force of gravity acting on the object on earth
Answer:
55.68
Explanation:
Before a collision, a 25 kg object is moving at 12 m/s to the right. After a collision with stationary box, the 25 kg object moves at 8 m/s to the right. What is the resulting momentum of the box?
The initial momentum of the 25 kg object is 25 kg * 12 m/s = 300 kgm/s. After the collision, the momentum of the 25 kg object is 25 kg * 8 m/s = 200 kgm/s. According to the conservation of momentum, the momentum lost by the 25 kg object is equal to the momentum gained by the box. Therefore, the resulting momentum of the box is 300 kgm/s - 200 kgm/s = 100 kg*m/s.
(Figure 1) is the potential-energy diagram for a 500 g particle that is released from rest at A . What is the particle's speed at B ?
Answer:
Explanation:
according to the graph at B the potential energy of the particle is 2J
therefore we can use the kinetic energy equation to calculate the particle's velocity or speed.
\(E_{k} =1/2mv^{2}\)
2J= 1/2*1/2kg*v^2
8=v^2
v= 2√2 ms-1
The once-popular LP (long-play) records were 12 in. in diameter and turned at a constant 33 1/3 rpm.
The angular speed of the long-play (LP) in rad/s is 3.49rads/s and its period in seconds is 1.80 seconds.
How is angular speed and period calculated?To convert the angular speed from revolution per minutes (rpm) to seconds the following formula is used:
ω = (k revolution/60 seconds) x (2π/1 revolution)
ω = 2πk/60seconds (k x 2π/T in seconds)
where, k = constant in rpm, 33 1/3rpm = 33.3333rpm
ω = 2π33.3333/60 = 209.4374/60
ω = 3.49rad/s
For period in seconds, ω = 2π/T
where, T = period and ω = angular speed, 3.49rad/s
To find T, T = 2π/ω
T = 2π/3.49rad/s
T = 1.80s
Using the formula, ω = 2πk/60seconds, the angular speed in rad/s of the LP is 3.49rad/s and the time period for the records using T = 2π/ω is 1.80s.
The full question is:
The once-popular LP (long-play) records were 12 inches in diameter and turned at a constant 33(1/3) rpm. Find the angular speed of the LP in rad/s and its period in seconds.
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how to reduce fluid friction
Answer:
By making the object sharp pointed
If a transverse wave passes from a flexible spring into a heavier stiffer spring where the wave has a greater speed, what will happen to the
reflected wave in the flexible spring?
When a transverse wave passes from a flexible spring into a heavier stiffer spring, the wave will have a greater speed in the stiffer spring due to the increased stiffness of the material. As a result, when the wave reaches the end of the stiffer spring, some of the energy from the wave will be reflected back towards the flexible spring.
The reflected wave in the flexible spring will be inverted, meaning that it will be flipped upside down. Additionally, the amplitude of the reflected wave will depend on the amount of energy that was transmitted into the stiffer spring.
If a large amount of energy was transmitted, then the reflected wave will have a higher amplitude. Conversely, if only a small amount of energy was transmitted, then the reflected wave will have a lower amplitude.
Overall, the reflected wave in the flexible spring will experience a change in both its orientation and amplitude due to the transmission and reflection of the initial wave in the stiffer spring.
When a transverse wave passes from a flexible spring into a heavier, stiffer spring where the wave has a greater speed, the reflected wave in the flexible spring will have a reduced amplitude and undergo a phase change of 180 degrees, resulting in an inverted waveform.
This occurs due to the impedance mismatch between the two springs, causing part of the wave energy to be reflected and part to be transmitted.
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Earth’s gravitational field is 7.51 N/kg at the altitude of the space shuttle. What is the size of the force of attraction between a student of mass 53 kg and Earth? Answer in units of N.
Explanation:
395.53 N
The force of attraction between a student of mass 53 kg and Earth is calculated using the equation F = G*m1*m2/r^2, where G is the gravitational constant (7.51 N/kg), m1 is the mass of the student (53 kg), m2 is the mass of Earth, and r is the distance between them (the altitude of the space shuttle). Plugging in these values, we get F = 7.51*53*5.97e24/((6.37e6 + 6.37e6)^2) = 395.53 N.
Problem 19: Oil with a density of 892 kg/m is flowing smoothly through a pipe, as
shown. In the lower portion, the oil is flowing at vi = 1.84 m/s, and the pressure gauge
indicates P, = 237 kPa. In the upper portion of the pipe, oil is flowing at V2 = 3.61 m/s at a
height that is 8.63 m above the lower portion.
The pressure of the oil with the given properties flowing in the horizontal pipe at the upper portion of the pipe is 157 kPa.
Pressure in the upper portion of the pipe
The pressure in the upper portion of the pipe is calculated by applying Bernoulli's equation,
P₁ + ¹/₂ρv₁² + ρgh₁ = P₂ + ¹/₂ρv₂² + ρgh₂
Given;
P₁ = 237 kPav₁ = 1.84 m/sh₁ = 0v₂ = 3.61 m/sh₂ = 8.63 mρ = 892 kg/m³P₂ = ?Susbtsitute the given parameters and solve the for the pressure in the upper portion of the pipe.
237,000 + ¹/₂(892)(1.84)² + (892)(9.8)(0) = P₂ + ¹/₂(892)(3.61)² + (892)(9.8)(8.63)
238,509.9776 = P₂ + 81,252.325
P₂ = 238,509.9776 - 81,252.325
P₂ = 157,257.65 Pa
P₂ ≅ 157 kPa
Thus, the pressure of the oil with the given properties flowing in the horizontal pipe at the upper portion of the pipe is 157 kPa.
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What is the acceleration of a ball with a mass of 0.40 kg is hit with a force of 16N?
Answer:
40 m/s^2
Explanation:
Mass= 0.40 kg
Force= 16 N
Therefore the acceleration can be calculated as follows
F = ma
16= 0.40 × a
16= 0.40 a
a= 16/0.40
a= 40 m/s^2
Hence the acceleration is 40 m/s^2
را Three forces acting at a point of 300N forces may varies but the angle between is always 40. Determine the value of the angle for which the value of the resultant force is parallel to BB.
The value of the angle for which the value of the resultant force is parallel to b-b is 6.35°. Therefore, the correct option is A.
What is an angle?An angle can be defined as the figure that is formed when two rays or lines shares a common endpoint.
Let the x and y-axis be shown here. If the resultant is directed along the x-axis, then its component in the y direction is zero.
0 = ΣFy = 300 sin θ + 300 sin (40 + θ) – 500 sin 30°
Therefore, sin θ+ sin (40+ θ)= 500 sin 30°÷ 300
⇒ sin θ + sin (40 + θ) = 0.833
⇒ 2 sin (40 + θ + θ)÷ 2 × cos 40 + θ—θ) ÷ 2 = 0.833
⇒ 2 sin (20 + θ) × cos (20) = 0.833
⇒ θ = 6.35°
Hence, the value of angle is 6.35°. Therefore, the correct option is A.
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The question is incomplete, but most probably the complete question is,
Three forces acting at a point are shown in Fig. The direction of the 300 N forces may vary, but the angle between them is always 40°. Determine the value of θ for which the resultant of the three forces is directed parallel to b-b.
A. 6.35°
B. 6.00°
C. 7.40°
D. 4.81°
To measure the vast distance in space ,scientists use a unit called a ______, which is how far travels in one year.
The closest star to Earth is ______.
(Look at the picture)
Answer:
"light year" - about 6 million million miles
The nearest star is alpha Centauri - about 24 million million miles away
26) Star A and star B are the same surface temperature. Star A has twice the radius as star B. How does the amount of energy radiated per unit area from the surface of A compare to surface of B?
Answer:
1/4
Explanation:
Total radiation energy depends on temperature, so \(E_A=E_B\)
Radius \(R_A=2R_B\)
so the surface area \(S_A=4 S_B\)
energy per unit area
\(\frac{E_A}{S_A}=\frac{1}{4}\frac{E_B}{S_B}\)
A boat on a river is moving with a steady speed. The engine is running.
What would happen if the engine was turned off?
If the engine of a boat on a river is turned off while the boat is moving with a steady speed, several things would happen Loss of propulsion,Drifting,Loss of steering control and Potential hazards.
Loss of propulsion: Without the engine running, the boat would lose its power source for propulsion. The boat would gradually slow down and eventually come to a stop unless other external forces, such as currents or wind, continue to move it.
Drifting: Once the boat comes to a stop, it would start to drift with the current of the river or be affected by wind forces. The direction and speed of the drift would depend on the strength and direction of the current or wind.
Loss of steering control: When the engine is turned off, the boat's steering mechanism, such as a rudder, would also lose power. Without the ability to steer, the boat would follow the course determined by the river's current or the wind direction.
Potential hazards: Depending on the surroundings and the current conditions, there could be potential hazards for a boat that is no longer under power. These hazards might include other vessels, obstacles, shallow areas, or strong currents. The boat's crew would need to take appropriate actions to ensure the safety of the boat and its occupants.
It's important to note that the specific behavior of the boat after the engine is turned off can vary depending on factors such as the size and design of the boat, the strength and direction of the current, and the presence of wind or other external forces.
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If Scobie could drive a Jetson's flying car at a constant speed of 160.0 km/hr across oceans and space, approximately how long would Scoobie take to drive around Earth's equator in days
Scobie will take 10 days to drive around Earth's equator.
To calculate the time that takes Scobie to drive around Earth's equator we need to find the distance, which is given by the equation of a circumference:
\( d = 2\pi r \)
Where:
r: is the Earth's radius = 6371 km
Then, the distance is:
\( d = 2\pi r = 2\pi*6371 km = 40030.2 km \)
Now, if we divide the above distance by the speed of the car we can find the time:
\( t = \frac{d}{v} = \frac{40030.2 km}{160.0 km/h} = 250.2 h*\frac{1 d}{24 h} = 10 d \)
Therefore, Scobie will take 10 days to drive around Earth's equator.
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