An object moving with constant velocity has an acceleration of zero and a net force acting on it also zero, as forces are balanced and it is in equilibrium according to Newton's second law.
Explanation:If an object moves with a constant velocity, the acceleration of the object is zero. This is because acceleration is defined as the change in velocity over time, and if the velocity is constant, there is no change. Therefore, according to Newton's second law, which states F = ma (force equals mass times acceleration), if the acceleration is zero, the net force acting on the object must also be zero. This scenario is true for all objects in equilibrium, whether they are moving at constant velocity or are stationary relative to Earth. Objects in equilibrium have net forces that are balanced, resulting in no acceleration.
As for the net force acting on the object, it would also be zero. When an object travels at constant velocity, it is not accelerating, which means the forces acting on it are in equilibrium—the sum of all forces equals zero. This could include situations such as a car moving at a steady speed on a flat road where the engine's thrust equals the frictional forces opposing it.
Which type of graph is best to show the percentage of gases in Earth's atmosphere?
circle graph
Explanation:
Which two elements have similar characteristics?
A: neon and fluorine
B: chlorine and sulfur
C: fluorine and chlorine
D: oxygen and chlorine
E: argon and fluorine
Which of the following statements about the tundra is not true? a. The tundra has limited vegetation. b. There are two main categories of the tundra biome. c. Temperatures in the tundra can get below -20° F. d. The tundra biome does not support animal life. Please select the best answer from the choices provided A B C D
The false statement about the tundra is D. The tundra biome does indeed support animal life, including polar bears, caribou, and mountain goats, among others.
The correct answer to which of the following statements about the tundra is not true is D. The tundra biome does not support animal life is the false statement. Although the tundra environment is extreme, the biodiversity, while limited, does include a variety of animal life that has adapted to the harsh conditions.
Arctic tundra and alpine tundra are the two main categories of the tundra biome. The arctic tundra can go below -20° F and has limited vegetation, primarily mosses, lichens, short grasses, and some flowering plants. The alpine tundra is found at high altitudes with a longer summer growing season compared to the arctic tundra.
Animals such as polar bears, caribou, and mountain goats, as well as birds that migrate to the arctic tundra in the summer, are adapted to survive in the tundra biome. These animals have special adaptations to handle the cold temperatures and scarce resources found in the tundra.
How to find initial velocity in projectile motion problems when you are not given the vlocity?
Sung rolls a marble off the edge of a table. She rolls the marble off the same table again but at a different speed. Which best describes the projectile motion of the marble as it leaves the table at the two different speeds?
Answer: Correct answer is "The marbles horizontal velocity changes, but its vertical velocity does not."
Explanation: I have had the same question on my High School Physics test, and got the answer correct.
What three things help determine ecoregions
Name two things that reduce friction
Answer:
Reducing the normal force against the ground will reduce the amount of friction force, you can do this reducing the weight or mass of the object that you want to move.
Smoothing down the surface, for example, on a wet floor, the water fills some of the little imperfections in the ground, so the coefficient of dynamic friction in the ground is reduced.
You've always wondered about the acceleration of the elevators in the 101 story-tall empire state building. one day, while visiting new york, you take your bathroom scale into the elevator and stand on them. the scales read 130 lb as the door closes. the reading varies between 110 lb and 160 lb as the elevator travels 101 floors. part a what is the magnitude of the acceleration as the elevator starts upward?
The acceleration of the elevator can be calculated using Newton's second law of motion and the recorded changes in weight on the scale while standing in the elevator. When the elevator begins its ascent, the apparent weight increases due to the increased force needed to move upward, which can be converted to calculate the acceleration.
Explanation:The subject of your question involves understanding the physics concepts of acceleration and force. When standing in an elevator that is accelerating upwards, the scale reads a higher weight because the scale must exert more force to move you upwards, resulting in a feeling of being heavier.
To determine the acceleration of the elevator, we first need to comprehend the concept of net force, which in this scenario, is the difference between your apparent weight (the reading on the scale) and your actual weight. From Newton's second law, we know that net force is equal to mass times acceleration (F=ma).
So, if your weight on the scale varies between 110 lb and 160 lb, then the difference is 50lb. Converting this weight difference to force (gravity being ~9.8 m/s^2), we get approximately 222.41 Newtons.
Given that your regular weight is 130lb, converting this to mass gives us approximately 59 kg (1lb ≈ 0.453592 kg). So, we can now find the acceleration (a) using the formula a = F/m. Substituting the values, we get the acceleration as approximately 3.76 m/s² as the elevator starts upward. Note that this is an ideal value, actual acceleration would be less due to factors like air resistance, friction, etc.
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A 3.75 kg ball is lifted from the floor to a height of 1.5 m above the floor. What is its increase in potential energy?
Will light pass through a pair of polaroids when the axes are aligned? when the axes are at right angles to each other? why?
Name the three basic parts of a spiral galaxy and describe what is found in each.
Give the reason why all atoms have total charge of zero
Which element does not contain any neutrons ?
Which hypothetical scenario would result in the moon not having different phases?
Approximately what percentage of the water on earth is freshwater (liquid and solid)
Most lakes _____. contain saltwater are found near the equator were formed from melting glaciers are recharged by evaporation
Formed from melting glaciers
Most lakes are formed from melting glaciers.
Explanation;The lakes that are formed from melted glaciers are called glacial lakes. These lakes are formed as a result of erosion of the land by a glacier, and which then melts and fills the space created.Glacial erosion are responsible for creation of depressions and holes on the surface of the earth. When these depressions are filled with water they result in formation of lakes. These glacial melt water act a steady source of water for communities all over the world.4: in the cabin of a jetliner that cruises at 600 km/h, a pillow drops from an overhead rack. does a passenger walking down the aisle have to worry about the pillow slamming into her and knocking her over? explain
Answer:
To answer this question, we need to recur to the law of inertia.
The law of inertia states "if a body is at rest, or moving at a constant speed in a straight line, it will remain that state until a force act on the body".
So, in this case, the jet is traveling at a speed of 600 km/h, which means that all bodies inside the jet are also traveling at this speed, because they are in it.
Also, if we apply the law of inertia here, imagine that somebody falls from the jet, that person will travel at a speed near to 600km/h, if we discard the wind, following a constant trajectory.
The same phenomenon happens with the pillow, when it falls from an overhead rack, if the passanger is passing under the trajectory of the pillow, it will slam her, because due to intertia, the pillow will continue its movement until something stop it, which can be the person's body.
Which of these would best represent the effect of an embargo on trade with China?
Answer:d
Explanation:
Look at the graph.
If the price of peaches increases, what can be expected?
Answer:
The correct answer is B) Fewer peaches will be sold.
Explanation:
Hope this helps! :)
If an atom has 7 valence electrons, how many electrons does it need to gain to achieve a stable electron configuration?
Food webs - transferring energy and matter from one level to another. Here you see four food webs. One or more are incorrect. Which food web(s) show the correct sequence of organisms, from start to top level consumer?
A) C Eliminate
B) D
C) B and C
D) A and B
So, what's the answer ?
An object is launched with an initial velocity of 50.0 m/s at a launch angle of 36.9∘ above the horizontal. part a determine x-values at each 1 s from t = 0 s to t = 6 s
Explanation :
It is given that :
Initial velocity, [tex]v_0=50\ m/s[/tex]
Launching angle, [tex]\theta_0=36.9^0[/tex]
The motion followed by this object is called as its projectile motion.
the horizontal component is given by:
[tex]x_t=x_0+v_0cos\theta_0t[/tex].......(1)
Put all values in equation (1)
For t = 0, [tex]x_t=0[/tex]
For t = 1 s, [tex]x_1=0+50\ cos(36.9)\ 1=39.95\ m[/tex]
For t = 2 s, [tex]x_2=0+50\ cos(36.9)\ 2=79.9\ m[/tex]
For t = 3 s, [tex]x_3=0+50\ cos(36.9)\ 3=119.85\ m[/tex]
For t = 4 s, [tex]x_3=0+50\ cos(36.9)\ 4=159.8\ m[/tex]
For t = 5 s, [tex]x_4=0+50\ cos(36.9)\ 5=199.75\ m[/tex]
For t = 6 s, [tex]x_3=0+50\ cos(36.9)\ 6=239.7\ m[/tex]
Hence, this is the required solution.
Give several examples of an object's motion is which a great distance is traveled but the displacement is zero
You slide a box of books at constant speed up a 30° ramp, applying a force of 200 n directed up the slope. the coefficient of sliding friction is 0.18. (a) how much work have you done when the box has risen 1 m vertically? (b) what's the mass of the box?
The box is moving at constant speed, so acceleration=0 and the equation of the forces acting on the box along the ramp becomes:
[tex]F-F_f-W_p = 0[/tex] (1)
where F=200 N is the force that pushes the box, Ff is the frictional force, and Wp is the component of the weight parallel to the ramp.
The frictional force can be written as
[tex]F_f = \mu m g cos \theta[/tex]
where [tex]\mu=0.18[/tex] is the coefficient of friction, m is the mass of the box, [tex]g=9.8 m/s^2[/tex] is the frictional force and [tex]\theta=30^{\circ}[/tex] is the angle of the ramp.
The component of the weight along the ramp is
[tex]W_p = mg sin \theta[/tex]
Substituting this into eq.(1), we have
[tex]F-\mu mg cos \theta- mg sin \theta=0[/tex]
and we can find the mass of the box:
[tex]m=\frac{F}{\mu g cos \theta + g sin \theta}=\frac{200 N}{(0.18)(9.8 m/s^2)(cos 30^{\circ})+(9.8 m/s^2)(sin 30^{\circ})}=31.1 kg[/tex]
Now we can also find the work done on the box. this is given by the gain in potential energy of the box, since there is no change in kinetic energy (the speed of the box is constant). Since the box has risen vertically by 1 m, the gain in potential energy (and the work done) is
[tex]W=\Delta U=mg \Delta h=(31.1 kg)(9.8 m/s^2)(1 m)=304.8 J[/tex]
So, the two answers are
(a) 304.8 J
(b) 31.1 kg
The physics concept of work and forces were used to answer this question. The work done was calculated to be 200 Joules and the mass was determined using the given force, angle, and coefficient of friction.
Explanation:Let's solve this problem using the concepts of physics. First, we need to find how much work was done. Work done (W) = Force (F) x Distance (d) x cos (theta), where theta is the angle between the force and the direction of displacement. Here, as the force was directed up the slope and the box moved up the slope too, the angle (theta) is 0 degrees. Cos 0 = 1, so the equation reduces to W = F x d.
Given, Force (F) = 200 N and Distance (the vertical height) (d) = 1 m (Remember, the work done is equal to the force multiplied by the distance moved in the direction of the force). So, the work done = 200 N x 1 m = 200 Joules.
Next, we need to determine the mass of the box. The forces acting on the body will be gravity (mg), friction (fr) and the applied force (F). According to the information given, the box moved up the slope at a constant speed. This means the net force acting on the box should be zero. Therefore, F - fr - mg*sin(theta) = 0. Given is the coefficient of friction as 0.18, so fr = mu*N = mu*mg*cos(theta), where N = mg*cos(theta) is the normal force. Substituting fr into the net force equation, we can solve for mass (m) = F/(g*(mu*cos(theta) + sin(theta))). Substituting the known values, we get the mass of the box.
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To obtain the sine of 35 degrees, use ________.
Far from any other masses, two masses, m1 and m2, are interacting gravitationally. The value for the mass of m1 suddenly doubles. What happens to the value of the gravitational force that mass m2 exerts on mass m1?
The air is less dense at higher elevations, so skydivers reach a high terminal speed. The highest recorded speed for a skydiver was achieved in a jump from a height of 39,000 m. At this elevation, the density of the air is only 4.3% of the surface density. Estimate the terminal speed of a skydiver at this elevation. Suppose that the density of air at sea level is 1.2 kg/m3, the mass of the skydiver is 75 kg , and the cross section area of the skydiver is 0.72 m2.
Answer: The terminal velocity of the skydiver is 281.30 m/s.
Explanation:
Mass of the sky diver = 75 kg
Density of the air at height of 39,000 m =4.3% of the surface density of the air
= [tex]1.2 kg/m^3\times \frac{4.3}{100}=0.0516 kg/m^3[/tex]
Cross sectional area of the sky diver , A = [tex]0.72 m^2[/tex]
Coefficient of drag ,C = 0.5
The terminal speed is given by the formula:
[tex]v_T=\sqrt{\frac{2mg}{c\rho A}}[/tex]
[tex]v_T=\sqrt{\frac{2\times 75 kg\times 9.8 m/s^2}{0.5\times 0.0516 km/m^3\times 0.72 m^2}}=281.30 m/s[/tex]
The terminal velocity of the skydiver is 281.30 m/s.
The terminal speed of the skydiver is [tex]\boxed{281.452{\text{ m/s}}}[/tex].
Further Explanation:
When a body falls from the sky it will accelerate continuously in the downward direction by the force of gravity results in increment in the acceleration. Another force is acting on the body is friction force due to air or the particles present in the atmosphere in the in the upward direction or opposite to the direction of motion also called drag force. The drag force is directly proportional to the square of the velocity of falling body. Therefore, with increase in the velocity the drag force also increasing which is proportional to equal to the square of the velocity of falling body. And time will come where the drag force is equal to the weight or the force of gravity results in zero resultant force means zero acceleration. At this point the speed become constant and reached its maximum velocity. This speed is called the terminal velocity.
Given:
The height of the jump is [tex]39000\text{ m}[/tex].
The density air is [tex]4.3\%[/tex] of the surface density.
The density of air at sea level is [tex]1.2{\text{ kg/}}{{\text{m}}^3}[/tex].
The mass of the skydiver is [tex]75\text{ kg}[/tex].
The cross sectional area of the skydiver is [tex]0.72{\text{ }}{{\text{m}}^2}[/tex].
Concept:
The expression for the terminal speed is:
[tex]\fbox{\begin\\{v_t} =\sqrt {\dfrac{{2mg}}{{\rho A{C_d}}}}\end{minispace}}[/tex] …… (1)
Here, [tex]{v_t}[/tex] is the terminal speed, [tex]m[/tex] is the mass of the skydiver, [tex]g[/tex] is the acceleration due to gravity, [tex]\rho[/tex] is the density, [tex]A[/tex] is the cross sectional area and [tex]{C_d}[/tex] is the drag coefficient.
The term [tex]g[/tex] is called acceleration due to gravity and it has a constant value of [tex]9.81{\text{ m/}}{{\text{s}}^2}[/tex] and the drag coefficient of a human in free fall is [tex]0.5[/tex].
The density of the air at the altitude of [tex]39000\text{ m}[/tex] is:
[tex]\rho=4.3\%\times{\rho _{air}}[/tex]
Here,[tex]\rho[/tex] is the density of the air at the altitude of [tex]39000\text{ m}[/tex] and [tex]{\rho _{air}}[/tex] is the surface density of the air.
Substitute [tex]1.2{\text{ kg/}}{{\text{m}}^3}[/tex] for [tex]{\rho _{air}}[/tex] in the above equation.
[tex]\begin{aligned}\rho&=4.3\%\times1.2{\text{ kg/}}{{\text{m}}^3}\\&=\frac{{4.3}}{{100}}\times1.2{\text{ kg/}}{{\text{m}}^3}\\&=0.0516{\text{ kg/}}{{\text{m}}^3}\\\end{aligned}[/tex]
Substitute [tex]0.0516{\text{ kg/}}{{\text{m}}^3}[/tex] for [tex]\rho[/tex], [tex]75\text{ kg}[/tex] for [tex]m[/tex], [tex]9.81{\text{ m/}}{{\text{s}}^2}[/tex] for [tex]g[/tex], [tex]0.72{\text{ }}{{\text{m}}^2}[/tex] for [tex]A[/tex] and [tex]0.5[/tex] for [tex]{C_d}[/tex] in equation (1).
[tex]\begin{aligned}{v_t}&=\sqrt{\frac{{2\left({75{\text{ kg}}}\right)\left( {9.81{\text{ m/}}{{\text{s}}^2}} \right)}}{{\left({0.0516{\text{ kg/}}{{\text{m}}^3}}\right)\left({0.72{\text{}}{{\text{m}}^2}} \right)\left( {0.5} \right)}}}\\ &=\sqrt{\frac{{1471.5}}{{0.018576}}}{\text{ m/s}}\\&=\sqrt{79215.116}{\text{ m/s}}\\&=281.452{\text{ m/s}}\\ \end{aligned}[/tex]
Therefore, the terminal speed of the skydiver is [tex]\boxed{281.452{\text{ m/s}}}[/tex].
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Answer Details:
Grade: High school
Subject: Physics
Chapter: Kinematics
Keywords:
Air, less, dense, higher, elevation, skydivers, reach high, terminal speed, recorded, achieved, jump, height, 39,000 m, density, air, 4.3%, surface density, estimate, sea, level, , mass, 75 kg, cross section, area, , 281.452 m/s.
You push an object 0.032 m by exerting 0.010 N in force. The work is the force times the distance. How much work have you done expressed in ergs?
A. 3200 ergs
B. 0.00032 ergs
C. 0.32 ergs
D. 32 ergs
Answer:
Work done, W = 3200 ergs
Explanation:
It is given that
Force acting on the object, F = 0.01 N
Distance covered by the object, d = 0.032 m
Let W is the work done by the object. The product of force and the distance is called the work done by the object. Mathematically, it is given by :
[tex]W=F\times d[/tex]
[tex]W=0.01\ N\times 0.032\ m[/tex]
W = 0.00032 Joules
Since, [tex]1\ joule=10^7\ ergs[/tex]
This gives, work done by the object, W = 3200 ergs. So, the correct option is (a).
Who Tried to take credit for newton's 500 page book on motion?
A stone is thrown upward from ground level. the initial speed is 352 feet per second. in how many seconds will the stone hit the ground?
Final answer:
Use the kinematic equation for projectile motion, S = ut + 0.5at², to find the time it takes for a stone thrown upward at 352 feet per second to hit the ground considering the acceleration due to gravity as 32.2 feet per second squared. Calculate the time to the peak, and then double it to find the total time.
Explanation:
The student has asked about the time it will take for a stone thrown upward at an initial speed of 352 feet per second to hit the ground. To find the time it takes for the stone to hit the ground, we can use the kinematic equations of motion that describe the motion of projectiles under the influence of gravity. Since the acceleration due to gravity is 32.2 feet per second squared downward, we can use the equation S = ut + 0.5at², where S is the displacement (which would be zero, as the stone returns to the ground level), u is the initial velocity, a is the acceleration due to gravity, and t is the time.
We rearrange the equation to solve for t. However, we need to consider that the stone will take as much time to reach the highest point as it will take to fall back down from that point. Therefore, we calculate the time to rise to the peak using the equation u = gt, and then double it. Plugging in the values, we get t = u/g, which gives us t = 352/32.2, and doubling this time gives us the total time to hit the ground.