Answer:
15.C Mucus reduces friction
what advantage does hovercraft have over a boar or a road vehicle?
Answer:
The advantages of Hovercraft:
They can travel over almost any non-porous surface.
They can operate to and from any unprepared beach or slipway.
They take fast, direct routes compared to a conventional marine vessel.
If you pitch a baseball with twice the kinetic energy you gave it in the
previous pitch, the magnitude of its momentum is
Answer:
the magnitude of momentum is √2≈ b
Explanation:
hope that helped
A car is moving at 10 m/s on a horizontal road with friction on a dry day. The car can travel around a traffic circle with a minimum radius of 4.8 meters. It rains and the car around a traffic circle with a minimum radius of 11.8 meters. What is the percentage of the coefficient of static friction on the rainy day compared to the dry day
Answer:
[tex]\mu_w=86\%[/tex]
Explanation:
From the question we are told that:
Velocity on Dry road [tex]V_d=10m/s[/tex]
Radius Dry [tex]r_d=4.8[/tex]
Radius wet [tex]r_w=11.8[/tex]
Generally the equation for coefficient of static friction on the dry day is mathematically given by
[tex]\mu mg=\frac{mv^2}{r_d}[/tex]
[tex]\mu g=\frac{v^2}{r_d}[/tex]
[tex]\mu_d 9.8=\frac{10^2}{4.8}[/tex]
[tex]\mu_d=2.125[/tex]
Generally the equation for the relationship between Radius & coefficient of static friction is mathematically given by
[tex]\frac{\mu_d}{\mu_w}=\frac{r_d}{r_w}[/tex]
[tex]\frac{\mu_w}{2.125}=\frac{4.8}{11.8}[/tex]
[tex]\mu_w=0.86[/tex]
Therefore
[tex]\mu_w=86\%[/tex]
In the chemical equation Zn+2HCI ZNCI+H the reaction are
Answer:
In the chemical equation Zn + 2HCL-> ZnCl2 + H2, the reactants are zinc and hydrochloric acid. In the chemical equation Zn + 2HCL-> ZnCl2 + H2, the reactants are zinc and hydrochloric acid.
Explanation:
this is the correct
Work and the Dot Product
A variable 1D force acts on an object of mass 2 kg, which is initially moving at 5 m/s to the right (along the positive x direction). The net force is given by:
F x = 20x2-10 i Newtons x
The force acts on the object as it displaced from x = 1 m to x = 4 m .
a) Findthespeedoftheobjectatx=4m.
b) Is there a gain or loss in kinetic energy or no loss in kinetic energy in the
displacement of the object? Explain.
Answer:
a) v_f = 5,06 m/s, b) GAIN in kinetic energy.
Explanation:
a) For this exercise we will use the relationship between work and kinetic energy
W = ΔK
Work is defined by
W = F. d
bold indicates vectors
the displacement is
d = x_f - x₀
d = 4 -1
d = 3i m
we calculate
W = 20 10⁻² 3 i.i
let's remember that
i.i = j.j = 1
i.j = 0
W = 6.0 10⁻¹ J
we substitute in the first equation
W = K_f - K₀
W = ½ m (v_f ² -v₀²)
v_f ² = [tex]\frac{2W}{m} + v_o^2[/tex]
let's calculate
v_f ² = 2 6.0 10⁻¹ /2 + 5²
v_f = √25.6
v_f = 5.06 m / s
b) we can see that the speed at the end of the movement is greater than the initial speed, therefore there is a GAIN in kinetic energy.
factors that favour mining in South Africa
Answer:
According to the data, factors influencing mining investment in South Africa's favour are the availability of labour and skills, the quality of the country's infrastructure, the quality of its geological database, and the State's environmental regulations.
If a person pulls back a rubber band on a slingshot without letting to go of it, what type of energy will the rubber band have? A. Potential energy B. Rotational energy C. Kinetic energy D. Translational energy
Answer:
Potential energy. Releasing it, the potential energy would convert into motion, kinetic energy.
Potential energy is when an object has some sort of potential eg. for motion such as in this example.
Answer:
A. Potential energy
Explanation:
Potential energy can be thought of as stored energy, which has the potential of becoming kinetic energy once it has been released.
What are 3 artificial and 2 natural sources of electromagnetic radiation?
Answer: its b bro
Explanation:
ajafa'jfbA'FJ
A kingfisher bird that is perched on a branch a few feet above the water is viewed by a scuba diver submerged beneath the surface of the water directly below the bird. Does the bird appear to the diver to be closer to or farther from the surface than the actual bird
Answer:
The bird appears farther
Explanation:
This is because as the light from the bird travels into the water which has a higher refractive index than air, light rays from the kingfisher bird bend towards the normal at the water surface and thus enter the eye of the scuba diver. Now, if we project the light rays from the eyes of the scuba diver into the air, we see that they appear to come from a point farther than that of the actual kingfisher bird perched on the branch.
So, the bird appears to the diver to be farther from the surface than the actual bird
1. What are Earth's natural climate cycles?
Find the weight of a man whose mass is 40 kg on earth.
(also
write complete data plus proper formula).
I am sure it help you with that much ☺️
Explanation:
pleasae give me some thanks please good morning sister
A compact car has a maximum acceleration of 3.0 m/s2 when it carries only the driver and has a total mass of 1300 kg. What is its maximum acceleration after picking up four passengers and their luggage, adding an additional 400 kg of mass?
Answer:
[tex]a_2=3.88m/s^2[/tex]
Explanation:
From the Question we are told that:
Initial Mass [tex]m_1=1300kg[/tex]
Final mass [tex]m_2=1300+400=>1700kg[/tex]
[tex]a_1=3.0m/s^2[/tex]
Generally the equation for Force is mathematically given by
[tex]F=m_1a_1[/tex]
[tex]F=1300*5[/tex]
[tex]F=6500N[/tex]
Generally the equation for Final acceleration is mathematically given by
[tex]F'=m_2*a_2[/tex]
[tex]a_2=\frac{6500}{1700}[/tex]
[tex]a_2=3.88m/s^2[/tex]
Two argon atoms form the molecule Ar2 as a result of a van der Waals interaction with U0 = 1.68×10-21 J and R0= 3.82×10 the frequency of small oscillations of one Ar atom about its equilibrium position.
Answer:
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
Explanation:
From the given information:
The elastic potential energy can be calculated by using the formula:
[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]
Making K the subject;
[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]
[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]
k = 2.3 × 10⁻² N/m
Now; the frequency of the small oscillation can be determined by using the formula:
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
where;
m = mass of each atom = 1.66 × 10⁻²⁶ kg
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
Jacie made a model to show the water cycle. The model she made is shown
below.
Which process in the model represents condensation?
A. As water vapor transfers heat to ice cubes, it forms clear droplets outside the
plastic wrap.
B. As water vapor gains heat from ice cubes, it forms clear droplets outside the
plastic wrap.
C. As water vapor transfers heat to ice cubes, it forms colored droplets inside the
plastic wrap.
D. As water vapor gains heat from ice cubes, it forms colored droplets inside the
plastic wrap
Answer:
option C
Explanation:
as water vapor transfer heat, colored drops are seen inside the wrap.
Question 2 of 32
A water-skier with a mass of 68 kg is pulled with a constant force of 980 N by
a speedboat. A wave launches him in such a way that he is temporarily
airbome while still being pulled by the boat, as shown in the image below.
Assuming that air resistance can be ignored, what is the vertical acceleration
that the water-skier experiences on his return to the water surface? (Recall
that g = 9.8 m/s2)
Rope Force
ODON
Weight
O A. - 18.1 m/s2
OB. - 15.6 m/s2
O C. -11.2 m/s2
OD. -9.8 m/s2
Answer:
OD. -9.8 m/s2
Explanation:
The only force vertical force that is acting on the skier is gravity and since its pulling him back it's a negative force down the y axis.
The food calorie, equal to 4186 J, is a measure of how much energy is released when food is metabolized by the body. A certain brand of fruit-and-cereal bar contains 160 food calories per bar.
Part A
If a 67.0 kg hiker eats one of these bars, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential energy?
Express your answer in meters.
Part B
If, as is typical, only 20.0 % of the food calories go into mechanical energy, what would be the answer to Part A? (Note: In this and all other problems, we are assuming that 100% of the food calories that are eaten are absorbed and used by the body. This is actually not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the rest are eliminated by the body. Metabolic efficiency varies considerably from person to person.)
Express your answer in meters.
Answer: 1 cal is 4.186 J, 1 kcal = 4186 J A : 1014 m , B 200 m
Explanation: A) Work done by climber is change in potential energy.
W = ΔEp = mgh = 67.0 kg· 9.81 m/s²· h = 160 kcal · 4186 J / kcal.
Solve h = 160 kcal · 4186 J / kcal /67.0 kg· 9.81 m/s² = 1014 m
B Energy is only 20 % : Then h = 0.20 ·160 kcal · 4186 J / kcal /67.0 kg· 9.81 m/s² = 200 m.
Actually, muscles also produce heat from most of the energy provided by food.
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery.
Complete question:
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.
Answer:
The value of its capacitance is 1.027 x 10⁻¹² F
Explanation:
Given;
area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²
separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m
voltage of the battery, V = 18 V
The value of its capacitance is calculated as;
[tex]C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F[/tex]
Therefore, the value of its capacitance is 1.027 x 10⁻¹² F
A two-slit interference experiment in which the slits are 0.200 mm apart and the screen is 1.00 m from the slits. The m = 1 bright fringe in the figure is 9.49 mm from the central bright fringe. Find the wavelength of the ligh
Answer:
λ = 1.90 10⁻⁶ m
Explanation:
The interference pattern for the two-slit case is
d sin θ = m λ
let's use trigonometry
tan θ = y / L
interference experiments angles are small
tan θ = sin θ /cos θ = sin θ
sin θ = y / L
we substitute
d y / L = m λ
λ = [tex]\frac{ d \ y}{m \ L}[/tex]
we calculate
λ = 0.2000 10⁻³ 9.49 10⁻³ / (1 1.00)
λ = 1.898 10⁻⁶ m
λ = 1.90 10⁻⁶ m
The wavelength of the light after calculation is find out to be λ = 1.90 *10⁻⁶ m
What is wavelength?
The distance between two successive troughs or crests is known as the wavelength. The peak of the wave is the highest point, while the trough is the lowest.The wavelength is also defined as the distance between two locations in a wave that have the same oscillation phase.
The interference pattern for the two-slit case is
d sin θ = m λ
let's use trigonometry
[tex]tan\theta=\dfrac{y}{L}[/tex]
interference experiments angles are small
[tex]sin\theta=\dfrac{y}{L}[/tex]
we substitute
[tex]\dfrac{dy}{L}=m\lambda[/tex]
[tex]\lambda=\dfrac{dy}{mL}[/tex]
we calculate
[tex]\lambda=\dfrac{0.2\times 10^{-3}\times 9.49\times 10^{-3}}{1\times 1}[/tex]
[tex]\lambda=1.90\times 10^{-6}\ m[/tex]
Hence the wavelength of the light after calculation is find out to be λ = 1.90 *10⁻⁶ m
To know more about wavelength follow
https://brainly.com/question/10728818
does net force stay the same when a massless pulley is replaced by a pulley with mass
12. A car is travelling at 30 m/s when the driver sees a red light in the distance and immediately applies the brakes. The car comes to a stop 1.5 s later. How far did the car move from the time the driver applied the brake to when it came to a stop?
Answer:
22.5 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 30 m/s
Time (t) = 1.5 s
Final velocity (v) = 0 m/s
Distance (s) =?
The distance to which the car move before stopping from the time the driver applied the brake can be obtained as follow:
s = (u + v)t/2
s = (30 + 0)1.5 / 2
s = (30 × 1.5) / 2
s = 45 / 2
s = 22.5 m
Thus, the car will move to a distance of 22.5 m before stopping from the time the driver applied the brake.
Two charged bees land simultaneously on flowers that are separated by a finite distance. For a few moments, the charged bees rest on the flowers. The charged bees both generate an electric field, and while the charged bees are resting on the flowers, the net electric field at some distance between them is zero.
Required:
Do the bees have the same or opposite signs of charge?
Answer:
the charge of the bees must be of the same sign
Explanation:
The electric field is given by the relation
E = k q / r²
This electric field has outgoing direction if the charge is positive and incoming towards the charge if it is negative.
The force generated by this field on a test charge is
F = q E
Since the charge is a scalar, the direction of the force is the same as the electric field.
In this case the two flowers are at a certain distance and the two charged bees land on them, so the force on a test charge is the vector sum of the force that each bee creates, so that this force is subtracted from the two bees must have the same charge sign.
The force created by the bee on the left goes to the right and the force created by the bee on the right goes to the left, so the forces are subtracted,
Consequently the charge of the bees must be of the same sign
6. A transverse periodic wave on a string with a linear density of 0.200 kg/m is described by the following equation: y = 0.08 sin(469t – 28.0x), where x and y are in meters and t is in seconds. What is the tension in the string? A) 3.99 N B) 32.5 N C) 56.1 N D) 65.8 N
Answer:
T = 56.11 N
Explanation:
Given that,
The equation of a wave is :
y = 0.08 sin(469t – 28.0x),
where x and y are in meters and t is in seconds
The linear mass density of the wave = 0.2 kg/m
The speed of wave is given by :
[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]
Also,
[tex]v=\dfrac{\omega}{k}[/tex]
We have,
[tex]k=469\ and\ \omega=28[/tex]
Put all the values,
[tex]\dfrac{\omega}{k}=\sqrt{\dfrac{T}{\mu}}\\\\(\dfrac{\omega}{k})^2=\dfrac{T}{\mu}\\\\T=(\dfrac{\omega}{k})^2\times \mu[/tex]
Put all the values,
[tex]T=(\dfrac{469}{28})^2\times 0.2\\\\T=56.11\ N[/tex]
So, the tension in the string is 56.11 N.
the magnitude of the electrical force acting between a +2.4x10-8c charge and 1+1.8x10-6 charge that are separated by 1.008m is
Answer:
3.83×10¯⁴ N
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +2.4x10¯⁸ C
Charge 2 (q₂) = +1.8x10¯⁶ C
Distance apart (r) = 1.008 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
The magnitude of the electrical force acting between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × 2.4x10¯⁸ × 1.8x10¯⁶ / (1.008)²
F = 0.0003888 / 1.016064
F = 3.83×10¯⁴ N
Thus the magnitude of the electrical force acting between the two charges is 3.83×10¯⁴ N
A toddler weighs 10 kg and raises herself onto tiptoe (on both feet). Her feet are 8 cm long with each ankle joint being located 4.5 cm from the point at which her feet contact the floor. While standing on tip toe:
(a) what is the upward normal force exerted by the floor at the point at which one of the toddler's feet contacts the floor?
(b) what is the tension force in one of her Achilles tendons? (c) what is the downward force exerted on one of the toddler's
ankle joints?
Answer:
a.49 n
b. 63 n
c. 112 n
Explanation:
a.10 times 9.8 from gravity/2 = 49 n
b. 49n times 4.5/8-4.5 = 63 n
c 49n + 63 n = 112 n
A certain light truck can go around a flat curve having a radius of 150m with a max speed if 26.5. What is the max speed it can go around a curve having a radius of 76.0m
Answer:
The maximum speed is 18.86 m/s.
Explanation:
initial radius, r = 150 m
maximum speed, v = 26.5 m/s
new radius, r' = 76 m
Let the new maximum speed is v'.
The formula of the maximum speed is
[tex]tan\theta = \frac{v^2}{rg}[/tex]
So,
[tex]\frac{v'^2}{v^2}=\frac{r'}{r}\\\\\frac{v'^2}{26.5\times 26.5}=\frac{76}{150}\\\\v=18.86 m/s[/tex]
A motorboat embarks on a trip, heading downstream in a river in which the current flows at a rate of 1.5m/s. After 30.0 minutes, the boat has traveled a distance of 24.3 km downstream. How long will it take the boat to travel upstream to its original point of embarkation
Answer:
[tex]t=2413s[/tex]
Explanation:
From the question we are told that:
Velocity [tex]v=1.5m/s[/tex]
Time [tex]t=30min=>30*60=>1800[/tex]
Distance [tex]d=24.3km[/tex]
Generally the Newton's equation for Speed going down the stream is mathematically given by
[tex]v + u = \frac{d}{t}[/tex]
[tex]1.5+v=frac{24300}{1800}[/tex]
[tex]v=12m/s[/tex]
Therefore
[tex]v + u = \frac{d}{t}[/tex]
[tex]t=\frac{24300}{12-1.5}[/tex]
[tex]t=2413s[/tex]
44.7
When Xavier places his hands near a light bulb, he notices that certain areas around the light bulb are warmer than
others. Which best explains this?
The areas to the sides of the light bulb are warmest because of conduction,
O The areas to the sides of the light bulb are warmest because of convection,
The area directly above the light bulb is warmest because of conduction,
The area directly above the light bulb is warmest because of convection.
Save and Exit
Submit
Mark this and retum
Nex
Answer:
The area directly above the light bulb is warmest because of convection.
Explanation:
if all the sides of the bulb are equally close to the light source inside the bulb, all area of the bulb would be equally heated by conduction. however, convection heating mainly heats up the surface above the light source. in convection heating, the air above the surface of the light source get heated by the light source and expands, casuing it to be less dense and rise to the top of the bulb. colder denser air at the top of the bulb sink to the light source adn gain heat and expands, becoming less dense. this process repeats and the surface above the light source becomes the warmest due to convection heating
violations in a responsible manner in a democratic society?
Activity 5:
Ending
From your findings, what conclusions and recommendations can you make on the
issue of human rights violations to:
5.1
Government
Answer:
Kindly check explanation
Explanation:
The trampling and violation of human rights individuals, groups and corporate organizations is really alarming ad as such, the government who are charged to protect the right and interest of its citizen. In other to curtail the trending issues of human right violation, it is imperative if sensitization programmes could be organiz d in other to keep people informed of the various ways in which people's right may be trampled upon. With these education, the ignorance can be expunged leaving only those who genuinely decides to relate and threaten the right of his fellow country person.
The laws on human right violation should be reviewed and capital punishment metted on violators in other to send a strong warning to those who still nurture the intention.
Three ice skaters, numbered 1, 2, and 3, stand in a line, each with her hands on the shoulders of the skater in front. Skater 3, at the rear, pushes forward on skater 2. Assume the ice is frictionless.
Identify all the action/reaction pairs of forces between the three skaters. Draw a free-body diagram for skater 2, in the middle.
Answer:
Following are the responses to the given question:
Explanation:
The mass (mg) is downwards, the typical [tex]N_2[/tex] pressure is upward. This pushing force of [tex]\vec{F}_{3 \to 2}[/tex] is pushed forward by skater 3 on skater 2. (considered as rightward direction). The strength of the slater 2 in reply is the slater [tex]\vec{F}_{2\to 3}[/tex] Skater three moves to the left.
Its push force [tex]\vec{F}_{2\to1}[/tex] imparted to the skateboarder 2 in relation those above forces The force[tex]\vec{F}_ {1\to 2}[/tex] on skater 2 by skater 1 is as a response, to a left. Skater 2's free body system is as follows:
Coherent light with wavelength 597 nm passes through two very narrow slits, and theinterference pattern is observed on a screen a distance of 3.00{\rm m} from the slits. The first-order bright fringe is adistance of 4.84 {\rm mm} from the center of the central bright fringe.
For what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?
Answer:
The required wavelength is 1.19 μm
Explanation:
In the double-slit study, the formula below determines the position of light fringes [tex]y_m[/tex] on-screen.
[tex]y_m = \dfrac{m \lambda D}{d}[/tex]
where;
m = fringe order
d = slit separation
λ = wavelength
D = distance between screen to the source
For the first bright fringe, m = 1, and we make (d) the subject, we have:
[tex]d = \dfrac{(1) \lambda D}{y_1}[/tex]
[tex]d = \dfrac{ \lambda D}{y_1}[/tex]
replacing the value from the given question, we get:
[tex]d = \dfrac{ (597 \ nm )\times (3.00 \ m)}{4.84 \ mm} \\ \\ d = \dfrac{ (597 \ nm \times (\dfrac{1 \ m}{10^9\ nm}) )\times (3.00 \ m)}{4.84 \ mm(\dfrac{1 \ m}{1000 \ mm })} \\ \\ d = 3.7 \times 10^{-4} \ m[/tex]
In the double-slit study, the formula which illustrates the position of dark fringes [tex]y_m[/tex] on-screen can be illustrated as:
[tex]y_m = (m+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]
The value of m in the dark fringe first order = 0
∴
[tex]y_0 = (0+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]
[tex]y_0 = (\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]
making λ the subject of the formula, we have:
[tex]\lambda = \dfrac{2y_o d}{D} \\ \\ \lambda = \dfrac{2(4.84 \ mm) \times \dfrac{1 \ m}{1000 \ mm} (3.7 \times 10^{-4} \ m) }{3.00 \ m}[/tex]
[tex]\lambda = 1.19 \times 10^{-6} \ m ( \dfrac{10^6 \mu m }{1\ m}) \\ \\ \lambda = 1.19 \mu m[/tex]
