Answer:
Generally, minors are not allowed to engage in contracts, except when minors seek basic necessities. Basic necessities include food, shelter and clothes (normal clothes, not designer clothes). If a minor engages in any contract regarding basic necessities, it is considered a valid contract. If a minor engages in contracts to purchase anything else not considered a basic necessity, the contracts are voidable.
A high diver of mass 74.0 kg jumps off a board 9.00 m above the water. If his downward motion is stopped 2.50 seconds after he enters the water, what average upward force did the water exert on him
Answer:
1120 N
Explanation:
The velocity with which he hits the water can be found with kinematics:
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-9.00 m)
v = -13.3 m/s
Or it can be found with conservation of energy.
PE = KE
mgh = ½ mv²
v = √(2gh)
v = √(2 × -9.8 m/s² × -9.00 m)
v = -13.3 m/s
Sum of forces on the diver after he hits the water:
∑F = ma
F − mg = m Δv/Δt
F − (74.0 kg) (9.8 m/s²) = (74.0 kg) (0 m/s − (-13.3 m/s)) / (2.50 s)
F = 1120 N
Title: Ohm’s Law using PhET Simulation The fundamental relationship among the three important electrical quantities current, voltage, and resistance was discovered by Georg Simon Ohm. The relationship and the unit of electrical resistance were both named for him to commemorate this contribution to physics. One statement of Ohm’s law is that the current through a resistor is proportional to the voltage across the resistor and inversely proportional to the resistance. In this experiment you will see if Ohm’s law is applicable by generating experimental data using a PhET Simulation:_____________.
Show us the experiment then
as the mass of the cart increases, the acceleration of the cart
Answer:
decrease
Explanation:
as the mass of the cart increase the acceleration of the cart decrease
Oil at 150 C flows slowly through a long, thin-walled pipe of 30-mm inner diameter. The pipe is suspended in a room for which the air temperature is 20 C and the convection coefficient at the outer tube surface is 11 W/m2 K. Estimate the heat loss per unit length of tube.
Answer:
1.01 W/m
Explanation:
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
From the above, we assumed that the pipe wall and the oil are in thermal equilibrium.
area of the pipe per unit length A = [tex]\pi r ^{2}[/tex] = [tex]7.069*10^{-4}[/tex] m^2/m
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130 = 1.01 W/m
The heat loss per unit length of tube should be considered as the 1.01 W/m.
Calculation of the heat loss:Since
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
Now
area of the pipe per unit length A should be
= πr^2
= 7.069*10^-4 m^2/m
Now
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130
= 1.01 W/m
hence, The heat loss per unit length of tube should be considered as the 1.01 W/m.
Learn more about heat here: https://brainly.com/question/15170783
An electromagnetic wave has a frequency of 1.0 < 1014 Hz. What is the
wavelength of the wave? Use the equation 1 = and the speed of light as
V
f
3.0 x 108 m/s.
O A. 3.3 x 10-8 m
B. 3.0 x 1022 m
C. 3.3 x 105 m
0
D. 3.0 x 10-6 m
Answer:
D. 3.0 × 10^-6 m
Explanation:
Wavelength is found by dividing the speed of light by the frequency:
λ = c/f = (3·10^8 m/s)/(1.0·10^14 Hz) = 3.0·10^-6 m
how does enormous energy get released from the sun
Answer:
By nuclear fission
Explanation:
The sun generates enormous energy through the process of nuclear fusion.
The core or the innermost part of the sun is characterized by high temperature and pressure. These two factors cause the separation of nuclei from electrons and the fusion of hydrogen nuclei to form a helium atom.
During the fusion process, energy is released.
Is overall loudness proportional to the number of speakers? F.E., if I use 4 identical sound devices instead of one, playing the same sound sample... Will they sound 4 times louder?
Answer:
a. No. The overall loudness is not proportional to the number of speakers.
b. No, it will not sound 4 times louder.
Explanation:
a. Loudness is the quality of how loud or soft the sound is. Since loudness is proportional to intensity and intensity proportional to the amplitude of the sound, the overall loudness of the sound would not be proportional to the number of speakers.
Since the sound from each speakers adds by superposition, their phases might be different but their amplitudes the same. So, the resultant wave from the four speakers has the same amplitude as the wave from one speaker. And, since loudness is proportional to amplitude, the loudness of the resultant wave is equal to the loudness of each individual speaker.
So, the loudness is not proportional to the number of speakers.
b. No, it will not sound 4 times louder.
This is because loudness is a function of amplitude.
Since the devices are identical and playing the same sound, their amplitudes are the same. They will thus add by superposition to generate a resultant wave which has an amplitude the same as each individual wave.
Since the amplitude of the resultant wave is the same and not an addition of each individual amplitude, the resultant wave does not sound 4 times louder.
helppppp me it's urgent please
the last one is the light will move in some manner not determined by the information here.
Light bends away from the normal, because it's moving from higher to lower refractive index.
Same bend-direction as when it goes from water into air.
A light, rigid rod is 51.4 cm long. Its top end is pivoted on a frictionless horizontal axle. The rod hangs straight down at rest with a small, massive ball attached to its bottom end. You strike the ball, suddenly giving it a horizontal velocity so that it swings around in a full circle. What minimum speed at the bottom is required to make the ball go over the top of the circle?
Answer:
4.489 m/s
Explanation:
We are given;
Length of rod;L = 51.4 cm = 0.514 m
When the ball is at the top end, it's velocity will be zero and it's kinetic energy will be transformed to potential energy.
Thus; ½mv² = mgh
m will cancel out to give;
v² = 2gh
Where:
g is acceleration due to gravity = 9.8 m/s²
h is the maximum height of the ball which is equal to twice the length of the rod.
Thus, h = 2(0.514) = 1.028 m
v² = 2 × 9.8 × 1.028
v = √20.1488
v = 4.489 m/s
SPEAR is a storage ring at the Stanford Linear Accelerator which has a circulating beam of electrons that are moving at nearly the speed of light (2.998 108 m/s). If a similar ring is about 80.0 m in diameter and has a 0.59 A beam, how many electrons are in the beam
Answer:
n = 3.1x10¹²
Explanation:
To find the number of electrons we need to find first the charge (q):
[tex] I = \frac{q}{\Delta t} \rightarrow q = I*\Delta t [/tex] (1)
Where:
I: is the electric current = 0.59 A
t: is the time
The time t is equal to:
[tex]v = \frac{\Delta x}{\Delta t} \rightarrow \Delta t = \frac{\Delta x}{v}[/tex] (2)
Where:
x: is the displacement
v: is the average speed = 2.998x10⁸ m/s
The displacement is equal to the perimeter of the circumference:
[tex] \Delta x = 2\pi*r = \pi*d [/tex] (3)
Where d is the diameter = 80.0 m
By entering equations (2) and (3) into (1) we have:
[tex]q = I*\Delta t = I*\frac{\Delta x}{v} = \frac{I\pi d}{v} = \frac{0.59 A*\pi*80.0 m}{2.99 \cdot 10^{8} m/s} = 4.96 \cdot 10^{-7} C[/tex]
Now, the number of electrons (n) is given by:
[tex] n = \frac{q}{e} [/tex]
Where e is the electron's charge = 1.6x10⁻¹⁹ C
[tex] n = \frac{q}{e} = \frac{4.96 \cdot 10^{-7} C}{1.6 \cdot 10^{-19} C} = 3.1 \cdot 10^{12} [/tex]
Therefore, the number of electrons in the beam is 3.1x10¹².
I hope it helps you!
Which option describes the most cost-effective way for a skateboard company to purchase plywood?
- a pallet of 100 sheets for $1,000
- a pallet of 200 sheets for $1,800
- a pallet of 500 sheets for $3,500
- a pallet of 1,000 sheets for $8,000
Answer:
a pallet of 500 sheets for $3,500
Explanation:
just did the test
Using the formula W = mg, how many milliliters of water with a density of 1g/mL are required to weigh 0.75 newtons and g = 9.81 m/s2? Round to the nearest tenth. (Note: The mass will be in kg in your answer, thus convert to g and then to mL.) Answer: mL
Given that,
Weight = 0.75 N
Acceleration due to gravity = 9.81 m/s²
Density of water = 1 g/ml
We need to calculate the volume of water
Using formula of weight
[tex]W=mg[/tex]
[tex]W=\rho Vg[/tex]
[tex]mg=\rho\times V\times g[/tex]
Where, V = volume
g = acceleration
m = mass
Put the value into the formula
[tex]0.076\times 9.8\times10^3=1\times V\times9.8[/tex]
[tex]V= \dfrac{0.076\times 9.8\times10^3}{1\times9.8}[/tex]
[tex]V=76\ ml[/tex]
Hence, The volume of water is 76 mL.
Answer:
76.5mL
Explanation:
w = mg
0.75N = m * 9.81
m = 0.75 / 9.81 = 0.0765 kg
The mass in grams is 0.0765 * 1000 = 76.5g
At a density of 1 g/mL,
76.5g ÷ 1g/ml = 76.5mL
Los primeros vuelos de prueba del transbordador espacial usaban un "planeador" (con masa de 980 kg incluyendo el piloto). Después de un lanzamiento horizontal a 480 km/h a una altitud de 3500 m, el planeador finalmente aterrizaba con una rapidez de 210 km/h ¿Cuál era la fuerza promedio que la resistencia del aire ejercía sobre él cuando entraba con un ángulo de planeo constante de 12° con respecto a la Tierra?
Answer:
La fuerza promedio que ejerce la resistencia del aire sobre el planeador cuando ingresa con un ángulo de planeo constante de 12 ° con respecto a la Tierra es 9613.8 N
Explanation:
Los parámetros dados son;
Masa del planeador = 980 kg.
Velocidad de lanzamiento horizontal = 480 km / h
Altitud de lanzamiento = 3500 m.
Velocidad de aterrizaje final = 210 km / h
Ángulo de deslizamiento, θ = 12 °
La ecuación para el equilibrio de fuerzas.
L · cos (θ) + D · sin (θ) = W
L · sin (θ) + D · cos (θ)
Dónde:
L = fuerza de elevación
D = fuerza de arrastre
W = Peso del "planeador"
W = Masa del "planeador" × Aceleración debido a la gravedad = 980 kg × 9.81 m / s²
W = 9613.8 N
Por lo tanto, tenemos;
L · cos (12) + D · sin (12) = 9613.8 ,,,,,,,,,,,,, (1)
L · sin (12) = D · cos (12) ,,,,,,,,,,,,, (2)
De la ecuación (2), tenemos;
L = D · (cos (12)/sin (12))
Sustituyendo el valor de L = D · (cos (12) / sin (12)) en la ecuación (1), se obtiene;
D · (cos (12)/sin (12)) · cos (12) + D · sin (12) = 9613.8
D = 9613.8/((cos (12)/sin (12)) · cos (12) + sin (12)) = 1998.82 N
L = D · (cos (12)/sin (12)) = 1998.82 · (cos (12)/sin (12)) = 9403.72 N
Por lo tanto, la fuerza da la fuerza promedio que la resistencia del aire ejerce sobre el planeador cuando ingresa con un ángulo de planeo constante de 12 ° con respecto a la Tierra;
F = √ (L² + D²) = √ (9403.72² + 1998.82²) = 9613.8 N.
An airplane on a runway has an initial speed of 15 meters/second. What is its final velocity when it travels 600 meters on the runway with a constant acceleration of 6 meters/second2?
Answer:
86.17m/s
Explanation:
Since the motion is of constant acceleration, we can use one of the equations of motion as follows;
v² = u² + 2as
Where;
v = final velocity of the airplane
u = initial velocity of the airplane
s = distance covered by the plane
a = acceleration
From the question,
u = 15m/s
s = 600m
a = 6m/s²
Substitute these values into the equation above as follows;
v² = 15² + 2(6)(600)
v² = 225 + 7200
v² = 7425
v = √7425
v = 86.17m/s
Therefore, the final velocity of the airplane is 86.17m/s
A car is parked at rest on a horizontal road. The upward force of the road on the car (the normal force) is the same size as the downward pull of gravity A car is parked at rest on a horizontal road. The upward force of the road on the car (the normal force) is the same size as the downward pull of gravity Because of Newton's first law. Because they are an action/reaction pair. Both of the above. None of the above. Some other reason.
Answer:
Option A - Because of Newton's first law of motion
Explanation:
This is as a result of Newton's first law of motion.
Newton's first law of motion which is also referred to as the law of inertia states that; an object at rest will remain at rest and an object in motion will remain in motion with the same speed and in the same direction unless it's acted upon by an unbalanced force.
In this question, the car is at rest and it doesn't move because the upward force of the car is balanced by the downward pull of gravity. It can only move when an external force is applied to it. This fulfills Newton's first law of motion.
If a car is travelling with a positive velocity, and after it collides with another car moving with a negative velocity, it is still moving with a positive velocity, in which direction could the second car be moving? (Select all that apply)
Answer: In the direction of the first car.
Explanation:
Let the initial velocity of the first car be equal to U1 and the initial velocity of the second car be equal to U2
Given that the car is travelling with a positive velocity, and after it collides with another car moving with a negative velocity, the momentum before collision will be
M1U1 - M2U2
If the first car is still moving with a positive velocity, the second car will also be moving in the direction of the first car if the collision is inelastic collision.
Therefore, the direction of the second car will be positive direction if the collision is inelastic collision.
Show all work and answer all 4 parts. Projectile motion. 20 points. Thank you.
Answer:
i took g = 9.8m/s
A. 1.16secs
B. 2.32secs
C. 6.57m
D. 57.91m
Explanation:
A. How long does the Missile take to reach ot peak?
Time taken (t) =( U²Sin (angle) )/g
u = initial velocity = 25m/s
angle given = 30°
g = acceleration due to gravity = 9.8m/s²
t = U² x Sin (angle) / g
t = 25² x Sin(30)/9.8
t = 1.61secs
B. How long is the missile in the air in total?
T = 2t
T = 2 x 1.61 = 2.32 secs
C. what maximum Height does the missile reach?
Maximum height = U²Sin²(angle) / 2gM.H =25² x Sin(30)² / 2 x 9.8M.H=6.57mMaximum height= 6.57mD. How far does the missile travel Horizontally?
Range = U²2Sin(angle)/gRange = 25² x 2 x Sin(30) / 9.8Range = 57.91mMt Everest is at the height of 8850 m; its base is at an elevation of 5400m. The temperature here drops at the rate of 1 degree per 100m. If temperature at the base is -5, what will be the temperature at the top?
Answer:
The temperature at the top is -39.5 degree
Explanation:
The height of Mt Everest = 8850 m
The base is at the elevation of = 5400 m
The drop in temperature = 1 degree per 100 m
At the base, the temperature = -5
Since every 100 meter rise in height results in the one degree drop of temperature. Therefore, first find the difference between the base and height. Multiply with negative one.
Difference in heights = 8850 – 5400 = 3450
The temperature at the top = (3450 / 100) + 5 = -39.5 degree
Since there is fall in temperature so we use negative sign.
A car is moving along an open road. Suddenly, a sheep walks into the road. The braking distance for the car is 35m. If the stopping distance is 50m, how far did the car travel before the driver put their foot on the brakes?
Answer:
15 m
Explanation:
The distance travelled by the driver before they put their foot on the brakes can be derived from the stopping distance and the braking distance of the car.
d = stopping distance - braking distance
Stopping distance = 50 m
Braking distance = 35 m
d = 50 - 35 m
d = 15 m
The distance travelled by the driver before they put their foot on the brakes is 15 m
Study the image of earths layer which statement correctly compares the thicknesses of earths layers
This question is incomplete because the options are missing; here is the complete question:
Study the image of the Earth's layer which statement correctly compares the thicknesses of earths layers
A. Earth’s mantle is thinner than its oceanic crust.
B. Earth’s outer core is thicker than its mantle.
C. Earth’s continental crust is thicker than its lithosphere.
D. Earth’s lithosphere is thinner than its asthenosphere.
The answer to this question is D. Earth’s lithosphere is thinner than its asthenosphere.
Explanation:
The image shows the different layers that are part of Earth, as well as the thickness of each layer, in kilometers. In this, the thickest layer is the Mantle that is almost 2900 kilometers; this is followed in thickness by the outer and the inner core.
Additionally, other layers such as the continental/oceanic crust, the asthenosphere, and the lithosphere that are near the surface are thinner. About this, it can be concluded the lithosphere is thinner than the asthenosphere because the lithosphere has a thickness of 100 km, while the asthenosphere thickness is 660km. This makes option D the correct.
whats the texture of copper wire
Copper is a reddish orange, soft metal with a bright metallic luster. It is an excellent conductor of heat and electricity. Copper surfaces when exposed to air gradually turns to a dull, brownish color.
A toaster oven draws 300.0 watts of power. If it is plugged into an outlet with a voltage of 115 volts, what current is in the toaster's circuit?
OA 3.45 10% amps
ОВ.
415 amps
Ос.
285 amps
OD
2.61 amps
Answer:
I = 2.61 A
Explanation:
It is given that,
Power of a toaster is 300 watts
It is plugged into an outlet with a voltage of 115 volts
We need to find the current in the toaster's circuit.
The electrical power is given by the formula as follows :
P = V I
I is electric current
[tex]I=\dfrac{P}{V}\\\\I=\dfrac{300}{115}\\\\I=2.61\ A[/tex]
So, 2.61 A of current is in the toaster's circuit.
Answer: D.
2.61 amps
Explanation: edmentum
Two teams are playing tug-of-war. Team A, on the left, is pulling on the rope with an effort of 5000 N. If the rope is moving at a constant velocity, how hard and in which direction is team B pulling?
A. 2500 N to the left
B. 5000 N to the right
C. 2500 N to the right
D. 5000 N to the left
Explanation:
If Team A is on the left, B is on the right
if the force is constant, it means that the effort applied is equal.
So Team B is pulling 5000N to the right.
A ball is moving up an inclined plane, when it's velocity changes from 15m/s to 7m/s in 3 seconds . What is the acceleration.
Explanation:
[tex]a = \frac{change \: in \: velocity}{change \: in \: time} [/tex]
[tex]a = \frac{v - u}{t} [/tex]
[tex]v = 7 {ms}^{ - 1} [/tex]
[tex]u = 15 {ms}^{ - 1} [/tex]
[tex]t = 3 \: seconds[/tex]
[tex]a = \frac{7 - 15}{3} [/tex]
[tex]a = \frac{ - 8}{3} [/tex]
[tex]a = - 2.67 \: {ms}^{ - 2} [/tex]
Question 1 of 10
2 Points
What is the SI unit for gravitational potential energy?
A. Newton
B. Joule
C. Meter
D. Watt
Answer:
Joule
Explanation:
As any unit of energy (kinetic, potential, etc.), the gravitational potential energy has SI units of Joule. This s the work done by a force of 1 Newton to displace an object 1 meter in the direction of application of the force.
If you travel at 60 miles per hour, how far will you travel in 2 hours?
A. not enough information
B. 45 miles
OC. 60 miles
D. 120 miles
If an object of 3.5 cm heitght is kept at a distance of 12 cm from the convex lens of focal length 8 cm. Find the position,nature and height of the height
Answer: The image is at a distance of 24 cm on the right of the lens, it will be an inverted, real image and has a height of 7 cm.
Explanation: To determine the position of the image of the object, use the Lens Equation, which is described by:
[tex]\frac{1}{f} = \frac{1}{u} +\frac{1}{v}[/tex]
where
f is focal length;
u is distance of the object from the lens;
v is distance of the image of the pbject from the lens;
Calculating the position of the image will be:
[tex]\frac{1}{v} =\frac{1}{f} - \frac{1}{u}[/tex]
[tex]\frac{1}{v} = \frac{1}{8} - \frac{1}{12}[/tex]
[tex]\frac{1}{v} = \frac{1}{24}[/tex]
v = 24 cm
Note that the distance of the image from the lens is positive. This means that the formed image is positioned at the right of the lens and is inverted.
For the height of the image, it can be used the Magnification (M) of an image:
M = [tex]\frac{h_{i}}{h_{o}} = \frac{v}{u}[/tex]
where:
[tex]h_{i}[/tex] is the height of the image
[tex]h_{o}[/tex] is the height of the object
For this image:
[tex]\frac{h_{i}}{3.5} = \frac{24}{12}[/tex]
[tex]h_{i} =[/tex] 2 * 3.5
[tex]h_{i}[/tex] = 7 cm
The image formed for this lens is 24 cm on the right of the lens, inverted related to the object and has height of 7 cm.
21. A toy car starts from rest and accelerates at 1.50 m/s’ [E] for 5.25 s. What is the final
velocity, V, of the car? Show your work.
Answer: The final Velocity, V, of the car is 7.9m/s
Explanation:
Given the following :
Toy car starts from rest, therefore,
Initial Velocity (u) = 0
Acceleration (a) = 1.5m/s^2 E for time, t = 5.25s
Final velocity (V) of the car =?
Using the motion equation:
v = u + at
Where v = final Velocity
u = Initial Velocity
a = acceleration
t = time
v = 0 + 1.5(5.25)
v = 0 + 7.875
v = 7.875m/s
v = 7.9m/s
Which two substances could be produced in a combustion reaction?
A. Carbon dioxide and oxygen
B. Carbon dioxide and water
C. Heptane and water
D. Heptane and oxygen
Answer: B. Carbon dioxide and water
Explanation: Combustion simply refers to burning. A combustion reaction may thus be explained as a chemical process which involves the reaction between a substance and an oxidizer usually oxygen gas. The combustion reaction in most cases involves the reaction between hydrocarbons which are combustible compounds and oxygen gas. The reaction between the combustible compound and oxygen gas is then followed by the release of energy, with the product being carbondioxid and water. Below is an example of combustion reaction.
Combustion the Methane CH4.
Here, CH4 reacts with oxygen gas(O2) to produce carbondioxide and water.
CH4 + 2O2 ------> CO2 + 2H2O
Answer:
Correct answer B
Explanation:
Just took quiz
What is the kinetic energy of a 200kg boat moving at 2.7m/s?
Answer:
729 JoulesExplanation:
Kinetic energy is 1/2mv²
where m is the mass of the object
v is the velocity
From the question
m = 200kg
v = 2.7m/s
Kinetic energy is
1/2 × 200 × 2.7²
= 729 Joules
Hope this helps you