Answer:
The wheels turn 101.27 radians in that time.
Explanation:
First, we need to find the angular velocity of the tires. We use the following formula for this purpose:
Δv = rω
ω = Δv/r
where,
ω = Angular Velocity of Tires = ?
Δv = change in linear velocity = 5.33 m/s - 0 m/s = 5.33 m/s
r = radius of tire = 0.33 m
Therefore,
ω = (5.33 m/s)/(0.33 m)
ω = 16.15 rad/s
Now, the angular displacement covered by tires can be found out by using the general formula of angular velocity:
ω = θ/t
θ = ωt
where,
θ = angular displacement = ?
t = time = 6.27 s
Therefore,
θ = (16.15 rad/s)(6.27 s)
θ = 101.27 radians
Four cubes of the same volume are made of different materials: lead (density 11,300 kg/m3), aluminum (density 2700 kg/m3), wood (density 800 kg/m3), and Styrofoam (density 50 kg/m3). You place the cubes in a large container filled with water.Rank the buoyant forces that the water exerts on the cubes from largest to smallest.Rank from largest to smallest. To rank items as equivalent, overlap them.
Answer:
Lead > Aluminium > Wood > Styrofoam
Explanation:
Buoyant force is described by the Archimedes's Principle, which states that buoyant force is equal to the weight of the fluid displaced by the submerged object. By Newton's Laws, the buoyant force is represented by the following equation of equilibrium:
[tex]\Sigma F = F_{D} - W_{cube} = 0[/tex]
[tex]F_{D} = W_{cube}[/tex]
[tex]F_{D} = \rho_{cube} \cdot g \cdot V_{cube}[/tex]
Where:
[tex]\rho_{cube}[/tex] - Density of the cube, measured in kilograms per cubic meter.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]V_{cube}[/tex] - Volume of the cube, measured in cubic meters.
Let suppose that volume of the cube is known. Given that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the buoyant force is computed for each material:
Lead ([tex]\rho_{cube} = 11,300\,\frac{kg}{m^{3}}[/tex])
[tex]F_{D} = \left(11,300\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]
[tex]F_{D} = 110,819.1V_{cube}[/tex]
Aluminium ([tex]\rho_{cube} = 2,700\,\frac{kg}{m^{3}}[/tex])
[tex]F_{D} = \left(2,700\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]
[tex]F_{D} = 26478.9V_{cube}[/tex]
Wood ([tex]\rho_{cube} = 800\,\frac{kg}{m^{3}}[/tex])
[tex]F_{D} = \left(800\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]
[tex]F_{D} = 7845.6V_{cube}[/tex]
Styrofoam ([tex]\rho_{cube} = 50\,\frac{kg}{m^{3}}[/tex])
[tex]F_{D} = \left(50\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]
[tex]F_{D} = 490.35V_{cube}[/tex]
Therefore, the buoyant forces that the water exerts on the cubes from largest to smallest corresponds to: Lead > Aluminium > Wood > Styrofoam.
You are to drive to an interview in another town, at a distance of 310 km on an expressway. The interview is at 11:15 a.m. You plan to drive at 100 km/h, so you leave at 8:00 a.m. to allow some extra time. You drive at that speed for the first 100 km, but then construction work forces you to slow to 42.0 km/h for 42.0 km. What would be the least speed needed for the rest of the trip to arrive in time for the interview
Answer:
Explanation:
Time to cover first 100 km = 1 hour.
time remaining = 3.15 - 1 = 2.15 hour .
Time to cover next 42 km = 1 hour .
Time remaining = 2.15-1 = 1.15 hour.
Distance to be covered = 310 - 142
= 168 km
least speed needed = distance remaining / time remaining
= 168 / 1.15
= 146.08 km / h .
Which best describes earth's magnetic field lines?
Answer:
The field lines go out of Earth near Antarctica, enter Earth in northern Canada, and are not aligned with the geographic poles.
Explanation:
Answer: the field lines go out of earth near the north pole, enter earth in the south pole, and are not aligned with the geographic poles.
Explanation: just took the test on edge 2020.
A natural force of attraction exerted by the earth upon objects, that pulls
objects towards earth's center is called
A natural force of attraction exerted by the earth upon objects, that pulls objects towards earth's center is called Gravitational force .
An object on a number line moved from x = 15 cm to x = 165 cm and then
moved back to x = 25 cm, all in a time of 100 seconds.
What was the average velocity of the object?
Answer:
v_avg = 2.9 cm/s
Explanation:
The average velocity of the object is the sum of the distance of all its trajectories divided the time:
[tex]v_{avg}=\frac{x_{all}}{t}[/tex]
x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm
Then, x_all = 150cm + 140cm = 290cm
The average velocity is, for t = 100s
[tex]v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}[/tex]
hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s
Answer:
The average velocity of the object is 0.1cm/s
Explanation:
Given that the object travels from point 15cm to 165cm and back to 25cm within 100 seconds
The average velocity is calculated as thus.
Average Velocity = ∆D/t
Where ∆D represent the displacement.
The displacement is calculated as follows.
∆D = End point - Start Point.
From the question, the end and start point are 25cm and 15cm respectively.
Hence,
∆D = 25cm - 15cm
∆D = 10cm.
t = 100 seconds
So, Average Velocity = 10cm/100s
Average Velocity = 0.1cm/s
Hence, the average velocity of the object is 0.1cm/s
A traffic light is weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper cable makes angles of 41° and 63° with the horizontal. Calculate the tension in each of the three cables.
Answer:
T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N
Explanation:
This is a balance exercise where we must apply the expressions for translational balance in the two axes
∑ F = 0
Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º
decompose the tension of the two upper cables
cos 41 = T₁ₓ / T1
sin 41 = T₁y / T1
T₁ₓ = T₁ cos 41
T₁y= T₁ sin 41
for cable gold
cos 63 = T₂ / T₂
sin 63 = [tex]T_{2y}[/tex] / T₂
We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.
Let's start by analyzing the point where the traffic light meets the vertical cable
T₃ - W = 0
T₃ = W
T₃ = 200 N
now let's write the equations for the single point of the three wires
X axis
- T₁ₓ + T₂ₓ = 0
T₁ₓ = T₂ₓ
T1 cos 41 = T2 cos 63
T1 = T2 cos 63 / cos 41 (1)
y Axis
[tex]T_{1y}[/tex] + T_{2y} - T3 = 0
T₁ sin 41 + T₂ sin 63 = T₃ (2)
to solve the system we substitute equation 1 in 2
T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W
T₂ (cos 63 tan 41 + sin 63) = W
T₂ = W / (cos 63 tan 41 + sin 63)
We calculate
T₂ = 200 / (cos 63 tan 41 + sin 63)
T₂ = 200 / 1,2856
T₂ = 155.6 N
we substitute in 1
T₁ = T₂ cos 63 / cos 41
T₁ = 155.6 cos63 / cos 41
T₁ = 93.6 N
therefore the tension in each cable is
T₁ = 93.6 N
T₂ = 155.6 N
T₃ = 200 N
Suppose there is a uniform magnetic field, B, pointing into the page (so your index finger will point into the page). If the velocityof a proton is straight up(thumb pointing up) then RHR2 shows that the force points to the left. What would the direction of the force be if the velocitywere a) down b) to the rightc) to the leftd) into the pagee) out of the page
Answer:
Explanation:
We shall show all given data in vector form and calculate the direction of force with the help of following formula
force F = q ( v x B )
q is charge , v is velocity and B is magnetic field.
Given B = - Bk ( i is right , j is upwards and k is straight up the page )
v = v j
F = q ( vj x - Bk )
= -Bqvi
The direction is towards left .
a ) If velocity is down
v = - v j
F = q ( - vj x - bk )
= qvB i
Direction is right .
b ) v = v i
F = q ( vi x - Bk )
= qvB j
force is upwards
c ) v = - vi
F = q ( -vi x - Bk )
= -qvBj
force is downwards
d ) v = - v k
F = q( - vk x -Bk )
= 0
No force will be created
e ) v = v k
F = q( vk x -Bk )
= 0
No force will be created
An atom with 19 protons and 18 neutrons is an
The potential in a region between x = 0 and x = 6.00 m is V = a + bx, where a = 10.6 V and b = -4.90 V/m. (a) Determine the potential at x = 0. 10.6 Correct: Your answer is correct. V Determine the potential at x = 3.00 m. V (b) Determine the magnitude and direction of the electric field at x = 0. magnitude 4.9 Correct: Your answer is correct. V/m direction +x Correct: Your answer is correct. Determine the magnitude and direction of the electric field at x = 3.00 m. magnitude V/m direction
Answer:
a) 10.6V
b) E = 4.9V/m, +x direction
c) E = 4.9V/m, +x direction
Explanation:
You have the following function:
[tex]V=a+bx[/tex] (1)
for the potential in a region between x=0 and x=6.00 m
a = 10.6 V
b = -4.90V/m
[tex]V=10.6V-4.90\frac{V}{m}x[/tex]
a) for x=0 you obtain for V:
[tex]V=10.6V-4.90\frac{V}{m}(0m)=10.6V[/tex]
b) The relation between the potential difference and the electric field can be written as:
[tex]E=-\frac{dV}{dx}=-b=-(-4.9V/m)=4.9V/m[/tex] (2)
the direction is +x
c) The electric field is the same for any value of x between x=0 and x=6m.
Hence,
E = 4.9V/m, +x direction
What is the radius of a nucleus of an atom?
Answer:
i think its 1-10 ´ 10-15 m
Explanation:
It is found that nuclear radii range from 1-10 ´ 10-15 m. This radius is much smaller than that of the atom, which is typically 10-10 m. Thus, the nucleus occupies an extremely small volume inside the atom. The nuclei of some atoms are spherical, while others are stretched or flattened into deformed shapes.
A box is being pulled to the right. What is the magnitude of the Kinect frictional force?
The so-called "force" of friction always acts exactly opposite to the direction in which an object is moving. So for this question, since the box is moving to the right, the "force" of friction is acting toward the left.
There's not enough information given in the question to determine its magnitude.
A ball is fired from a toy gun with an initial speed of 7.00 m/s with the gun aimed at 300 upward with respect to the horizontal. The point of firing is 1.32 m above the horizontal floor. What is the horizontal component of the ball's initial velocity?
Answer:
6.0621 m/s
Explanation:
we all know that horizontal component of the velocity in projectile motion is
= u cosα
u= initial speed of 7.00 m/s of the gun.
α = angle of the initial velocity with the horizontal = 30°
therefore, horizontal component of the ball's initial velocity = 7×cos30°
[tex]7\times\frac{\sqrt{3} }{2} \\=6.0621 \text{m/s}[/tex]
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t2. (a) What is the maximum height reached by the ball? ft (b) What is the velocity of the ball when it is 384 ft above the ground on its way up? (Consider up to be the positive direction.) ft/s What is the velocity of the ball when it is 384 ft above the ground on its way down?
Answer:
a) h_max = 400ft
b) v = 1024 ft/s
Explanation:
The general equation of a motion is:
[tex]s(t)=v_ot-\frac{1}{2}gt^2[/tex] (1)
You have the following equation of motion is given by:
[tex]s(t)=160t-16t^2[/tex] (2)
You compare both equations (1) and (2) and you obtain:
vo: initial velocity = 160ft/s
g: gravitational acceleration = 32ft/s
a) The maxim height reached by the ball is given by:
[tex]h_{max}=\frac{v_o^2}{2g}=\frac{(160ft/s)^2}{64ft/s^2}=400ft[/tex]
b) The velocity is given by:
[tex]v^2=v_o^2-2gh\\\\v=\sqrt{(160ft/s)^2-2(32ft/s^2)(384ft)}=1024ft/s[/tex]
Answer:
a) Smax = 400 ft
b) v = 32 ft/s
c) v = - 32 ft/s
Explanation:
(a)
The function given for the height of ball is:
s = 160 t - 16 t²
Therefore, in order to find the time to reach maximum height (in-flexion point), we must take the derivative with respect to t and set it equal to zero:
Therefore,
160 - 32 t = 0
32 t = 160
t = 160/32
t = 5 sec
Therefore, maximum distance is covered at a time interval of 5 sec.
Smax = (160)(5) - (16)(5)²
Smax = 800 - 400
Smax = 400 ft
b)
First we calculate the time at which ball covers 384 ft
384 = 160 t - 16 t²
16 t² - 160 t + 384 = 0
Solving Quadratic Equation:
Either:
t = 6 sec
Or:
t = 4 sec
Since, the time to reach maximum height is 5 sec. Therefore, t < 5 sec
Therefore,
t = 4 sec
Now, we find velocity at 4 sec by taking derivative od s with respect to t at 4 sec:
v = 160 - 32 t
v = 160 - (32)(4)
v = 32 ft/s
c)
Since, t = 6 s > 5 s
The second value of t = 6 sec must correspond to the instant when the ball is 384 ft above the ground while traveling downward.
Hence, velocity at that time will be:
v = 160 - (32)(6)
v = -32 ft/s
Negative sign due to downward motion.
A projectile is fired with an initial speed of 37.7 m/s at an angle of 41.2° above the horizontal on a long flat firing range. Determine the maximum height reached by the projectile.
Answer:
h = 31.46 m
Explanation:
We have,
Initial speed of a projectile is 37.7 m/s
It was projected at an angle of 41.2° above the horizontal on a long flat firing range.
It is required to find the maximum height reached by the projectile. The formula used to find it is given by :
[tex]h=\dfrac{u^2\sin^2\theta}{2g}[/tex]
Plugging all the known values,
[tex]h=\dfrac{(37.7)^2\times \sin^2(41.2)}{2\times 9.8}\\\\h=31.46\ m[/tex]
So, the maximum height reached by the projectile is 31.46 m.
You are pushing a cart across the room, and the cart has wheels at the front and the back. Your hands are placed on top of the cart at the center (left to right) of the top edge, pushing horizontally. There is friction between the wheels and the floor. Is the normal force between the floor and the front wheels greater than, smaller than, or equal to the normal force between the floor and the rear wheels? a) The force on the front wheels must be smaller than the force on the rear wheels. b) The force on the front wheels must be equal to the force on the rear wheels. c) The force on the front wheels must be greater than the force on the rear wheels.
Answer:
The normal force is the force that the floor does as a reaction of the gravitational force that an object does against the floor (is the resistance that objects have when other objects want to move trhough them, and the force comes by the 3rd Newton's law, and this is specially used in cases where the first object is fixed, like walls or the floor). With this in mind, the point in where the normal force will be greater is the point that is closer to the center of mass of the object (the point with more mass)
If the wheels are in the extremes of the object, and the center of mass is in the middle of the object, the normal force will be equal. Now if for example, you put a little mass in one end of the object, now the center of weight displaces a little bit and is not centered, and the side is where you put the weight on will receive a bigger normal force from the floor than the other side.
The force on the front and rear wheels can be determined by finding the
force acting under equilibrium conditions.
The normal reaction between the floor and the front wheel is larger. The
correct option is c) The force on the front wheels must be greater than the
force on the rear wheels.
Reasons:
The forces acting on the wheels are;
Vertical forces acting on the cart:
The weight of the cart acting on the floor
The normal reaction of the floor on the wheels
The horizontal acting on the cart:
Pushing force acting horizontally
Friction force acting in the reverse direction
At equilibrium, the cart does not tip over
Sum of moment about the rear wheel = [tex]\mathbf{W \times \dfrac{d}{2} + F \times h - N_{front} \times d} = 0[/tex]
Where;
h = The height of the cart
d = depth of the cart
[tex]N_{front}[/tex] = The normal reaction at the front wheels
Therefore;
[tex]\displaystyle N_{front} = \frac{W \times \dfrac{d}{2} + F \times h}{d} = \mathbf{ \frac{W}{2} + \frac{F \times h}{d}}[/tex]
Sum of moment about the front wheel = [tex]\mathbf{ F \times h + N_{rear} \times d - W \times \dfrac{d}{2} } = 0[/tex]
Therefore;
[tex]\displaystyle N_{rear} = \frac{W \times \dfrac{d}{2} - F \times h}{d} = \mathbf{ \frac{W}{2} - \frac{F \times h}{d}}[/tex]
Which gives;
[tex]\displaystyle N_{front} = \mathbf{ \displaystyle N_{rear} + 2 \times \frac{F \times h}{d}}[/tex]
[tex]\displaystyle 2 \times \frac{F \times h}{d} > 0[/tex]
Therefore;
[tex]\displaystyle N_{front} > \displaystyle N_{rear}[/tex]
The correct option is c) The force on the front wheels must be greater than
the force on the rear wheels.
Learn more here:
https://brainly.com/question/17574217
with one example each , state and explain five outputs derieved from a business policy
A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.540 m/s. The block has a head-on elastic collision with a 40.0 gram block that is initially at rest. Since the collision is head-on, all velocities lie along the same line, both before and after the collision. (a) What is the speed of the 20.0 gram block after the collision
Answer:
0.180 m/s
Explanation:
Solving the equations for conservation of momentum and energy for an elastic collision gives ...
v₁' = ((m₁ -m2)v₁ +2m₂v₂)/(m₁ +m₂) . . . . v₁' is the velocity of m₁ after collision
__
Here, we have (m₁, m₂, v₁, v₂) = (20 g, 40 g, 0.540 m/s, 0 m/s).
Substituting these values in to the equation for v₁', we have ...
v₁' = ((20 -40)(0.540) +2(40)(0))/(20 +40) = (-20/60)(0.540)
v₁' = -0.180 . . . m/s
The speed of the 20 g block after the collision is 0.180 m/s to the left.
18. Compared to its weight on Earth, a 5 kg object on the moon will weigh A. the same amount. B. less. C. more.
Answer:
B. less
Explanation:
acceleration due to gravity on Earth, g = 9.8 m/s²
acceleration due to gravity on Moon, g = 1.6 m/s²
Given mass of the object as, m = 5 kg
Weight of an object is given as, W = mg
Weight of the object on Earth, W = 5 x 9.8 = 49 N
Weight of the object on Moon, W = 5 x 1.6 = 8 N
Therefore, the object weighs less on the moon compared to its weight on Earth.
The correct option is "B. less"
Common static electricity involves charges ranging from nanocoulombs to microcoulombs. (a) How many electrons are needed to form a charge of −7.50 nC? electrons (b) How many electrons must be removed from a neutral object to leave a net charge of 0.580 µC? electrons
Answer:
a) 4.681*10^10 electrons
b) 3.67*10^12 electrons
Explanation:
The amount of electrons in a charge of 1C is:
[tex]1C=6.2415*10^{18}\ electrons[/tex]
You use the previous equality as a conversion factor.
a) The sing of the charge is not important in the calculation of the number electrons, so, you use the absolute value of the charge
[tex]7.50nC=7.50*10^{-9}C*\frac{6.2415*10^{18}}{1C}=4.681*10^{10}\ electrons[/tex]
In 7.50nC there are 4.61*10^18 electrons
b)
[tex]0.580\mu C=0.580*10^{-6}C*\frac{6.2415*10^{18}}{1C}=3.67*10^{12}\ electrons[/tex]
To obtain a charge of 0.580 µC in a neutral object you need to take out 3.67*10^12 electrons
Calculate the speed of a wave on a guitar string which has a mass of 55 g and a length of 30 cm if the tension in the string is 200 N
Answer:
33 m/s
Explanation:
v = √(T / (m/L))
where v is the velocity,
T is the tension,
and m/L is the mass per length.
v = √(200 N / (0.055 kg / 0.30 m))
v = 33 m/s
To Da
The radius of a circular track is 7 m. What will be the speed of a scooter, if it takes 20 seconds to complete one round of the
track?
published yet
04
published yet
A: 2.2 m/s
published yet
B: 22 m/s
02
published yet
C: 0.22 m/s
D: 022m/s
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Answer:
A
Explanation:
Firstly, recall the equation for the circumference of a circle (C=2*pi*r). Then apply the formula speed = distance/time.
Answer:
i see they did already i am just supporting them
Answer:
A
Explanation:
Firstly, recall the equation for the circumference of a circle (C=2*pi*r). Then apply the formula speed = distance/time.
The kinetic energy of a ball with a mass of 0.5 kg in a velocity of 10 meterss
Answer:
The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s is J.
Explanation:
Thus the kinetic energy of the ball is 25 J. The unit Joule (J) is the same as kgm^2/s^2 and is the SI unit for kinetic energy.
Hope this helps.
A 58.0 kg snow skier is on the top of 351 m high hill. After she has gone down a vertical distance of 142 m, what is her mechanical energy? Explain your answer (CER)!
Answer:
Explanation:
Initially skier is at a height of 351 m . Her kinetic enery will be nil because she is at rest . Her potential energy will be calculated as follows
potential energy = mgh where m is mass , h is height and g is acceleration due to gravity
potential energy = 58 x 9.8 x 351
= 199508 .4 J
Total mehanical energy = potential energy + kinetic energy
= 199508.4 J
According to conservation of mechanical energy , at the height of 142 m also total mechanical energy will be same . At this height some potential energy will be converted into kinetic energy but total of potential and kinetic energy will be same.
As part of an exercise program, a woman walks south at a speed of 2.00 m/s for 60.0 minutes. She then turns around and walks north a distance 3000 m in 25.0 minutes (a) What is the woman's average velocity during her entire motion?
Answer:
0.824m/s
Explanation:
To calculate the average velocity we have to find the total distance and the total time
We have to find the distance and time in each motions
FIRST MOTION
The values given are
Speed= 2m/s , t = 60minutes
The time has to be converted to seconds. 60×60 = 3600seconds
Distance= speed×time
= 2× 3600
= 7200m
In the first motion the distance is 7200m and the time is 3600seconds
SECOND MOTION
The values given are
Distance= 3000m
Time= 25 mins to seconds
= 25×60
= 1500 seconds
In the second motion distance is 3000m and the time is 1500 seconds
The total distance can be calculated by applying the formular ( d1-d2) since she moved in an opposite direction
Total distance= 7200-3000
= 4200m
The total time (t1+t2) = 1500+3600
= 5100 seconds
Therefore, average velocity is calculated by applying the formular
Total distance/ Total time
= 4200/5100
= 0.824m/s South
Hence the average velocity is 0.824m/s South.
Answer:
The woman's average velocity during her entire motion is 2 m/s
Explanation:
Given;
initial speed of the woman, u = 2.00 m/s
initial time taken, t₁ = 60 minutes = 3600 seconds
initial displacement of the woman, x₁ = ?
final displacement of the woman, x₂ = 3000 m north
final time taken , t₂ = 25.0 minutes = 1500 seconds
The woman's average velocity during her entire motion:
initial displacement of the woman, x₁ = u x t₁ = 2.00 m/s x 3600 seconds
= 7200 m South
[tex]Average \ velocity = \frac{\delta X}{\delta t} = \frac{X_1-X_2}{t_1-t_2} \\\\V_{avg.} = \frac{7200-3000}{3600-1500} = \frac{4200}{2100} = 2 \ m/s[/tex]
Therefore, the woman's average velocity during her entire motion is 2 m/s
Adiabatic Equation of State Derive the adiabatic equation of state (2.3.19) using particle conservation (2.3.7) and energy conservation (2.3.21), by assuming that the heat flow vector q and all collision terms in these equations are zero.
p=Cn^{\gamma } Equation 19
\frac{\partial n}{\partial t}+\bigtriangledown (nu)=G-L Equation 7
\bigtriangledown (\frac{3}{2}pu)=\frac{\partial }{\partial t}(\frac{3}{2}p)\mid c Equation 22
Answer:
Explanation:
jhgfhmhvmghmm
A plane travels 2.5 KM at an angle of 35 degrees to the ground, then changes direction and travels 5.2 km at an angle of 22 degrees to the ground. What is the magnitude and direction of the planes total displacement??
Answer:
7.7 km 26°
Explanation:
The total x component is:
x = 2.5 cos(35°) + 5.2 cos(22°) = 6.87
The total y component is:
y = 2.5 sin(35°) + 5.2 sin(22°) = 3.38
The magnitude is:
d = √(x² + y²)
d = 7.7 km
The direction is:
θ = atan(y/x)
θ = 26°
The decrease of PE for a freely falling object equals its gain in KE, in accord with the conservation of energy. By simple algebra, find an equation for an object's speed v after falling a vertical distance h. Do this by equating KE to its change of PE.
Answer:
[tex]v = \sqrt{20h} [/tex]
Explanation:
The potential energy (PE) we are looking here is gravitational potential energy (GPE).
GPE= mgh,
where m is the mass of an object,
g is the gravitational field strength
h is the height of the object
KE= ½mv²,
where m is the mass and v is the velocity
loss in GPE= gain in KE
mgh= ½mv²
gh= ½v² (divide by m throughout)
Assuming that the object is on earth, then g= 10N/Kg
½v²= 10h (substitute g=10)
v²= 20h (×2 on both sides)
v= √20h (square root both sides)
At the same temperature, two wires made of pure copper have different resistances. The same voltage is applied at the ends of each wire. The wires may differ in:________
a. length.b. cross-sectional area.c. resistivity.d. amount of electric current passing through them
Answer:
A and B are true
Explanation:
C and D are false
Answer:
The answer is the length, cross-sectional area, and the amount of electric current passing through them.
So,
a
b
d
are the correct answers.
Explanation:
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This answer is 100% correct
If an object is in equilibrium, which of the following statements is not true? A) The velocity is constant. B) The object must be at rest. C) The net force acting on the object is zero. D) The acceleration of the object is zero. E) The speed of the object remains constant.
When the object is in equilibrium so the statement i.e. not true should be that object should be at rest.
Newton second law of motion:
In the case when the net force of an object should be zero at the time when it is in equilibrium. In the case when the net force should be zero the acceleration should also be zero. And, if the acceleration if zero so the speed and velocity should remain the same. So for maintaining the equilibrium, the object should not have to be only at rest.
learn more about newton here: https://brainly.com/question/18453410
ISC 152
Describe the
two major sources of energy and
give two
examples
each
HEYA!!
Here's your Answer:
Energy is the power we use for transportation, for heat and light in our homes and for the manufacture of all kinds of products. There are two sources of energy: renewable and non-renewable energy
RENEWABLE ENERGY: Renewable energy, often referred to as clean energy, comes from natural sources or processes that are constantly replenished. For example, (sunlight or wind keep shining and blowing), even if their availability depends on time and weather.
NON-RENEWABLE ENERGY:Non-renewable energy comes from sources that will run out or will not be replenished in our lifetimes—or even in many, many lifetimes.
Most non-renewable energy sources are (fossil fuels: coal, petroleum, and natural gas).
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