Answer:
I think the sun is the right answer
Name the type of relationship between current and potential difference for a resistor at constant temperature. [
The current flowing through a resistor at a constant temperature is directly proportional to the potential difference across it. This is called Ohm's law
1. Imagine a bowling ball with a mass of 5,0 kg that is dropped from a height of 100m,
a. At the moment the bowling ball is dropped it is at a height of 100m, What is its
potential energy?
b. At the moment the bowling ball is dropped, it is not yet moving. Therefore, its
kinetic energy is what?
c. What is the total of both potential and kinetic energy?
a.
Given,
height (h) = 100m
mass (m) = 5kg
acceleration due to gravitation (g) = 9.8ms^-2
Potential energy
= mgh
= 5kg × 9.8ms^-1 × 100m
= 4900 kgm²s^-2
= 4900 J
b.
Since the ball is not moving yet, its kinetic energy is 0.
c.
The total of potential and kinetic energy at every point of the journey is same, i.e., 4900 J.
6
A light bulb changes
???? energy into
and ??? energy
The
??? energy is useful energy, and the heat energy is ??
energy
Explanation:
electrical energy into heat energy
electrical , thermal
Two stones are dropped from the edge of a 60m cliff , the second stone 1.6secon after the first . How far below the top of the cliff is the second stone when the separation between the two stone is 36m?
Answer:
The separation between the two stones is 36 m, when the second stone is approximately 10.9 m below the top of the cliff
Explanation:
The given parameters are;
The height of the cliff from which the stones are dropped, h = 60 m
The time at which the second stone is dropped = 1.6 seconds after the first
The distance below the top of the cliff when the distance between the two stones is 36 m = Required
We have;
The kinematic equation of motion that can be used is s = u·t - (1/2)·g·t²
For the first stone, we have, s₁ = u·t₁ - (1/2)·g·t₁²
For the second stone, we get; s₂ = u·t₂ - (1/2)·g·t₂²
t₁ = t₂ + 1.6
g = The acceleration due to gravity ≈ 9.81 m/s²
s = The distance below the cliff top
The initial velocity of the stones, u = 0
Let t represent the time from which the second stone is dropped at which the distance between the two stones is 36 m, we have;
s₁ = u·(t + 1.6) + (1/2)·g·(t + 1.6)²
s₂ = u·t + (1/2)·g·t²
u = 0
∴ s₁ - s₂ = 36 = (1/2)·g·(t + 1.6)² - (1/2)·g·t²
2 × 36/(g) = (t + 1.6)² - t² = t² + 3.2·t + 2.56 - t² = 3.2·t + 2.56
2 × 36/(9.81) = 3.2·t + 2.56
t = (2 × 36/(9.81) - 2.56)/3.2 = ≈ 1.49 s
t ≈ 1.49 s
s₂ = (1/2)·g·t²
∴ s₂ = (1/2) × 9.81 × 1.49² ≈ 10.9
The distance below the top of the cliff of the second stone when the the separation between the two stones is 36 m, s₂ ≈ 10.9 m.
Suppose the charged sphere is made from a conductor, rather than an insulator. Do you expect the magnitude of the force between the point charge and the conducting sphere to be greater than, less than, or equal to the force between the point charge and an insulating sphere
Answer:
* Point charge outside the radius of the sphere r> R, the force in the two systems is the same
* Point charge inside the sphere r <R, therefore the force in the system with the insulating sphere is greater
Explanation:
To answer this question let's use the relation
F = q E
with q being the point charge and E the electric field created by the sphere.
If we use Gauss's law
The electric field flux is proportional to the wax charge within the surface.
Let's analyze our situation.
* Point charge outside the radius of the sphere
r> R
where R is the radius of the sphere and r the distance from the center of the sphere to the point charge
in this case the waxed charge for the insulating and conducting sphere is the same, therefore the force in the two systems is the same
* Point charge inside the sphere
r <R
conductive sphere.
As the charges are mobile, they are located on the surface of the sphere and there is no waxed charge within a Gaussian surface that passes through the point charge, therefore the electric field is zero and consequently the force
F = 0
insulating sphere
Charges cannot move therefore there is a fraction of charge within a surface that passes through the point charge, consequently the electric field is different from zero
Fe> 0
for this second position the force on the conducting sphere is zero
therefore the force in the system with the insulating sphere is greater
7) A ball is thrown upward at an initial velocity of 8.2 m/s, from a height of 1.8 meters above the ground. The height of the ball h, in metres can be represented, after t seconds, is modelled by the equation h = –4.8t² + 8.2t + 1.8. (a) Determine the height of the ball after 1.7 seconds.
Answer:
8392
Explanation:
d=s/t
What acceleration is produced on a mass of 200g, when a force of 10N is exerted on it?
Answer:
f=ma......10N=0.2a....=50m/s
Answer:
here,
force = mass× acceleration
10 = 0.2 kg × a
or, 10/0.2=a
or, a = 50km/h^2
is the required ans .
Help quick
How does the ramp produce mechanical advantage?
A it reduces the amount of input force needed to do a certain amount of work
B it reduces the distance over which the input force needs to be applied
C it reduces the amount of useful work done on objects move it up the ramp
D what is the overall amount of work done on object moving up the ram
C it reduces the amount of useful work done on objects move it up the ramp
Is velocity ratio of a machine affected by applying oil on it?Explain with reason.
Answer:
It depends upon SAE No. of oil. (SAE means Society of Automotive Engineers). However, it usually does protect against friction.
Explanation:
If we use very viscous oil, it does not reach all the parts. Very thin oil will flows away easily and gets wasted. Grease is used in such cases. It is generally used around ball-bearing. Normal grease or oil is never used where there is high pressure, high temperature and high speed. Special lubricants are used in such cases. In cold season the oil becomes thick and in hot season it becomes thin. Therefore selection of lubrication also depends on the season. It is always advisable to refer operating manual of the equipment before selecting the lubricant.
The intensity of friction depends on following factors:
i) The area involved in friction.
ii) The pressure applied on the surfaces.
Force = Pressure ´ Area Frictional force will increase, if the area of contact will increase or if pressure applied on the surface increased.
why a person feel weightlessness in a spacecraft orbiting around a heavenly body
Answer:
The orbital velocity an aircraft orbiting around a heavenly body is found as follows;
At the orbital velocity, [tex]F_G[/tex] = [tex]F_C[/tex]
Where;
[tex]F_G[/tex] = The gravitational force = [tex]\dfrac{G \cdot M \cdot m}{R_E^2}[/tex]
[tex]F_C[/tex] = The centripetal force = [tex]\dfrac{m \cdot v_0^2}{R_E}[/tex]
Therefore
[tex]v_0 = \sqrt{\dfrac{G \cdot M}{R_E} }[/tex]
Therefore, at the orbital velocity of the spacecraft, the centripetal force attracting the person away from the central region heavenly body is equal to the gravitational force pulling the person towards the center of the heavenly body (which was felt as her or his weight), and the person feels weightless while inside the orbiting spacecraft
Explanation:
how is one standard kilogram defined in SI system?
Answer:
One standard kilogram is define as the mass of platinium iridium cylinder having equal diameter and height kept at the particular condition of international bureo of weigh and measure sevre near paris.
You observe that you see more mockingbirds in small trees and more hawks in large trees. Which of the following is an appropriate scientific question based on this observation?
How does the size of a tree affect the bird species that prefer to live in it?
How do birds fly?
What type of food do birds eat?
What time of year to birds mate?
Answer:
A) How the size of a tree affect the bird species that prefer to live in it
Explanation:
I took the quiz
Use the KMT to explain what happens to water vapor when it encounters a
cold glass of water.
Answer:
The postulates of the Kinetic Molecular Theory, KMT, are;
(1) In an ideal gas the molecules are in constant motion
(2) The collisions between molecules of gases are perfectly elastic
(3) The volume occupied by the molecules are negligible
(4) The temperature of the gas is directly proportional to its kinetic energy
(5) The intermolecular forces in the gas are negligible
According to the KMT, gaseous water vapor molecules are in constant motion and move at a speed that depends on their temperature. The intermolecular forces between the molecules are negligible and when they collide with the cold glass, they lose temperature to the glass, thereby reducing their temperature, kinetic energy and therefore, their speed is reduced.
The increasingly temperature of the water vapor coming in contact with the cold glass gives rise to reduced speed of the cooled gas molecules, thereby causing them to move closer together after having elastic collisions and to cluster with tiny particles in the air, to form tiny droplets
The rapid cooling on the cold glass surface causes the droplets to form rapidly on the cold glass surface which makes them visible as condensed water on the surface of the cold glass of water
Explanation:
What is the decay constant for Oxygen-19 if it has a half-life of 26.5s?
A)0.0262/s
B)18.4
C)0.0377/s
C)38.2/s
Answer:
Option A.
Explanation:
We define the half time T as the time such that an initial quantity A reduces to its half.
So we can model the quantity as a function of time like:
P(t) = A*e^(-k*t)
Then for the half time, T, we will have:
P(T) = A/2 = A*e^(-k*T)
solving for k, we get:
A/2 = A*e^(-k*T)
1/2 = e^(-k*T)
ln(1/2) = ln( e^(-k*T)) = -k*T
-ln(1/2)/T = k
Here we know that the half time is T = 26.5s
if we input that in the above equation, we get:
-ln(1/2)/26.5s = k = 0.0262 s^-1
Then the correct option is A
An object is thrown with velocity 7.1 m s-1 vertically upwards on the Moon. The
acceleration due to gravity on the Moon is 1.62 ms? What is the time taken for the object to return to its starting point?
A 3.5 s
B 4.4 s
C 6.5 s
D 8.8 s
Answer:D
Explanation:
a car moves at a speed of 40km/h. it is stopped by applying brake which produce a uniform acceleration of-0.5m/s^2. how much distance will it move before coming to stop ? plz its urgent
Answer:
Explanation:
I answered this in question #24313516
b) A force is represented in magnitude and direction as (6N, 250degrees. Find both the vertical and horizontal components of the force.
Answer:
Explanation:
To find the horizontal component, the x component specifically, use the formula:
[tex]V_x=Fcos\theta[/tex] and for the vertical component, the y component, use the formula:
[tex]V_y=Fsin\theta[/tex]
where F is the magnitude of the force and theta is the angle in degrees.
For the x-component:
[tex]V_x=6cos250[/tex] so
[tex]V_x=-2.1[/tex] and depending upon whether this is a displacement vector or a velocity vector, the label would be meters/feet or m/s, respectively.
For the y-component:
[tex]V_y=6sin250[/tex] so
[tex]V_y=-5.6[/tex]
A tightrope walker is walking between two buildings holding a pole with length L=14.0 m, and mass mp=17.5 kg. The daredevil grips the pole with each hand a distance d=0.595 m from the center of the pole. A bird of mass mb=560 g lands on the very end of the left‑hand side of the pole. Assuming the daredevil applies upward forces with the left and right hands in a direction perpendicular to the pole, what magnitude of force Fleft and Fright must the left and right hand exert to counteract the torque of the bird?
Answer:
F = 32.28 N
Explanation:
For this exercise we must use the rotational equilibrium relation
Σ τ = 0
In the initial configuration it is in equilibrium, for which all the torque and forces are compensated. By the time the payment lands on the bar, we assume that the counter-clockwise turns are positive.
W_bird L / 2 - F_left 0.595 - F_right 0.595 = 0
we assume that the magnitude of the forces applied by the hands is the same
F_left = F_right = F
W_bird L / 2 - 2 F 0.595 = 0
F = [tex]\frac{m_{bird} \ g L} { 4 \ 0.595}[/tex]
we calculate
F = 0.560 9.8 14.0 /2.38
F = 32.28 N
If the power of a working system(machine) is 2088.8 watts then its power in horsepower will be: A. 1hp B. 2.7hp C. 2.9hp D. 2.8hp
Answer:
D. 2.8 hp
Explanation:
1 hp=746 watts
2088.8÷746= 2.8
Make a tree diagram based on the topic motion which includes all the concept like uniform & non uniform motion accelerated motion equation of motion motion etc
I think it is a education tips
1) The position of an object to the north of a flagpole is given by x(t) = bt2 – c , where b and c are constants.
a) What is v(t), the velocity of the object as a function of time?
b) What is a(t), the acceleration of the object as a function of time?
c) At some time t the object is located at the flagpole. What is the velocity of the
object at that instant?
Answer:
a) The velocity of the object as a function of time, v(t) is 2·b·t
b) The acceleration of the function of time, a(t) is 2·b
c) The time at which the object is at the flagpole is t = √(c/b)
Explanation:
The function that gives the position of the object north of the flagpole, x(t) is presented as follows;
x(t) = b·t² - c (b and c are constants)
a) The velocity of the object as a function of time, v(t), is derived as follows
v(t) = x'(t) = d(b·t² - c)/dt = 2·b·t
The velocity of the object as a function of time, v(t) = 2·b·t
b) The acceleration of the function of time, a(t) = v'(t) = d(2·b·t)/dt = 2·b
c) The time at which the object is at the flagpole is given by the x-intercept of the function, where x(t) = 0, as follows;
At the x-intercept, we have, x(t) = 0 and x(t) = b·t² - c
∴ 0 = b·t² - c, which gives
b·t² = c
t² = c/b
t = ±√(c/b), we reject the negative value to get;
The time at which the object is at the flagpole, t = √(c/b).
A 2.0kg object is dropped from a height of 30m.
After it drops for 2.0 seconds, what is its kinetic
energy and what is its potential energy?
(Assume no air resistance.)
Answer:
1) The kinetic energy of the object after it drops for 2.0 seconds is approximately 384.9 Joules
2) The potential energy of the object after it drops for 2.0 seconds is approximately 204 J
Explanation:
1) The given mass of the object, m = 2.0 kg
The height from which the object is dropped, h = 30 m
The kinetic energy of the object after it drops for 2.0 seconds = Required
Kinetic energy, K.E. = (1/2)·m·v²
Where;
v = The velocity of the object
The kinematic equation for finding the velocity of the object is presented as follows;
v = u + g·t
Where;
u = The initial velocity of the object = 0
g = The acceleration due to gravity of the object ≈ 9.81 m/s²
t = The time of motion of the object = 2.0 seconds
∴ The velocity after 2 seconds, v ≈ 0 + 9.81 m/s² × 2 s = 19.62 m/s
The kinetic energy, K.E. after 2 seconds as the object drops is given as follows;
[tex]K.E._{(after \ two \ seconds)}[/tex] = (1/2) × 2.0 kg × (19.62 m/s)² = 384.9444 J ≈ 384.9 J
2) The total energy, M.E. of the object at the top, h = 30 m, u = 0, is given as follows;
The total mechanical energy, M.E. = P.E. + K.E.
M.E. = m·g·h + (1/2)·m·u²
∴ M.E. = 2.0 kg × 9.81 m/s² × 30 m + 0 = 588.6 J
M.E. = 588.6 J
Given that the total mechanical energy, M.E., is constant, we have;
At 2.0 seconds, M.E. = 588.6 J , K.E. ≈ 384.9 J, P.E. = M.E. - K.E.
∴ P.E. = 588.9 J - 384.9 J ≈ 204 J
The potential energy after it drops 2.0 seconds, P.E. ≈ 204 J
Lúc 7g bạn an đi từ nhà đến trường với tóc độ trung bình là 20km/h . Bạn đến trường lúc 7g20. Tính khoảng cách từ nhà tới trường?
Answer:
Distance = 6.667 kilometres
Explanation:
Given the following data;
Speed = 20 km/h
Departure time = 7:00
Arrival time = 7:20
Time taken = 20 minutes
To calculate the distance travelled from home to school;
First of all, we would have to convert the value of time in minutes to hours.
Conversion:
60 minutes = 1 hour
20 minutes = X hours
Cross-multiplying, we have;
X = 20/60 = 1/3 hours
Mathematically, the distance travelled by an object is calculated by using the formula;
Distance = speed * time
Distance = 20 * 1/3
Distance = 20/3 =
Distance = 6.667 kilometres
The radius of the base of a wooden cylinder 2m and its altitude is 7m. What is its mass?
Answer:
88 m ^2
Explanation:
A object of mass 200kg is pushed from rest by a force of 500N along a horizontal plane for 5.0 seconds. Calculate the acceleration of the object
Answer:
force=mass×acceleration
hence
acceleration is given by force÷mass
(500÷200)*5=12.5
Calculate the relative atomic mass of MgO
Answer:
MgO relative atomic mass is 40
Explanation:
Mg=24
O=16
a. What is the physical quantity in which force is divided by area called?
Any guess 50 pts + brainlist if correct
Answer:
pressure
Explanation: pressure=force/area
Answer:
The quantities that can be measured are called physical quantities.....
a small object is placed between two plane mirrors inclined at an angle of 60° to each other in a dark room how many images are seen explain
Answer:
nothing
Explanation:
bocouse of darkness
A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.45 m above the bottom of the chute with an initial speed of 1.23 m/s . The toy animal emerges horizontally from the bottom of the chute and continues sliding along a horizontal surface with a coefficient of kinetic friction of 0.231 . How far from the bottom of the chute does the toy zebra come to rest? Assume g=9.81 m/s2 .
Answer:
The answer is "4.97 m".
Explanation:
[tex]u = 1.23\ \frac{m}{s}\\\\[/tex]
[tex]H= 1.45 \ m\\\\[/tex]
[tex]\mu = 0.231\\\\[/tex]
The law of conservation tells us that heat energy at the top with kinetic energy at the top equals kinetic energy at the base.
[tex]mgh+\frac{1}{2}mu^2=\frac{1}{2}mv^2\\\\2gh +u^2 =v^2\\\\v=\sqrt{u^2+2gh}[/tex]
[tex]v=\sqrt{(1.23\ \frac{m}{s})^2+2(9.81 \frac{m}{s^2}) +(1.45\ m)[/tex]
[tex]=\sqrt{1.5129+19.62 +1.45}\\\\=\sqrt{22.5829}\\\\=4.75\ \frac{m}{s}[/tex]
Friction force is given by the formula
[tex]f=-\mu mg \\\\ma= -\mu mg\\\\a=-\mu g\\\\[/tex]
[tex]= -(0.231) \ (9.81\ \frac{m}{s^2})\\\\=-2.26611 \ \frac{m}{s^2}[/tex]
Now by using an equation of motion as
[tex]v^2-u^2= 2as[/tex]
From the above the distance traveled is
[tex]S=\frac{v^2-u^2}{2a}[/tex]
[tex]S=\frac{(0)^2-(4.75\ \frac{m}{s})^2}{2(-2.26611\ \frac{m}{s^2})}\\\\[/tex]
[tex]=\frac{-(4.75\ \frac{m}{s})^2}{2(-2.26611\ \frac{m}{s^2})}\\\\=\frac{-22.5625}{-4.53222}\\\\=4.97[/tex]
In other words, the distance from the bottom of the chute to the point where the toy zebra comes to rest is [tex]s = 4.97\ m[/tex]
a bus starts to move from rest if its velocity becomes 90 km per hour after 8 speed calculate its acceleration.
write with full process.
answer=3.13ms–²
Answer:
Acceleration, a = 3.13 m/s²
Explanation:
Given the following data;
Initial velocity = 0 m/s (since it's starting from rest).
Final velocity = 90 km/h
Time = 8 seconds
Conversion:
90 km/h = (90 * 1000)/3600 = 25 m/s
To find the acceleration, we would use the first equation of motion;
V = U + at
Where;
V is the final velocity.
U is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Substituting into the equation, we have;
25 = 0 + a*8
25 = 8a
Acceleration, a = 25/8
Acceleration, a = 3.125 ≈ 3.13 m/s²
