Answer:
The golf ball would be revolving around the baseball while the baseball would be revolving around the basketball.
Explanation:
The basketball would be represented as the Sun, the baseball would be represented as the Earth and the golf ball would be represented as the moon. Since the Moon revolves around the Earth and the Earth revolving around the Sun makes a perfect explaination for this question.
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
In a lunar experiment, a 950-g aluminum (920 J/(°Ckg)) sphere is dropped from the space probe while is 75 m above the Lunar ground. If the sphere’s temperature increased by 0.11°C when it hits the ground, what percentage of the initial mechanical energy was absorbed as thermal energy by the aluminum sphere?
Answer:
13.759 % of the initial mechanical energy is lost as thermal energy.
Explanation:
By the First Law of Thermodynamics we know that increase in internal energy of the object ([tex]U[/tex]), in joules, is equal to the lost amount of the change in gravitational potential energy ([tex]U_{g}[/tex]), in joules:
[tex]\frac{x}{100} \cdot \Delta U_{g} = \Delta U[/tex] (1)
Where [tex]x[/tex] is the percentage of the energy loss, no unit.
By definition of the gravitational potential energy and internal energy, we expand this equation:
[tex]\frac{x\cdot m \cdot g \cdot h}{100} = m\cdot c\cdot \Delta T[/tex] (1b)
Where:
[tex]m[/tex] - Mass of the object, in kilograms.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]h[/tex] - Initial height of the object above the lunar ground, in meters.
[tex]c[/tex] - Specific heat of aluminium, in joules per degree Celsius-kilogram.
[tex]\Delta T[/tex] - Temperature increase due to collision, in degree Celsius.
If we know that [tex]m = 0.95\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]h = 75\,m[/tex], [tex]c = 920\,\frac{J}{kg\cdot ^{\circ}C}[/tex] and [tex]\Delta T = 0.11\,^{\circ}C[/tex], then the percentage of energy loss due to collision is:
[tex]x = \frac{100\cdot c\cdot \Delta T}{g\cdot h}[/tex]
[tex]x = \frac{100\cdot \left(920\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (0.11\,^{\circ}C)}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (75\,m)}[/tex]
[tex]x = 13.759\,\%[/tex]
13.759 % of the initial mechanical energy is lost as thermal energy.
A plane mirror produces a _____.
virtual image
refracted image
real image
Answer:
Explanation:
A plane mirror is the kind you look into when you look into a "regular" mirror. The image you see is right-side-up. These images are virtual. Real images are always upside down and are made by mirrors that are "parabolic" in shape. Virtual images are always right-side-up.
The atomic bomb dropped on Hiroshima converted about 7.00x10-4kg of mass to energy. How much energy did that bomb produce?
A)2.10x10^5J
B)7.78x10^-21J
C)6.30x10^13J
D)2.10x10^61J
Answer:
[tex] \sf \: given \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bf \: mass \: \: \: m \: = 7.00 \times {10}^{ - 4} \: kg \\ \\ \bf \: E=mc^2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = > E=7.00 \times {10}^{ - 4} \times ({3 \times {10}^{8} })^{2} \\ \\ = > \green{ \boxed{ E = 6.3 \times {10}^{13} \: J}}[/tex]
1. Find the temperature when the degrees of the Celsius scale will be one fifth of the corresponding degrees of the Fahrenheit scale
2. How much heat is necessary to warm 500g of water from 20°C to 65°C?
Answer:
F = 9/5 C + 32 conversion from C to F
F = 9/5 * F/5 + 32
25 F = 9 F + 800
16 F = 800
F = 50
Check:
C = 5/9 ( F - 32) = 5/9 (50 - 32) = 10 as requested
Q = c m change in temp
Q = 1 cal/gm-deg C * 500 gm * 45 deg C = 22,500 calories
50 Fahrenheit heat required.
How much heat required?Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy between physical systems.
F = 9/5 C + 32 conversion from C to F
F = 9/5 * F/5 + 32
25 F = 9 F + 800
16 F = 800
F = 50
The answer is 50 Fahrenheit.
Learn more about heat transfer
https://brainly.com/question15086449
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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
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
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
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]
The radius of the base of a wooden cylinder 2m and its altitude is 7m. What is its mass?
Answer:
88 m ^2
Explanation:
We reduce friction in machines? why
Answer:
friction reduces the efficiency of machines, thus we must reduce the friction force that is acting upon it.
Answer:
Because it causes a lot of wear and tear in machine parts that move against each other. It erodes the surfaces and destroys their symmetries
Explanation:
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
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:
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:
A red car has a head-on collision with an approaching blue car with the same magnitude of momentum. A green car driving with the same momentum as the other cars collides with an enormously massive wall. Which of the three cars will experience the greatest impulse
All three cars experience the same impulse.
Impulse is equal to change in momentum.
Each car starts with the same amount of momentum and ends up with zero, so the magnitudes of all three changes are equal.
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
Calculate the relative atomic mass of MgO
Answer:
MgO relative atomic mass is 40
Explanation:
Mg=24
O=16
please answer these diagrammatic questions ASAP and please no spam answers
Answer:
i. The pressure of due to the water, P, is given according to the following equation;
P = ρ·g·h
Where;
ρ = The density of the water (a constant) = 997 kg/m³
g = The acceleration due to gravity = 9.81 m/s²
h = The height of the water (minimum h = h₁, maximum h = h₂)
The pressure is directly proportional to the water height, and we have;
The pressure, P, will be maximum when the water height, h, is maximum or h = h₂, which is the level DC
ii. The thrust = The force acting on the body = Pressure × Area
The maximum areas exposed to the water are on side AB and DC
However, the pressure at level DC, which is the location of the maximum pressure, is larger than the pressure at level AB, therefore, the maximum thrust will be at the level DC
Explanation:
84. Three resitors each of value 30 respectively are connected in a parallel
combination across a 10 V battery the current through each resitor is
Answer:
each resistor draws 1/3 of an amp or 0.33333 amps
Explanation:
V = I * R
V = 10 volts
R = 30 ohms
10 = I * 30 Divide by 30
10/30 = I
I = 0.33333
Discuss the role of globalization in the development of sI unit
Answer:
Sharing of informationExplanation:
The development of SI unit has helped in the sharing of scientific as well as techical information internationally.Answer:
It was created during the French Revolution in 1799 and has enabled for the international exchange of scientific and technical information. Calculating with SI units is also a lot easier than using the English system.
the atomic number of phosphorus is
Answer:
The atomic number of phosphorus is 15.
Explanation:
It’s found after Si(Silicon) and before S(sulphur)
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).
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 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 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
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.
If the velocity of a motorcycle increases from 30 mis too 50m/s in 10 seconds what will be the acceleration of motorcycle?
Answer:
2 m/s^2
Explanation:
a = ∆v / ∆t
a = (50 - 30) / 10
a = 20 / 10
a = 2 m/s^2
acceleration = u-v/t
50-30/10
=2m/s
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
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