Answer:
[tex]$E = m{c^2}$[/tex]
Explanation:
• The explosion caused by chemical reaction involves conversion of chemical energy into heat and sound, but the total mass of atoms participating in the reaction remains constant.
• In the nuclear fusion reaction, the one happening in the core of the sun, mass is directly converted into energy by the equation [tex]E=mc^{2}[/tex]
Hence, mass remains constant, it just converts in to velocity.
Learn more about nuclear fusion here:
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____can be used to transmit information
A heat
B patterns
c senses
D digital
Answer:
D. digital
Explanation:
Digital signals are transmitted through electromagnetic waves.
I'm not sure
Electromagnetic radiation from a 8.25 mW laser is concentrated on a 1.23 mm2 area. Suppose a 1.12 nC static charge is in the beam, and moves at 314 m/s. What is the maximum magnetic force it can feel
Answer:
The maximum magnetic force is 2.637 x 10⁻¹² N
Explanation:
Given;
Power, P = 8.25 m W = 8.25 x 10⁻³ W
charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C
speed of the charge, v = 314 m/s
area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²
The intensity of the radiation is calculated as;
[tex]I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2[/tex]
The maximum magnetic field is calculated using the following intensity formula;
[tex]I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T[/tex]
The maximum magnetic force is calculated as;
F₀ = qvB₀
F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)
F₀ = 2.637 x 10⁻¹² N
Give an example of a physical entity that is quantized. State specifically what the entity is and what the limits are on its values.
Answer:
A charge is a physical entity that has been quantized. The limits on its values are the value of a charged particle quantized in the state where 'n'...
Explanation:
One example of a physical entity that is quantized is:
The amount of money in your pocket.
The amount can't have any fraction of 1 cent.
Its value must be an integer-multiple of cents, or 0.01 dollar.
When it increases or decreases, it jumps from one integer number of cents to the next integer number. It doesn't "slide" from one to the next. It can never have a value between two integer numbers of cents.
1. Calculate the density of a 5.00 g marble which, when placed in a
graduated cylinder containing 60 milliliters of water, raised the volume of
the water to 68 milliliters.
2. An elevator with a mass of 1000kg is lifted 20 meters. How much work
was done on the elevator?
3. Calculate the potential energy of a 12 kg cat at the top of a 42 meter-
high hill.
4. Calculate the kinetic energy in joules of a 1500 kg car that is moving at a
speed of 10 m/s.
Answer:
1. 0.625 g/mL
2. 2×10⁵ J
3. 5040 J
4. 75000 J
Explanation:
1. Determination of the density.
We'll begin by calculating the volume of the marble. This can be obtained as follow:
Volume of water = 60 mL
Volume of water + Marble = 68 mL
Volume of marble =?
Volume of marble = (Volume of water + Marble) – (Volume of water)
Volume of marble = 68 – 60
Volume of marble = 8 mL
Finally, we shall determine the density of the marble. This can be obtained as follow:
Mass of marble = 5 g
Volume of marble = 8 mL
Density of marble =?
Density = mass / volume
Density of marble = 5 / 8
Density of marble = 0.625 g/mL
2. Determination of the work done.
Mass (m) = 1000 Kg
Height (h) = 20 m
Acceleration due to gravity (g) = 10 m/s²
Workdone (Wd) =?
The work done can be obtained as follow:
Wd = mgh
Wd = 1000 × 10 × 20
Wd= 2×10⁵ J
3. Determination of the potential energy.
Mass (m) = 12 Kg
Height (h) = 42 m
Acceleration due to gravity (g) = 10 m/s²
Potential energy (PE) =?
The potential energy can be obtained as follow:
PE = mgh
PE = 12 × 10 × 42
PE = 5040 J
4. Determination of the kinetic energy.
Mass (m) = 1500 Kg
Velocity (v) = 10 m/s
Kinetic energy (KE) =?
The kinetic energy can be obtained as follow:
KE = ½mv²
KE = ½ × 1500 × 10²
KE = 750 × 100
KE = 75000 J
Consider the following possibilities and select the correct choice.
1. Tx Ty > Tz
2. Tx Ty < Tz
3. Tx Ty = Tz
Answer:
Tx not but mybe
Explanation:
for that reason its just trying to help
A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the following distances from the axis of the rod, where distances are measured perpendicular to the rod's axis.
Answer:
Explanation:
From the question;
We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.
We are to calculate the following task, i.e. to determine the electric field at the distances:
a) at 4.75 cm
b) at 20.5 cm
c) at 125.0 cm
Given that:
the charge (q) = 33.3 nC/m
= 33.3 × 10⁻⁹ c/m
radius of rod = 5.75 cm
a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.
Then, the electric field will be zero.
b) The electric field formula [tex]E = \dfrac{kq }{d}[/tex]
[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}[/tex]
E = 1461.95 N/C
c) The electric field E is calculated as:
[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}[/tex]
E = 239.76 N/C
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.
The peak value of the electric field component of an electromagnetic wave is E. At a particular instant, the intensity of the wave is of 0.020 W/m2. If the electric field were increased to 5E, what would be the intensity of the wave?
Answer:
[tex]I_2=0.50 w/m^2[/tex]
Explanation:
From the question we are told that:
initial Intensity [tex]I_1=0.020 w/m^2[/tex]
Final Electric field [tex]E_2=5E[/tex]
Generally the equation for Relation ship between intensity and Electric field is mathematically given by
[tex]\frac{I_1}{I_2}= \frac{E_1^2}{E_2^2}[/tex]
Therefore
[tex]I_2=\frac{I_1}{ \frac{E_1^2}{E_2^2}}[/tex]
[tex]I_2=\frac{0.020}{ \frac{E^2}{5E^2}}[/tex]
[tex]I_2=0.50 w/m^2[/tex]
sanaysay tungkol sa pangangalaga ng mga endangered animals
Kinetic energy is the energy an object has due to its
die hele
Answer:
There are a couple of interesting things about kinetic energy that we can see from the equation.
Kinetic energy depends on the velocity of the object squared. This means that when the velocity of an object doubles, its kinetic energy quadruples. A car traveling at 60 mph has four times the kinetic energy of an identical car traveling at 30 mph, and hence the potential for four times more death and destruction in the event of a crash.
Kinetic energy must always be either zero or a positive value. While velocity can have a positive or negative value, velocity squared is always positive.
Kinetic energy is not a vector. So a tennis ball thrown to the right with a velocity of 5 m/s, has the exact same kinetic energy as a tennis ball thrown down with a velocity of 5 m
In June 1985, a laser beam was sent out from the Air Force Optical Station on Maui and reflected back from the shuttle Discovery as it passed by 354 km overhead. The diameter of the central maximum of the beam at the shuttle position was 500 nm. What is the effective diameter of the circular laser aperture at the Maui ground station
This question is incomplete, the complete question is;
In June 1985, a laser beam was sent out from the Air Force Optical Station on Maui and reflected back from the shuttle Discovery as it passed by 354 km overhead. The diameter of the central maximum of the beam at the shuttle position was said to be 9.1 m, and the beam wavelength was 500 nm.
What is the effective diameter of the circular laser aperture at the Maui ground station
Answer:
the effective diameter of the circular laser aperture at the Maui ground station is 4.747 cm
Explanation:
Given the data in the question;
Separation between observer and point L = 354 km = 354000 m
Linear separation D = 9.1 m
wavelength λ = 500 nm = 500 × 10⁻⁹ m
Now, for small angles;
θ = D / L
θ = 9.1 m / 354000 m
θ = 2.57 × 10⁻⁵ rad
For a circular aperture;
sinθ = ( 1.22 × λ ) / d
for small angles;
θ = ( 1.22 × λ ) / d
so
θ = 2 × θ
θ = 2 × [( 1.22 × λ ) / d]
we substitute
2.57 × 10⁻⁵ = 2 × [( 1.22 × 500 × 10⁻⁹ ) / d]
2.57 × 10⁻⁵ = 0.00000122 / d
d = 0.00000122 / 2.57 × 10⁻⁵
d = 0.04747 m
d = ( 0.04747 × 100 )m
d = 4.747 cm
Therefore, the effective diameter of the circular laser aperture at the Maui ground station is 4.747 cm
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]
An LC circuit is built with a 20 mH inductor and an 8.0 PF capacitor. The capacitor voltage has its maximum value of 25 V at t = 0 s.(a)How much time does it take until the capacitor is fully discharged for the first time? (b)What is the inductor current at that time?
Answer:
a) the required time is 0.6283 μs
b) the inductor current is 0.5 mA
Explanation:
Given the data in the question;
The capacitor voltage has its maximum value of 25 V at t = 0
i.e V[tex]_m[/tex] = V₀ = 25 V
we determine the angular velocity;
ω = 1 / √( LC )
ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )
ω = 1 / √( 1.6 × 10⁻¹³ )
ω = 1 / 0.0000004
ω = 2.5 × 10⁶ s⁻¹
a) How much time does it take until the capacitor is fully discharged for the first time?
V[tex]_m[/tex] = V₀sin( ωt )
we substitute
25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )
25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )
divide both sides by 25 V
sin( 2.5 × 10⁶ × t ) = 1
( 2.5 × 10⁶ × t ) = π/2
t = 1.570796 / (2.5 × 10⁶)
t = 0.6283 × 10⁻⁶ s
t = 0.6283 μs
Therefore, the required time is 0.6283 μs
b) What is the inductor current at that time?
[tex]I[/tex](t) = V₀√(C/L) sin(ωt)
{ sin(ωt) = 1 )
[tex]I[/tex](t) = V₀√(C/L)
we substitute
[tex]I[/tex](t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )
[tex]I[/tex](t) = 25 × 0.00002
[tex]I[/tex](t) = 0.0005 A
[tex]I[/tex](t) = 0.5 mA
Therefore, the inductor current is 0.5 mA
The time taken for the capacitor to fully discharge is 6.28 x 10⁻⁷ s.
The current in the inductor at the given time is 0.0005 A.
Angular velocity of circuitThe angular velocity of the circuit is calculated as follows;
[tex]\omega = \frac{1}{\sqrt{LC} } \\\\\omega = \frac{1}{\sqrt{20 \times 10^{-3} \times 8 \times 10^{-12} } } \\\\\omega = 2.5 \times 10^6 \ rad/s[/tex]
Time for the capacitor to fully dischargeV = V₀(sinωt)
sinωt = V/V₀
sinωt = = 25/25
sin(ωt) = 1
ωt = sin⁻¹ (1)
ωt = π/2
2.5 x 10⁶ t = π/2
t = 1.57 / (2.5 × 10⁶)
t = 6.28 x 10⁻⁷ s
Inductor current at the given timeThe current in the inductor at the given time is calculated as follows
[tex]I(t) = V_0 \sqrt{\frac{C}{L} } \\\\I = 25 \times \sqrt{\frac{8\times 10^{-12}}{20 \times 10^{-3}} } \\\\I = 0.0005 \ A[/tex]
Learn more about inductor current here: https://brainly.com/question/4425414
A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if the length of the runway is 2.00 km.ii) At this acceleration, how much time would the plane need from starting to takeoff. iii) What force must the engines exert to attain this acceleration
Answer:
i) the minimum acceleration to take off is 22500 km/h²
ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs
iii) required force that the engine must exert to attain acceleration is 625 kN
Explanation:
Given the data in the question;
mass of plane m = 360,000 kg
take of speed v = 300 km/hr = 83.33 m/s
i)
What should be the minimum acceleration to take off if the length of the runway is 2.00 km
from Newton's equation of motion;
v² = u² + 2as
we know that a plane starts from rest, so; u = 0
given that distance S = 2 km
we substitute
(300)² = 0² + ( 2 × a × 2 )
90000 = 4 × a
a = 90000 / 4
a = 22500 km/h²
Therefore, the minimum acceleration to take off is 22500 km/h²
ii) At this acceleration, how much time would the plane need from starting to takeoff.
from Newton's equation of motion;
v = u + at
we substitute
300 = 0 + 22500 × t
t = 300 / 22500
t = 0.0133 hrs
Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs
iii) What force must the engines exert to attain this acceleration
we know that;
F = ma
acceleration a = 22500 km/hr² = 1.736 m/s²
so we substitute
F = 360,000 kg × 1.736 m/s²
F = 624960 N
F = 625 kN
Therefore, required force that the engine must exert to attain acceleration is 625 kN
A 3.7-kg object is acted on by two forces. One of the forces is 11 N acting toward the
east. What is the other force if the acceleration of the object is 1.0 m/s2 toward the east?
Answer:
F₂ = -7.3 N
Explanation:
Given that,
The mass of an object, m₁ = 3.7 kg
First force, F₁ = 11 N
The net acceleration of the object is 1 m/s².
We know that,
F₁+F₂ = ma
11+F₂ = (3.7)(1)
F₂ = 3.7-11
F₂ = -7.3 N
so, the other force is 7.3 N and it is acting in west direction.
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]
Lab: Energy Transfer Instructions Click the links to open the resources below. These resources will help you complete the assignment. Once you have created your file(s) and are ready to upload your assignment, click the Add Files button below and select each file from your desktop or network folder. Upload each file separately. Your work will not be submitted to your teacher until you click Submit. Documents Descriptive Lab Report Guide Descriptive Lab Report Rubric
Answer:
The second one says cant open file............
What is Velocity in physics
Answer:
hii
Explanation:
i hope this helps you
Answer:
The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. ... Velocity is a physical vector quantity; both magnitude and direction are needed to define it
Explanation:
hope it helps
pls maek me as brainliest thanks❤
How does gravity affect your ability to live on a planet?
Two resistors, A and B, are connected in parallel across a 6.0-V battery. The current through B is found to be 2.0 A. When the two resistors are connected in series to the 6.0- V battery, a voltmeter connected across resistor A measures a voltage of 4.0 V. Find the resistances of A and B
Answer:
The resistance of A is 6 ohms and the resistance of B is 3 ohms
Explanation:
Step 1: For the first connection (parallel connection), the resistance of B will be calculated.
Note: in a parallel connection, the voltage through each resistor is the same.
[tex]V = I_AR_A = I_BR_B\\\\R_B = \frac{V}{I_B} = \frac{6}{2} = 3 \ ohms[/tex]
Step 2: The resistance of A will be calculated from the second connection (series connection)
Note: in series connection, the current flowing in each resistor is the same
[tex]V = V_A + V_B\\\\V = IR_A + IR_B\\\\The \ voltage \ drop \ in \ B; \ V_B = V- V_A\\\\V_B = 6 - 4 = 2 \ V\\\\IR_B = 2\ V\\\\I = \frac{2 \ V}{R_B}= \frac{2}{3} \ A\\\\The \ resistance \ of \ A \ is \ calculated \ as ;\\\\IR_A = 4 \ V\\\\R_A = \frac{4}{I} = \frac{4 \times 3}{2} = 6 \ ohms[/tex]
Hold a small piece of paper (e.g., an index card) flat in front of you. The paper can be thought of as a part of a larger plane surface. What single line could you use to specify the orientation of the plane of the paper (i.e., so that someone else could hold the paper in the same, or in a parallel, plane)?
Answer:
NORMAL
Explanation:
In mathematics to define a plane you need a line perpendicular to the plane called NORMAL and the point of application of this line at a point on the plane.
Based on this definition you only need to specify the normal plane is perpendicular to this line, it should be noted that this does not define a single plane but the whole family of plane is contains the normal.
Consequently only the NORMAL is needed
A 1.2-kg mass suspended from a spring of spring constant 22 N.m-1 executes simple harmonic motion of amplitude 5 cm. Assuming that the mass is at the equilibrium posiiton at t = 0, what is its displacement at t = 1.0 s?
Answer:
[tex]d =3.7*10^{-3} m[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=1.2kg[/tex]
Spring constant [tex]\mu=22Nm^{-1}[/tex]
Amplitude [tex]A=5cm=0.05m[/tex]
Generally the equation for displacement d is mathematically given by
[tex]d = Asin(\omega t)[/tex]
Where
[tex]\omega=angular\ velocity[/tex]
[tex]\omega=\sqrt{k/m}[/tex]
[tex]\omega=\sqrt{22/1.2}[/tex]
[tex]\omega=4.2817rads^{-1}[/tex]
Therefore
[tex]d = 0.05*sin(4.2817*1)[/tex]
[tex]d =3.7*10^{-3} m[/tex]
1. What different types of shots are taken on the basketball court?
Answer:
Here are a few commonly used types of shooting in basketball.
Jump Shot. A jump shot is most frequently used for a mid to long-range shots, including shooting beyond the arc. ...Hook Shot. A hook shot is when the shot is made while your body is not directly facing the basket. ...Bank Shot. ...Free Throw. ...Layup. ...Slam Dunk.To calculate the final enthalpy of the overall chemical equation, which step must occur?
Answer:
Explanation:
Reverse the second equation, and change the sign of the enthalpy
As part of a safety investigation, two 1300 kg cars traveling at 17 m/s are crashed into different barriers. Find the average forces exerted on:
a. the car that hits a line of water barrels and takes 1.5 s to stop
b. the car that hits a concrete barrier and takes 0.10 s to stop.
Answer:
a. F = 14,733.33 N
b. F = 221,000 N
Explanation:
Given;
mass of the cars, m = 1300 kg
velocity of the cars, v = 17 m/s
time taken for the first car to stop after hitting a barrier, t = 1.5 s
time taken for the second car to stop after hitting a barrier, t = 0.1 s
The average forces exerted on each car is calculated as follows;
a. the car that hits a line of water barrels and takes 1.5 s to stop
[tex]F = ma = m\times \frac{v}{t} = 1300 \times \frac{17}{1.5} = 14,733.33 \ N\\\\F = 14,733.33 \ N[/tex]
b. the car that hits a concrete barrier and takes 0.10 s to stop
[tex]F = ma = m\times \frac{v}{t}= 1300 \times \frac{17}{0.1} = 221,000 \ N\\\\F = 221,000 \ N[/tex]
You are using a constant force to speed up a toy car from an initial speed of 6.5 m/s
to a final speed of 22.9 m/s. If the toy car has a mass of 340 g, what is the work
needed to speed this car up?
By the work-energy theorem, the total work done on the car is equal to the change in its kinetic energy:
W = ∆K
W = 1/2 (0.34 kg) (22.9 m/s)² - 1/2 (0.34 kg) (6.5 m/s)²
W ≈ 82 J
Why can’t we see the molecules moving in a solid or liquid?
Answer:
Because molecules are to small for us people to see with the bare eye, unless you use a telescope.
Explanation:
Explanation:
Because in solid form, the molecules stood in place, not moving. In liquid state, the molecules move slowly but not as fast as air molecules.
: Ánh nắng mặt trời có cường độ đồng đều với bước sóng nằm trong vùng khả khiến 430nm-690nm đến đập vuông góc với một bản mỏng nước có bề dày 320nm, chiết suất 1,33 lơ lửng trong không khí. Tìm bước sóng thích hợp để ánh sáng phản từ bản mỏng là sáng nhất đối với người quan sát
20 points and brainliest‼️‼️‼️‼️
A 4.88 x 10-6 C charge moves 265 m/s
parallel (at 0°) to a magnetic field of
0.0579 T. What is the magnetic force
on the charge?
Answer:
0 N
Explanation:
Applying,
F = qvBsin∅................. Equation 1
Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.
From the question,
Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°
Substitute these values into equation 1
F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)
Since sin0° = 0,
Therefore,
F = 0 N
When a mass of 3.0-kg is hung on a vertical spring, it stretches by 0.085 m. Determine
the period of oscillation of a 4.0-kg object suspended from this spring.
Answer:
the period of oscillation of the given object is 0.14 s
Explanation:
Given;
mass of the object, m = 3 kg
extension of the spring, x = 0.085 m
The spring constant is calculated as follows;
[tex]F = mg = \frac{1}{2} ke^2\\\\2mg = ke^2\\\\k = \frac{2mg}{e^2} \\\\k = \frac{2\times 3 \times 9.8}{(0.085)^2} \\\\k = 8,138.41 \ N/m[/tex]
The angular speed of a 4 kg object is calculated as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\frac{2\pi }{T} = \sqrt{\frac{k}{m} } \\\\T= 2\pi \sqrt{\frac{m}{k} } \\\\T = 2\pi \sqrt{\frac{4}{8138.41} }\\\\T = 0.14 \ s[/tex]
Therefore, the period of oscillation of the given object is 0.14 s