Question
A radio station transmitted a radio wave with frequency of 25MHz the value of the frequency is equivalent to? Leave answer in Hz and in standard form.
Answer:
2.5 x 10⁷Hz
Explanation:
The measured frequency is 25MHz
To convert this to Hz, we simply change the M, which means mega, to its equivalent value which is 10⁶.
Therefore, we have;
25MHz = 25(10⁶) Hz
=> 25MHz = 25 x 10⁶ Hz
Now express this value in standard form as follows;
25 x 10⁶ Hz = 2.5 x 10¹ x 10⁶
=> 25 x 10⁶ Hz = 2.5 x 10⁷ Hz
Therefore, the value of 25MHz is equivalent to 2.5 x 10⁷Hz
The value of the frequency of the radio wave is equivalent to 2.5 × 10⁷ Hertz.
Given the data in the question;
Frequency of radio wave; [tex]f = 25MHz[/tex]
Frequency is the number of occurrences of a repeating event per unit of time. The SI unit for frequency is the hertz (Hz)
We know that;
[tex]1MHz\ ( Megahertz) = 1000000 Hz\ ( Hertz )[/tex]
so
[tex]25MHz = x[/tex]
We cross multiply
[tex]x = \frac{25MHz\ *\ 1000000Hz}{1MHz} \\\\x = 2.5 * 10^7 Hz[/tex]
Therefore, the value of the frequency of the radio wave is equivalent to 2.5 × 10⁷ Hertz.
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How can scientific phenomena be used in design?
CAN SOMEONE PLS HELP ME ILL GIVE YOU BRAINLIEST!!!!
A 7000 N braking force is used to slow a car down. The car was traveling at 25 m/s before the
brakes were applied and was traveling at 5 m/s after 6 seconds. What is the mass of the car? (ANS:
2100 kg)
Answer:
the answer is already there lol
Explanation:
Answer:
answer is there you can see from there lol
Explanation:
A certain AM radio wave has a frequency of 24Hz. Given that radio waves travel at 3 m/s, what is the wavelength of this radio wave?
Answer:
Wavelength of AM radio wave = 0.125 meter
Explanation:
Given:
Frequency of AM radio wave = 24 Hz
Velocity of AM radio wave = 3 m/s
Find:
Wavelength of AM radio wave
Computation:
Wavelength = Velocity / Frequency
Wavelength of AM radio wave = Velocity of AM radio wave / Frequency of AM radio wave
Wavelength of AM radio wave = 3 / 24
Wavelength of AM radio wave = 0.125 meter
Determine the force exerted by two particles that have a charge of +0.35 Coulombs each and are 1 meter apart
The electric force between two charges is:
F = (9 x 10⁹) Q₁ Q₂ / D²
F is the force, in Newtons
Q₁ and Q₂ are the two charges, in Coulombs
D is the distance between them, in meters
For these two particles:
F = (9 x 10⁹) (0.35) (0.35) / (1)²
F = (9 x 0.35 x 0.35 x 10⁹) / (1)
F = 1.10 x 10⁹ Newtons
Thatsa lotta force . . . like 124 thousand tons !
The reason it's so big is because the charges in this question are so big ... 0.35 Coulombs each. 1 Coulomb is a huge amount of charge.
Each of the particles feels the same force, pushing it away from the other particle. (The electric force between two charges is always the same in both directions, just like the gravitational force between two masses.)
Acc. to columb's law;
F = k × Q1Q2/r²Clearly, from the question you have:
Charge on both particles = +0.35 columbsDistance between them = 1 meter = 1 × 10-¹⁰ ÅNow just put them up in the formula:
F = k × Q1Q2/r²
F = 9 × 10⁹ × 0.35 × 0.35/1 × 10-¹⁰
F = 9 × 10⁹ × 35 × 35/ 10-¹⁰ × 100 × 100
F = 9 × 10⁹ × 35 × 35/ 10-¹⁰ × 10⁴
F = 9 × 10⁹ × 35 × 35/ 10-⁶
F = 9 × 10⁹+⁶ × 1225
F = 11025 × 10¹⁵
The diameter of a copper wire is thought to be approximately 0.3mm. Which instrument should be used to obtain a more accurate measurement of the diameter of the wire? A measuring tape B metre rule C micrometer D ruler 2
Answer:
C
Explanation:
The correct answer should be the micrometer.
The micrometer is an instrument that is capable of measuring the component of an object to very high precision. It is used to obtain accurate measurements in engineering.
In the case of the copper wire, the diameter of 0.3 mm was an approximation and a more accurate measure of the diameter would be obtained by using the micrometer.
The correct option is C.
Some people can roll the tongue into a tube shape, while others cannot. Practicing is not likely to help because tongue-rolling is an inherited trait. The allele for rolling the tongue (R) is dominant to the allele for not being able to roll the tongue (r). Look at the Punnett square. It shows a cross between two parents with at least one R allele. Drag each option to its spot to correctly finish the Punnett square.
Hope this helps you.
Calculate the force that is applied on this 0.25 kg ball, as it goes down to hit the ground.
(given gravity= 9.8 m/s2).
Answer:
F = 2.45 N
Explanation:
m = 0.25 kg
a = 9.8 m/s^2
F = ma
F = (0.25)(9.8)
F = 2.45 N
Can someone help me :)
Answer:
undergone a chemical change
Explanation:
Answer:
undergone a chemical change
A rock from the top of a hill is falling from rest.
Object name?
Starting energy form?
Conversion?
Final energy form?
Non usable form of energy?
Object name:- Rock
Starting energy:- Kinetic energy
Conversion:- Potential to kinetic energy
Final energy form:- Potential energy
Non usable form if energy:- Nil [ As it has potential & kinetic both]
HELP ME ILL MARK U THE BRAINLIEST
the crane takes 2.5s to raise the box 3.0 m. Calculate the power output of the crane
Answer:
Power = 1800 W
Explanation:
Power = work / time
Power = 4500 / 2.5
Power = 1800 W
define pressure snd state its SI units to
Answer:
Pressure (P) is the amount of force applied perpendicular to the surface of an object per unit area. The SI unit of pressure is the pascal (Pa), which is equivalent to N/[tex]m^{2}[/tex] (newtons per meter squared).
A 0.2 kg block sliding on a horizontal table slows down from 25 m/s to 20 m/s. How much energy does the block lose due to friction?
Answer:
the kinetic energy lost due to friction is 22.5 J
Explanation:
Given;
mass of the block, m = 0.2 kg
initial velocity of the block, u = 25 m/s
final velocity of the block, v = 20 m/s
The kinetic energy lost due to friction is calculated as;
[tex]\Delta K.E= K.E_f - K.E_i\\\\\Delta K.E= \frac{1}{2}mv^2 - \frac{1}{2}mu^2\\\\\Delta K.E= \frac{1}{2}m(v^2 -u^2)\\\\\Delta K.E= \frac{1}{2} \times 0.2 (20^2 - 25^2)\\\\\Delta K.E= -22.5 \ J[/tex]
Therefore, the kinetic energy lost due to friction is 22.5 J
Answer:
22.5 J
Explanation:
It loses the difference in kinetic energy between the starting and ending speeds and that is (1/2)(M)(V1)^2 - (1/2)(M)(V2)^2 = (1/2)(M)((V1)^2 -(V2)^2) = (1/2)(0.2 kg)((25)^2 - (20)^2) = 22.5 Joules.
Your friends sit in a sled in the snow. If you apply a force of 95N to them they have an acceleration of 0.8 m/s2. What is their mass
Mrs. Botts applies brakes on a car help it to decelerate at the rate of -0.80m/s2. What distance is required to stop the car when it is moving 17 m/s.
Answer:
Explanation:
vf=0
vi = 17
a = -0.8
Δx = ?
vf^2 = vi^2 + 2 a Δx
Δx = (vf^2 - vi^2) / 2a
Δx = (0-17^2) / 2 (-0.8)
Δx = 180.625 m
Courtney wants to practice doing science. Which example best illustrates her
using inquiry?
A. Courtney decides to drop several different objects off her
apartment's porch to see which falls the fastest.
B. Courtney pays close attention in her physical science class to see
if her teacher explains the speed of falling objects.
C. Courtney reads about the history of the study of gravity and learns
about why scientists first studied falling objects.
D. Courtney finds a calculation that explains the speed of falling
objects. She tries to understand the calculation by writing it down.
Answer:
A
Explanation:
because she wants to practice science and A is the practical option.
hich of these best describes Earth’s mantle?
Most dense layer, consists of the outer and inner parts
Thinnest under the oceans and thickest under continents
Middle layer, density increases with depth as pressure increas
Answer:
Middle layer, density increases with depth as pressure increases
Explanation:
The Earth's mantle is the middle layer of the earth. It is the largest layer of the earth consisting of about 83 % of the earth's volume and continues to a depth of 2900 km. Also, the mantle is hot and made of solid rock. It is hot since it is close to the core and its pressure and density increases with depth causing it to be more solid.
So, the Earth's mantle is the middle layer, density increases with depth as pressure increases.
A hockey player uses a hockey stick to hit a puck such that the stick provides an applied force on the puck The puck travels for distance of 0.85 m while experiencing the force from the stick
The puck leaves the stick with a speed of 8.0 m/s and travels at constant speed in a straight line along the horizontalice for a distance of 12 m. The frictional force between the puck and the ice
surface is negligible
Which of the following statements best describes the force or forces exerted on the puck after it is hit by the stick and slides on the ice?
(A) The applied force from the stick is still exerted on the puck during this time.
Since the puck's acceleration is zero, there are no forces exerted on the puck during this time.
Since there is no frictional force exerted on the puck, a normal force is not exerted on the puck, but the gravitational force is exerted on the puck
There is no frictional force exerted on the puck and the gravitational force and normal force are exerted on the puck and are equal in magnitude
Answer:
Explanation:
Let's analyze the situation presented in order to know which answer is correct.
When the stick collides with the puck, it exerts a force for a certain time and discants. / After this time the horizontal force decreases to zero and the disk continues to move by the action of the initial velocity on the x axis and the acceleration of gravity on the y axis.
Therefore, after the collision, the only force that acts on the disk is the gravitational attractive force (WEIGHT), directed on the axis and in a negative direction.
The correct answer is:
C) Since there is no frictional force exerted on the puck, a normal force is not exerted on the puck, but the gravitational force is exerted on the puck
The speed of a car cruising along a highway was 90.0km h^-1. What is this value in meters per second?
Answer:
25 m/s
Explanation:
1 km/h = .277778 m/s, so 90 x .277778 = 25
Pls help it’s really late <3
Answer:
22.4%
Explanation:
The plane is 300000 kg
It is 11000 meters about the earth's surface.
g = 9.81
PE = mgh
PE = 300000 * 9.81 * 11000
PE = 3.24* 10^10
The Kinetic Energy
KE = 1/2 m * v^2
m = 300000
v = 250 m/s
KE = 1/2 300000 * (250 m/s)^2
KE = 9.375 * 10^9
Total Energy = 3.24 * 10^10 + 9.375 * 10^9
Total Energy = 4.17 * 10^10
9.375 * 10^9
=========== * 100 % = 22.44 %
4.17 * 10^10
Craig is modeling the discovery of electromagnetism. Which procedure should he use?
A. moving a magnet into a coil of wire in a closed circuit
B. moving a magnet into a coil of wire in an open circuit
C. bringing a compass near a wire that has no electric current
D. bringing a compass near a wire that has an electric current
Answer:
A
Explanation:
moving a magnet into a coil of wire in a closed circuit
Craig, move a magnet into a coil of wire in a closed circuit for the discovery of electromagnetism. Option A is correct.
What is electromagnetism ?The physical interplay of electric charges, magnetic moments, and the electromagnetic field is known as electromagnetism.
The electromagnetic field can be static, moving slowly, or forming waves.
Craig, move a magnet into a coil of wire in a closed circuit for the discovery of electromagnetism.
Hence, option A is correct.
To learn more about the electromagnetism, refer to the link;
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A steel railroad track has a length of 79 when the temperature is 2◦C. What is the increase in the length of the rail on a hot day when the temperature is 35◦C? The linear expansion coefficient of steel is 11 × 10−6 (◦C)−1.
Answer:
We know that the length is 79 (there are no units of length in the question, so i will leave it as it is written, without units) when the temperature is 2°C.
We want to find the increase in length when the temperature is 35°C, knowing that the linear expansion coefficient is 11*10^(-6) °C^-1
The change in length is just given by the equation:
[tex]\Delta L = \alpha*L_0* (T_1 - T_0)[/tex]
where:
α = 11*10^(-6) °C^-1
T₀ is the initial temperature, so T₀ = 2°C
T₁ is the final temperature, so T₁ = 35°C
L₀ is the initial length, so L₀ = 79
Replacing these in the equation we get:
ΔL = ( 11*10^(-6) °C^-1)*79*(35°C - 2°C)
= 0.28677
So the length of the railroad will increase by 0.28677
Describe how a wind turbine generates energy.
when two capacitors are connected in series, the effective capacitance is 2.4muF and when connected in parallel, the effective capacitance is 10muF. calculate the individual capacitances.
Answer:
let one capacitor be x and other be y
[tex]in \: parallel \: connection : \\ \frac{x + y}{xy} = 10 \times {10}^{ - 6} - - - (a) \\ in \: series \: connection : \\ x + y = 2.4 \times {10}^{ - 6} - - - (b) \\ in \: (a) : \\ \frac{2.4 \times {10}^{ - 6} }{xy} = 10 \times {10}^{ - 6} \\ xy = 0.24 - - - (c) \\ from \: (b) : \\ y = (2.4 \times {10}^{ - 6} ) - x \\ \therefore \: in \: (c) : \\ x(2.4 \times {10}^{ - 6} - x) = 0.24 \\ 2.4 \times {10}^{ - 6} x - {x}^{2} = 0.24 \\ {x}^{2} - 2.4 \times {10}^{ - 6} x - 0.24 = 0 \\ x = 0.5 \\ 0.5y = 0.24 \\ y = 0.48 \\ { \boxed{ C _{1} = 0.5 \: farads}} \\ { \boxed{C _{2} = 0.48 \: farads}}[/tex]
2. A student drew the diagram below to model the movement of an object orbiting the Sun. Which object was she most likely modeling? a meteor a planet a comet an astroid
Answer:
A comet
Explanation:
The picture is of an elliptical orbit of a comet. You can tell it is a comet because of the tail. Also, I just took the test and it was a comet.
as light passes from water into air it has a refracted angle of 35 degrees. calculate the incident angle of the light.
A first aid provider is trained to do the following except:
A) diagnose the victims condition
B) recognize, asses and prioritize the need for first aid
C) seek professional medical assistance when necessary
D) provide appropriate first aid care
Explique cómo es que las estrellas emiten una radiación, específicamente qué es lo que sucede a nivel atómico
Answer:
Las estrellas se forman debido a la unión de pequeños elementos y partículas como el hidrógeno y el helio que experimentan una fusión nuclear en el núcleo debido a la intensa presión para formar un nuevo átomo (como dos hidrógeno que se combinan para formar un átomo de helio) que es un proceso. que requiere una gran cantidad de energía para comenzar, por lo que cada reacción de fusión produce una gran cantidad de energía que puede mantener la reacción y equilibra la fuerza gravitacional que atrae las masas de las partículas hacia el centro de la estrella
A medida que la energía de la estrella se irradia, la superficie de la estrella cede la energía en forma de calor y luz de manera que debido a la alta temperatura de las partículas de los átomos de los elementos se mueven muy rápida y energéticamente para emitir fotones rápidamente. (alta frecuencia) que se ve como el alto brillo de la estrella
Explanation:
Do only number 3 and thank
Charge = 0.2 Ah is the correct answer...
how do icebergs form
Answer:
They break off large ice sheets found at the North and South poles.
Answer:
The answer is D
Explanation:
They break off large ice sheets found at the poles.
Most ozone is found in a region of Earth’s
atmosphere between 10 and 20 miles above
Earth’s surface. This temperature zone of the
atmosphere is known as the
(1) thermosphere (3) stratosphere
(2) mesosphere (4) troposphere
Answer:
It’s stratosphere
Explanation:
Most ozone is found in a region of Earth’s atmosphere between 10 and 20 miles above Earth’s surface. This temperature zone of the atmosphere is known as the stratosphere.
What is ozone?Ozone is an allotrope of oxygen. It is blue in color and it is found in Earth's atmosphere. It protects the Earth from ultraviolet rays from the Sun.
Most ozone is found in the following layer of the Earth's atmosphere:The ozone subcaste is centered at an altitude between 10-15 country miles (15-25 km). Close to 90% of the ozone found in the atmosphere is in the stratosphere. Ozone that is found in this region is about 10 parts per million by volume (ppmv) to the troposphere's 0.04 ppmv.
Therefore, we have determined that Most ozone is found in a region of Earth’s atmosphere between 10 and 20 miles above Earth’s surface. This temperature zone of the atmosphere is known as the stratosphere.
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