Suppose a popular FM radio station broadcasts radio waves with a frequency of 96. MHz. Calculate the wavelength of these radio waves. Be sure your answer has the correct number of significant digits.

Answers

Answer 1

Answer:

3.125 meters.

Explanation:

(3.0*10^8)/(96*10^6)

= 3.125 meters.

Hope this helped!


Related Questions

Which of the following is the main idea of Thomas Paine's "Common Sense"?

Answers

D. The American colonists should focus their efforts on getting representation in the
British Parliament

A simple pendulum has a period T on Earth. If it were used on Planet X and found to have a period of only 71% of T, what would be the free fall acceleration due to gravity on this planet X?

Answers

Answer:

gₓ = 1.4 g = 13.8 m/s²

Explanation:

The time period of simple pendulum on earth is given as:

T = 2π√(L/g)

where,

L = Length of pendulum

g = acceleration due to gravity on the surface of earth = 9.8 m/s²

T = Time Period on Earth

but, on the other planets, it becomes:

Tₓ = 2π√(L/gₓ)

where,

Tₓ = Time Period on Surface of Other Planet

gₓ = acceleration due to gravity on other planet  = ?

It is given that:

Tₓ = 71% of T

Tₓ = 0.71 T

using values:

2π√(L/gₓ) = (0.71) 2π√(L/g)

√(1/gₓ) = (0.71)√(1/g)

squaring both sides:

1/gₓ = 0.71/g

gₓ = g/0.71

gₓ = 1.4 g = 13.8 m/s²

You are a guest magician in a circus. One of your tricks is to place a football on an inclined plane without the football rolling over. How can you achieve this?
Determine if the weight of the football is needed for you to prevent the football from rolling over. The coefficient of static friction of the plane is 0.067.

Answers

Answer:

You are a guest magician in a circus. One of your tricks is to place a football on an inclined plane without the football rolling over is explained below in details.

Explanation:

spinning ball halts after traveling some range due to friction energy act different direction of movement of the ball. you can observe in the figure.

Let any rolling ball of mass (m ) is traveling with velocity v ,

common effect on ball (N) = mg

because of motion, friction energy develops on the contact exterior and begins to resist the movement of the rolling ball.

hence,

fr = uN = umg act on communicating exterior, so, after any time due to friction energy rolling ball gets to rest.

SI unit differ from one country to another . true or false​

Answers

Answer:

false ..........false

Answer:

FALSE                                            

Explanation:

A ball is kicked off the ground reaching a maximum height of 60m and lands 80m away. Calculate the initial speed and the angle above the horizontal of the ball when it was kicked​

Answers

Answer:

36.87°

Explanation:

Given

Maximum height = 60m

Horizontal distance (range) = 80,m

Required

Initial speed U

Angle of launch

To get the speed, we will use the range formula;

R = U √2H/g

80 = U√2(60)/9.8

80 = U√12.25

80 = 3.5U

U = 80/3.5

U = 22.86m/s

Get the angle of launch

Using the formula

Theta = tan^-1(y/x)

y is the vertical distance

x is the horizontal distance

Theta = tan^-1(60/80)

Theta = tan^-1(0.75)

Theta = 36.87°

Hence the angle of launch is 36.87°

Find an expression for the kinetic energy of the car at the top of the loop.Express the kinetic energy in terms of m, g, h, and R.

Answers

Answer:

K.E₂ = mg(h - 2R)

Explanation:

The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:

K.E₁ + P.E₁ = K.E₂ + P.E₂

where,

K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)

P.E₁ = Initial Potential Energy = mgh

K.E₂ = Final Kinetic Energy at the top of the loop = ?

P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)

Therefore,

0 + mgh = K.E₂ + mg(2R)

K.E₂ = mg(h - 2R)

The kinetic energy of the car is expressed as K.E = mg(h - 2R). The energy of the object by the integrity of its movement is understood as kinetic energy.

What is kinetic energy?

The energy of the body by the virtue of its motion is known as the kinetic energy of the body. It is defined as the product of half of mass and square of velocity.

The following data is observed from the figure;

K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0

P.E₁ = Initial Potential Energy = mgh

K.E₂ = Final Kinetic Energy at the top of the loop =?

P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)

According to the law of conservation of energy, total energy is defined as the sum of kinetic energy and potential energy.

Total energy = kinetic energy+potential energy

On applying the law  of conservation of energy for top and bottom of the loop;

K.E₁ + P.E₁ = K.E₂ + P.E₂

0 + mgh = K.E₂ + mg(2R)

K.E₂ = mg(h - 2R)

Hence the kinetic energy of the car is expressed as K.E = mg(h - 2R).

To learn more about kinetic energy refer to the link;

brainly.com/question/999862

A radio station can be heard if the receiver is tuned to a frequency of 6x10^5 Hz. What is the wavelength of the radio waves

Answers

Answer:

5 * 10² m

Explanation:

We are Given:

Frequency of the wave = 6 * 10⁵ Hz

Wavelength of the wave:

We know the relation:

c = λν  

[where λ (Lambda) is the wavelength ,ν (nu) is the frequency and c is the speed of light ]

3 * 10⁸  m/s= (6 * 10⁵ )* λ                          [replacing known values]

λ = [tex]\frac{3 * 10^{8}}{6 * 10^{5}}[/tex]                                                     [dividing both sides by 6 * 10⁵]

λ = 1/2 * 10³  

λ = 1/2 * 10 * 10²                                        [10³ can be rewritten as 10 * 10²]

λ = 5 * 10² m

Therefore, the wavelength of the wave is 5 * 10² m

5.

State the number of significant digits in each measurement and

express the value in scientific notation.

a. 110 m

f. 20.40 m/s

b. 0.2°C

g0.71 ns

c. 0.090 11 kg

h. 0.06 kg

d. 52.5 X 10 - 3

7040 m

600.0 N

וורן

e.

Answers

Answer:

Tanong mo po sa guro mo welcome pasensya need ko po ng point

a 1000kg car uses a breaking force of 10,000N to stop in two second. What impulse acts on the car?

Answers

Answer:

5,000

Explanation:

Vf = Vi + a * t

If the speed of an object in uniform circular motion is constant and the radial distance is doubled, by what factor does the magnitude of the radial acceleration decrease?

Answers

Answer: Half

Explanation:

Given

The object is in uniform circular motion with constant speed.

Radial acceleration of the object is given by

[tex]\Rightarrow a_r=\dfrac{v^2}{r}[/tex]

Where, [tex]a_r=\text{Radial acceleration}[/tex]

[tex]v=\text{speed}\\r=\text{distance from the axis of rotation}[/tex]

If the radial distance [tex]r[/tex] is doubled, i.e. [tex]2r[/tex]

Radial acceleration is

[tex]\Rightarrow a'_r=\dfrac{v^2}{2r}\\\\\Rightarrow a'_r=\dfrac{a_r}{2}[/tex]

Radial acceleration becomes half of the initial value.

Learn more: https://brainly.in/question/21217248

Two speeding lead bullets, one of mass 15.0 g moving to the right at 295 m/s and one of mass 7.75 g moving to the left at 375 m/s, collide head-on, and all the material sticks together. Both bullets are originally at temperature 30.0°C. Assume the change in kinetic energy of the system appears entirely as increased internal energy. We would like to determine the temperature and phase of the bullets after the collision. (Lead has a specific heat of 128 J/(kg K), a melting point of 327.3°C, and a latent heat of fusion of 2.45 104 J/kg.)

Answers

Answer:

The final temperature of the bullets is 327.3 ºC.

Explanation:

Let suppose that a phase change does not occur during collision and collided bullets stop at the end. We represent the phenomenon by the First Law of Thermodynamics:

[tex]K_{A, o} + K_{B, o}-K_{A}-K_{B}+U_{A,o} + U_{B,o}-U_{A}-U_{B} = 0[/tex] (1)

Where:

[tex]K_{A,o}[/tex], [tex]K_{A}[/tex] - Initial and final translational kinetic energies of the 15-g bullet, measured in joules.

[tex]K_{B,o}[/tex], [tex]K_{B}[/tex] - Initial and final translational kinetic energies of the 7.75-g bullet, measured in joules.

[tex]U_{A,o}[/tex], [tex]U_{A}[/tex] - Initial and final internal energies of the 15-g bullet, measured in joules.

[tex]U_{B,o}[/tex], [tex]U_{B}[/tex] - Initial and final internal energies of the 7.75-g bullet, measured in joules.

By definitions of translational kinetic energy and sensible heat we expand and simplify the equation above:

[tex]\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T) = 0[/tex] (2)

Where:

[tex]m_{A}[/tex], [tex]m_{B}[/tex] - Masses of the 15-g and 7.75-g bullets, measured in kilograms.

[tex]v_{A,o}[/tex], [tex]v_{A}[/tex] - Initial and final speeds of the 15-g bullet, measured in meters per second.

[tex]v_{B,o}[/tex], [tex]v_{B}[/tex] - Initial and final speeds of the 7.75-g bullet, measured in meters per second.

[tex]c[/tex] - Specific heat of lead, measured in joules per kilogram-Celsius degree.

[tex]T_{o}[/tex], [tex]T[/tex] - Initial and final temperatures of the bullets, measured in Celsius degree.

Now we clear the final temperature of the bullets:

[tex](m_{A}+m_{B})\cdot c \cdot (T-T_{o}) = \frac{1}{2}\cdot [m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})][/tex]

[tex]T-T_{o} = \frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}[/tex]

[tex]T= T_{o}+\frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}[/tex] (3)

If we know that [tex]T_{o} = 30\,^{\circ}C[/tex], [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 295\,\frac{m}{s}[/tex], [tex]v_{B,o} = 375\,\frac{m}{s}[/tex], [tex]v_{A} = 0\,\frac{m}{s}[/tex], [tex]v_{B} = 0\,\frac{m}{s}[/tex] and [tex]c = 128\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the final temperature of the collided bullets is:

[tex]T = 30\,^{\circ}C+\frac{(15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right)}[/tex]

[tex]T = 852.534\,^{\circ}C[/tex]

Given that found temperature is greater than melting point, then we conclude that supposition was false. If we add the component of latent heat of fussion, then the resulting equation is:

[tex]\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)-U = 0[/tex] (4)

[tex]U=\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)[/tex]

If we know that [tex]T_{o} = 30\,^{\circ}C[/tex], [tex]T = 327.3\,^{\circ}C[/tex], [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 295\,\frac{m}{s}[/tex], [tex]v_{B,o} = 375\,\frac{m}{s}[/tex], [tex]v_{A} = 0\,\frac{m}{s}[/tex], [tex]v_{B} = 0\,\frac{m}{s}[/tex] and [tex]c = 128\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then latent heat received by the bullets during impact is:

[tex]U =\frac{1}{2}\cdot (15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] + \frac{1}{2}\cdot (7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right) \cdot (30\,^{\circ}C-327.3\,^{\circ}C)[/tex][tex]U = 331.872\,J[/tex]

The maximum possible latent heat ([tex]U_{max}[/tex]), measured in joules, that both bullets can receive during collision is:

[tex]U_{max} = (m_{A}+m_{B})\cdot L_{f}[/tex] (5)

Where [tex]L_{f}[/tex] is the latent heat of fusion of lead, measured in joules per kilogram.

If we know that  [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex] and [tex]L_{f} = 2.45\times 10^{4}\,\frac{J}{kg}[/tex], then the maximum possible latent heat is:

[tex]U_{max} = (15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(2.45\times 10^{4}\,\frac{J}{kg} \right)[/tex]

[tex]U_{max} = 557.375\,J[/tex]

Given that [tex]U < U_{max}[/tex], the final temperature of the bullets is 327.3 ºC.

A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant deceleration, find the value of acceleration (in ft/sec2) that accomplishes this.

Answers

Answer:

The value of acceleration that accomplishes this is 8.61 ft/s² .

Explanation:

Given;

maximum distance to be traveled by the car when the brake is applied, d = 450 ft

initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s

final velocity of the car when it stops, v = 0

Apply the following kinematic equation to solve for the deceleration of the car.

v² = u² + 2as

0 = 88.02² + (2 x 450)a

-900a = 7747.5204

a = -7747.5204 / 900

a = -8.61 ft/s²

|a| = 8.61 ft/s²

Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .

A peregrine falcon dives at pigeon . the falcon start downward from rest with free-fall acceleration . if the pigeon is 76.0 m below the initial position of the falcon , how long the falcon take to reach the pigeon? Assume that the pigeon remains at rest .

Answers

Answer:

3.94 s

Explanation:

Take down to be positive.  Given:

Δy = 76.0 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

76.0 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 3.94 s

hi! hina can you help me with those 2 questions ​

Answers

Question: Differentiate between how fossil fuels and nuclear fuels heat. How can the heat be released by both types of fuels be put to good use?

Answer: Fossil fuels and nuclear fuels both can be released and be put in to good use because both produce green house gasses that help the environment by creating more heat that helps the atmosphere.

Question: Explain why it is important for us to find alternatives for fossil fuels to meet our energy needs?

Answer: It’s important for us to find alternatives for fossil fuels energy needs because more renewable energy can lower the prices and need more natural gas and coal by increasing competition, our energy supplies, and when being used it can increase renewable energy that protects consumers when using fossil fuels.

Two horizontal forces, 230 N and 120 N, are exerted in opposite direction on a crate. What is the horizontal acceleration of the crate if the mass of the crate is 20 kg?
5.5 m/s2
20 m/s2
17.5 m/s2
72 m/s2

Answers

Answer:

a = 5.5 [m/s²]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the force is equal to the product of mass by acceleration.

ΣF =m*a

where:

F = Force [N] (units of Newtons)

m = mass = 20 [kg]

a = acceleration [m/s²]

Let's take the force of 230 [N] as positive, in this way the other force will be negative, by pointing in the opposite direction.

[tex]230 - 120 = 20*a\\110 = 20*a\\a=5.5 [m/s^{2} ][/tex]

Aluminum wire with a diameter of 0.8650 mm is wound onto a spool. The wire is insulated, but you have access to both ends. The resistivity of aluminum at 20.0 °C is 2.65 x 10^-8 Ω-m. You measure the resistance of the wire at that temperature, and it is 2.48 Ω. What is the length of the wire?

a. 8.10 x 10^4 m
b. 22.0 m
c. 5.68 m
d. 0.111 m
e. 55.0 m

Answers

Answer:

e. 55.0 m

Explanation:

Given;

diameter of the aluminum wire, d = 0.865 mm

radius of the wire, r = d/2 = 0.4325 mm = 0.4325 x 10⁻³ m

resistivity of the wire, ρ = 2.65 x 10⁻⁸ Ω-m

resistance of the wire, R = 2.48 Ω

The resistance of a wire is given by;

[tex]R = \frac{\rho \ L}{A} \\\\[/tex]

where;

L is length of the wire

A is area of the wire = πr² = π(0.4325 x 10⁻³ )² = 5.877 x 10⁻⁷ m²

Substitute the givens and solve for L,

[tex]L = \frac{RA}{\rho} \\\\L = \frac{(2.48)(5.877*10^{-7})}{2.65*10^{-8}}\\\\L = 55.0 \ m[/tex]

Therefore, the length of the wire is 55.0 m

A potential difference of 5v is used to produce a current of 4A for 4 hours through a heating coil. What is the heat produced.
a 80j
b 4.8kj
c 20j
d 4800kj​

Answers

Answer:

Explanation:

Heat generated by the coil is expressed as;

Heat generated = Power ×Time

Heat generated = IVt

I is the current = 4A

V is the voltage= 5V

t is the time = 4hrs

Convert to secs

t = 4(3600) = 14400secs

Substitute into the formula

Heat generated = 4×5×14400

Heat generated = 20×14400

Heat generated = 288000Joules

Heat generated =288KJ

A 10.0 kg ball weighs 98.0 N in air and weighs 65.0 N when submerged in water. The volume of the ball is:_________.A) 0.00245 m3. B) 0.00337 m3. C) 0.00457 m3. D) 0.00766 m3 E) 0.00980 m3

Answers

Answer: B) 0.00337 m3.

Explanation:

Given data:

Mass of the ball = 10kg

Weight of the ball in air = 98N

Weight of the ball in water = 65N

Solution:

To get the Volume of the ball when submerged in water, we divide the weight of the ball in water with the difference in apparent weight by 9.8m/s^2.

= 98 - 65 / 9.8

= 33 / 9.8

= 3.37kg

The volume of the ball is 3.37kg

The density of water is 1kg per Liter.

So 3.37 kg of water would have a volume of 3.37 Liters.

Therefore the ball would have a volume of 3.37 Liters (or 0.00337 cubic meters).

Which current is produced in homes

Answers

Batteries Produce DC current while homes produce AC current.
Hope this helped xxxx

Answer:

answer is C on edge 2021

Explanation:

A large pizza is cut into 8 even slices. A person orders 4 large pizzas from a restaurant. How many total slices of pizza did the person order?

Answers

Answer:

32 slices

Explanation:

Step one:

given data

we are told that 1 large pizza can be cut into 8 even slices

Required

we want to find how many slices are there in 4 large pizzas

Step two:

so if 1 pizza has 8 slices

       4 pizza will have x

cross multiply we have

x= 8*4

x=32 slices

Which of the following statements describe a perfectly in elastic collision
A) two billiard balls bounce off each other
B) a wad of chewing gum is thrown and sticks to a moving truck
C)Two ice skaters hit each other and fall over in opposite directions
D) a car crashes into a tree and rebounds in the other direction

Answers

Answer:

B

Explanation:

How safe is Wind Energy? Use in your own words.

Answers

Answer:

wind energy is safe

Explanation:

because it just is

Answer:

Wind energy isn't perfect, but better than the alternatives. There is a risk of falling off a turbine in maitenence and there are instances where birds fly into the blades, however there are far more casualties as a result of coal and other fossil fuel extraction.

Explanation:

Heidi (39 kg) is walking her tiny chihuahua, Chaxi (5.60 kg), on the sidewalk. To encourage Chaxi along, Heidi pulls forward with a force of 9.55 N. Identify the correct reaction force in response to Heidi’s action force.

a. The friction is less than 660 N since the beam can be moved at a constant velocity.
b. There is no friction acting on the beam since it is accelerating.
c. The friction is equal to 660 N since the beam is not accelerating.
d. The friction is greater than 660 N since the beam is not in equilibrium.

Answers

Answer:

The correct reaction force in response to Heidi's action force is:

c. The friction is equal to 660 N since the beam is not accelerating.

Explanation:

Heidi's action force does not affect the beam.  Since friction resists the sliding or rolling of one solid object over another, there is no friction acting on the beam, in this respect.  The reaction force is what makes the dog to move because it acts on it.  According to Newton's Third Law of Motion, forces always come in action-reaction pairs.  This Third Law states that for every action force, there is an equal and opposite reaction force.  This means that the dog exerts some force on Heidi, as he pulls it "forward with a force of 9.55 N."

The the figure shows a famous roller coaster ride. You can ignore friction. If the roller coaster leaves Point Q from rest, what is its speed at the top of 25 m peak(Point s

Answers

Answer:

22 m/s

Explanation:

PEf +KEf =PE0 +KE0 →PE0 −PEf =KEf

−mgΔy= 1 mv2 →v= −2gΔy = −2(9.8 m/s2)(−25 m)=22 m/s

A rock dropped from a 5 m height accelerates at 10 m/s and strikes the ground 1 s later. If the rock is dropped

from a height of 2,5 m, its acceleration of fall is

Answers

Answer:

10 m/s²

Explanation:

The above question simply indicates motion under gravity.

The acceleration due to gravity (i.e acceleration of free fall) has a constant value of 10 m/s².

Whether the rock is dropped from a height of 5 m or 2.5 m, it will accelerate at 10 m/s² before striking the ground. The only thing that will be different is the time taken for the rock to strike the ground when released from both 5 m and 2.5 m.

Thus, the rock will have a constant acceleration of 10 m/s² irrespective of the height to which it was released.

Since acceleration due to gravity is a constant,  the acceleration of the rock dropped from the 5 m height is the same as that dropped from the 2.5 m height and is equal to 10 m/s²

What is the acceleration due to gravity?

Acceleration due to gravity is the acceleration a body falling freely from a height above the earth surface which a body experiences due to the gravitational force of attraction of the earth on the body.

Acceleration due to gravity has a constant value which is equal to 10 m/s².

Therefore, the acceleration of the rock dropped from the 5 m height is the same as that dropped from the 2.5 m height and is equal to 10 m/s².

Learn more about acceleration due to gravity at: https://brainly.com/question/11873969

Measurements of the radioactivity of a certain isotope tell you that the decay rate decreases from 8255 decays per minute to 3110 decays per minute over a period of 4.50 days.
What is the half-life (T1/2) of this isotope?

Answers

Answer:half-life (T1/2) of this isotope =

Explanation:

The number of nuclei of any radioactive substance at a given time is expressed by

Nt = N0e⁻kt

Nt=decay of material  at a time t, =3110 decays per minute

N=decays at t=0, 8255 decays per minute

k=constant

Nt=N0e−kt

3110= 8255 e⁻k(4.50)

3110/ 8255=e−k(4.50)

0.3767 =  e−k(4.50)

In 0.3767  =   -k (4.50)

0.976=-4.5k

k=0.976/4.5

=0.2159

Also we know that t 1/2= time that it takes half the original material to decay.it is  related to the rate constant by

T₁/₂=ln  2 / k

Therefore half-life (T1/2) of this isotope

T₁/₂=ln  2/0.2159

T₁/₂=3.12 days

Bob rides his bike with a constant speed of 10 miles per hour. How long will he take to travel a distance of 15 miles?

Answers

[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]

➡ 150hrs.

[tex]{\underline{\pink{\textsf{\textbf{Explanation : }}}}}[/tex]

➡ Time = distance × speed

➡ Time = 15*10

➡ Time = 150hrs ans.

Consider a block sliding down a ramp whose motion is opposed by frictional forces. The total energy of this system is modeled by the equation:
Etotal = 1/2mv^2 + mgh + Ff(f is underscore)d

Which part of the equation represents the amount of energy converted to thermal energy?

A. mg
B. Ffd
C. mgh
D. 1/2 mv^2

Answers

b because it’s the most expla

Answer:

Do you have the answers for the unit 6 test?

Explanation:

PLZ

(iii) Calculate the distance travelled by the car in part Q.
Use the equation
distance travelled = average speed x time
(2)
distance travelled = ....... m

Answers

Answer:

distance travelled = 3000m

Explanation:

distance travelled = average speed x time

                               =30m/s*100s

                                =3000m

If the knee has coordinates of (1.2,2.7) and the ankle has coordinates of (1.5,2.1), with the shank weighing 3.2kg, what is the moment of inertia of the shank about the proximal end point? (%k=0.528)

Answers

Answer:

The value is [tex]I = 0.48 \ kg m^2[/tex]

Explanation:

From the question we are told that

   The coordinate of the knee is  [tex]knee = (1.2,2.7)[/tex]

    The coordinate of the ankle is [tex]ankle = (1.5,2.1)[/tex]

    The mass of the shank is  [tex]m = 3.2 \ kg[/tex]

 Generally the length between the knee and the ankle is mathematically represented as

      [tex]D = \sqrt{( 1.5 - 1.2)^2+ [ 2.1- 2.7]^2 }[/tex]

=>  [tex]D =0.67 \ m[/tex]

Generally the  moment of inertia of the shank about the proximal end point is mathematically represented as

         [tex]I= m * \frac{D^2}{3}[/tex]

=>       [tex]I = 3.2 * \frac{0.67^2}{3}[/tex]

=>       [tex]I = 0.48 \ kg m^2[/tex]

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