is paper an insulator or a conductor?

Answer:

if the frequency of the wave if tripled then period of wave gets tripled

When the wave frequency is tripled, the period of the wave becomes one-third of its original.

The frequency (f) of a wave is inversely proportional to the period (T) of the wave.

Mathematically,

\($f \propto \frac{1}{T}$\)

Thus, when frequency increases, the period decrease and when frequency decreases, the period of the wave increases.

So when frequency of a wave increases by three time, the period of the wave decreases by three times.

Hence, when the frequency is tripled, the period of the wave becomes one-third of its original value.

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The gravitational force of the robotic helicopter of mass 1.8 kg on Mars is 6.68 N.

Gravitational force is the force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface.

To calculate the helicopter's gravitational force, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the helicopter's gravitational force on Mars is 6.68 N.

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The equation that relates the forces acting on person to the acceleration of the person in elevator is as : N - mg = ma

An influence that causes the motion of object with mass to change it's velocity is called force.

The equation that relates forces acting on the person to acceleration of the person in elevator is: N - mg = ma

N is normal force exerted on the person by the elevator floor, m is mass of the person, g is acceleration due to gravity, and a is acceleration of the elevator.

This equation is derived from Newton's second law of motion, which states that net force acting on object is equal to its mass times its acceleration. In this case, net force acting on the person is the difference between normal force N and force of gravity mg, which is equal to ma, the product of person's mass and acceleration of the elevator.

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Answer:

1950 kpa

Explanation:

9 ÷ .6 = 15 times the speed. so 15 times the pressure 130 x15 = 1950 kpa

Answer:

the students where dided

Answer:

Assumption: there is no object between this point charge and the observer.

Explanation:

The electric field of a point charge is inversely proportional to the square of the distance from that point charge.

Let \(k\) denote Coulomb's constant (\(k \approx 8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-1}\).) Let the magnitude of that point charge be \(q\). At a distance of \(r\) from this charge, the electric field due to this charge would be:

\(\displaystyle E = \frac{k \cdot q}{r^{2}}\).

Convert the magnitude of the point charge in this question to standard units:

\(q = 1\; \rm nC = 10^{-9}\; \rm C\).

Apply that equation to find the magnitude of the electric field due to this point charge:

\(r = 1\; \rm m\):

\(\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(1\; \rm m)^{2}} \\ &\approx 9.0\; \rm N \cdot C^{-1}\end{aligned}\).

\(r = 2\; \rm m\):

\(\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(2\; \rm m)^{2}} \\ &\approx 2.2\; \rm N \cdot C^{-1}\end{aligned}\).

\(r = 3\; \rm m\):

\(\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(3\; \rm m)^{2}} \\ &\approx 1.0\; \rm N \cdot C^{-1}\end{aligned}\).

The direction of the electric field at a point is the same as the direction of a force from this field onto a positive point charge at this point.

Because the \((+1\; \rm nC)\) point charge here is positive, the electric field of this charge would repel other positive point charges. Hence, the electric field around this \((+1\; \rm nC)\!\) point charge at any point in the field would point away from this charge.

The answers are 4. The distance from where the ball was kicked is 38.06 meters, 5. The speed of the ball 2.5 seconds after it was kicked is 13.82 m/s, and  6. The ball is 21.88 meters above the ground 2.5 seconds after it is kicked.

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True, The ISO 14000 series identifies standards for environmental performance, in contrast to the ISO 9000 series. Around 570 international standards were created with their assistance.

The key distinction between ISO 9000 and ISO 14000 is that ISO 9000 is a Quality Management system that aids in the creation of high-quality products. ISO 14000, on the other hand, is an environmental performance management system. They support maintaining a healthy workplace. The ISO 14000 series of standards, as previously mentioned, consists of one standard (ISO 14001) that enterprises must adhere to and additional standards that serve as guidelines and support firms' compliance with ISO 14001. The foundation for creating an EMS is laid out in ISO 14001. Instead, it offers a mechanism for keeping an eye on, reining in, and enhancing performance with relation to those demands.

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If a 40 N block is resting on a rough horizontal table with a coefficient of static friction is 12 N.

Static friction is a force that resists the motion of two objects that are in contact with each other. It is caused by the forces of attraction between the two objects and is usually greater than the force of kinetic friction. The forces of static friction oppose the movement of the two objects and can be overcome by applying a force greater than the static friction.

The maximum force the block can withstand before it starts to move is 40 N multiplied by the coefficient of static friction.

The coefficient of static friction between the block and the table is determined by the materials of the block and the table and the surface roughness of the table.

If the coefficient of static friction is 0.3, then the maximum force the block can withstand before it starts to move is 40 N × 0.3 = 12 N.

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8.9 m/s

Explanation:

Note that the jumper launched from and then landed also on the ground. This means that the jumper's vertical displacement y is zero. From this information, we can determine the amount of time it took for him to make the jump using the equation

\(y = v_{0y}t - \frac{1}{2}gt^2 = (v_0\sin{45°})t - \frac{1}{2}gt^2\)

Since y = 0, then the equation above becomes

\(0 = (v_0\sin{45°})t - \frac{1}{2}gt^2\)

Solving for t, we get

\(t = \dfrac{2v_0\sin{45°}}{g}\)

We also know that

\(x = v_{0x}t = (v_0\cos{45°})t\)

Using the expression for t we got earlier into the equation above, we get

\(x = (v_0\cos{45°})\left(\dfrac{2v_0\sin{45°}}{g}\right)\)

\(\;\;\;\;=\dfrac{v_0^2}{g}(2\sin{45°}\cos{45°})\)

\(\;\;\;\;= \dfrac{v_0^2}{g}\sin{2(45°)°}\)

\(\Rightarrow x = \dfrac{v_0^2}{g}\)

Solving for \(v_0,\) we get

\(v_0^2 = gx \Rightarrow v_0 = \sqrt{gx}\)

Putting in the given values, we find that the final velocity is

\(v_0 = \sqrt{(9.8\:\text{m/s}^2)(8.0\:\text{m})} = 8.9\:\text{m/s}\)

Note: You can also solve this problem by using the range equation:

\(R = \dfrac{v_0^2}{g}\sin{2\theta}\)

0.82 m/s²

Though its gravity is weak compared to earth, Eris has the strongest gravity out of all the dwarf planets: 0.82 m/s².

Eris is the largest dwarf planet in the Solar System, and the ninth largest body orbiting our Sun. The discovery of Eris was so important because it was a celestial body larger than Pluto, which forced astronomers to consider, for the first time in history, what the definition of a planet truly is!

Hope this helps you!! :D

Answer:

1 ohm is equal to one volt (V)/ one ampere (1A) 1 Ohm is defined as the resistance of a conductor with a potential difference of 1 volt applied to the ends through which 1-ampere current flows. Ohms is the SI unit of electrical resistance.

Explanation:

I hope it's help u

(a) The amplitude of the wave is 0.2 m.

(b) The period of the wave is  4 s.

(c) The wavelength of the wave is 100 m.

(a) The amplitude of the wave is the maximum displacement of the wave.

amplitude of the wave = 0.2 m

(b) The period of the wave is the time taken for the wave to make one complete cycle.

period of the wave = 5.5 s - 1.5 s = 4 s

(c) The wavelength of the wave is calculated as follows;

λ = v / f

where;

f = 1/t = 1 / 4s = 0.25 Hz

λ = ( 25 m/s ) / 0.25 Hz

λ = 100 m

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Answer:

(2).........

Explanation:

(2)..........

An astronaut measure the period of a mass spring system on Earth.

The period of a mass spring system on the moon would be longer than the period on Earth. This is because the period of a mass spring system is dependent on the square root of the ratio of the mass to the spring constant, and the acceleration due to gravity. Since the acceleration due to gravity on the moon is only 1/6th of that on Earth, the restoring force on the mass will be weaker, resulting in a longer period. Therefore, the astronaut would measure a longer period for the same mass spring system on the moon than on Earth.

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\(v = 2.45×10^3\:\text{m/s}\)

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

\(F = \dfrac{d{p}}{d{t}}\) (1)

Assuming that the velocity remains constant then

\(F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}\)

Solving for \(v,\) we get

\(v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)\)

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust F is

\(F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}\)

The exhaust rate dm/dt is

\(\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}\)

\(\;\;\;\;\;= 1.36×10^4\:\text{kg/s}\)

Therefore, the velocity at which the exhaust gases exit the engines is

\(v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}\)

\(\;\;\;= 2.45×10^3\:\text{m/s}\)

Answer:

\( \boxed{ \sf{see \: below}}\)

Explanation:

The unit of speed is metre per second or m / s. It clearly depends on two fundamental units ; unit of length ( metre ) and unit of time ( second ). Hence , m/s is a derived unit.

Hope I helped!

Best regards! :D

The unit of speed is m/s. The two terms metre and second are fundamental quantities.

As,m/s is derived from these two fundamental quantities. Therefore, Unit of speed is a derived unit.

Answer:

effort arm mean the use of any work by using your hand force motion or by hand power

Recall the definition of average acceleration:

a = (v - u)/∆t

where u and v are the initial and final velocities, respectively.

So we have

2.2 m/s² = (v - 9.3 m/s) / (5.1 s)

v - 9.3 m/s = (2.2 m/s²) * (5.1 s)

v = 9.3 m/s + (2.2 m/s²) * (5.1 s)

v = 20.52 m/s ≈ 21 m/s

Answer:

A. The angular acceleration of the disk is -1.047 radians per square second.

B. The disk turns 4.715 radians while stopping.

C. The disk did 0.750 revolutions while stopping.

Explanation:

A. In this case, the disk is deceleration at a constant rate. Hence, the angular acceleration experimented by the object (\(\alpha\)), in radians per square second, can be found by means of this kinematic expression:

\(\alpha = \frac{\omega-\omega_{o}}{t}\) (1)

Where:

\(\omega_{o}\) - Initial angular speed, in radians per second.

\(\omega\) - Final angular speed, in radians per second.

\(t\) - Time, in seconds.

If we know that \(\omega_{o} \approx 3.142\,\frac{rad}{s}\), \(\omega = 0\,\frac{rad}{s}\) and \(t = 3\,s\), then the angular acceleration of the disk is:

\(\alpha = \frac{\omega-\omega_{o}}{t}\)

\(\alpha = -1.047\,\frac{rad}{s^{2}}\)

The angular acceleration of the disk is -1.047 radians per square second.

B. The change in position of the disk (\(\Delta \theta\)), in radians, is determined by the following kinematic formula:

\(\Delta \theta = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \alpha}\) (2)

If we know that \(\omega_{o} \approx 3.142\,\frac{rad}{s}\), \(\omega = 0\,\frac{rad}{s}\) and \(\alpha = -1.047\,\frac{rad}{s^{2}}\), then the change in position is:

\(\Delta \theta = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \alpha}\)

\(\Delta \theta = 4.715\,rad\)

The disk turns 4.715 radians while stopping.

C. A revolution equals 2π radians, then, then number of revolutions done by the disk while stopping is found by simple rule of three:

\(\Delta \theta = 4.715\,rad \times \frac{1\,rev}{2\pi\, rad}\)

\(\Delta \theta = 0.750\,rev\)

The disk did 0.750 revolutions while stopping.

Answer:

Part 1:

Option B is correct (It will remain unchanged).

Part 2:

Option C is correct (It will increase by a factor of √ 2)

Part 3:

Option E is correct (It will be half as fast/long.)

Part 4:

Option C is correct (It will increase by a factor of √ 2.)

Explanation:

Formula we are going to use:

V=f*λ

Where:

V is the speed of Sound

f is the frequency of wave

λ is the wavelength.

The speed of wave , tension and linear density have following relation:

\(V=\sqrt{F/\rho}\)

Where:

V is the speed of Sound (Initial)

F is the tension in string (Initial)

\(\rho\) is the linear density of string (Constant)

Terms:

V' is the new speed

f' is the new frequency

λ' is the wavelength

Solution:

Part 1:

From \(V=\sqrt{F/\rho}\):

Speed of Sound is independent of the frequency of shaking so speed well remain unchanged.

Option B is correct (It will remain unchanged)

Part 2:

If F'=2F then

\(V=\sqrt{F/\rho}\)

\(V'=\sqrt{F'/\rho}\\V'=\sqrt{2F/\rho}\\V'= \sqrt{2} * \sqrt{F/\rho}\\V'=\sqrt{2}V\)

Option C is correct (It will increase by a factor of √ 2)

Part 3:

Formula we are going to use:

V=f*λ

Given f'=2f,

Even though frequency is doubled we will keep velocities same. V=V' in order to find the changing wavelength.

V'=f'*λ'

f*λ=f'*λ'

f*λ=2f*λ'

Solving above Equation:

λ'=λ/2

Option E is correct (It will be half as fast/long.)

Part 4:

T'=2T means \(V'=\sqrt{2}V\) (From Part 1)

f'=f

Now:

V'=f'*λ'

\(\sqrt{2}f*\lambda=f'*\lambda '\\\sqrt{2}f*\lambda=f*\lambda '\\ \lambda '=\sqrt{2}*\lambda\)

Option C is correct (It will increase by a factor of √ 2.)

Answer:

Explanation:

The period of oscillation will remain unchanged because the period of oscillation of a pendulum does not depend upon the mass of the bob  . Here monkey along with bunch of banana represents bob .

When the monkey and banana were at height h /2 , they have potential energy as well as kinetic energy . banana is separated from the system . It carried its total energy along with it . But the energy of monkey remained intact with it . So it will keep on moving as usual . So it will attain the same maximum height as before .

Hence the amplitude of oscillation too will remain unchanged .

Answer:

Explanation:

Assuming you mean a = 16 m/s²

s = ½at²

t = √(2s/a)

t = √(2(201)/16)

t = 5.012484...

t = 5.0 s

Answer:

(a) 6.67 m/s  (b) 3.33 m/s

Explanation:

if i understood your question correctly, the jogger goes 150m east before going in the opposite direction west for 50m. (see diagram)

in avg speed, we use the total distance (150m + 50m = 200m)  and total time (30s)

v = d/t =200/30 =6.67 m/s

in avg velocity, we use the total displacement  (150m - 50m = 100m) and total time (30s)

v= d/t = 100/30 = 3.33 m/s

Answer:

The wave is travelling in the ±z-axis direction.

Explanation:

An electromagnetic wave has an oscillating magnetic and electric field. The electric and magnetic field both oscillate perpendicularly one to the other, and the wave travels perpendicularly to the direction of oscillation of the  electric and magnetic field.

In this case, if the magnetic field is in the +x-axis direction, and the electric field is in the +y-axis direction, we can say with all assurance that the wave will be travelling in the ±z-axis direction.

Given

Suppose a monochromatic light incident on a single slit produces a diffraction

pattern.

To find

(a) How would the pattern differ if you decrease the wavelength of the light by

half?

(b) How would the pattern differ if you tripled both the wavelength and slit width

at the same time?​

Explanation

The fringe width is given by

where D is the distance between the slit and the screen, d is the width of the slit.

a. If the wavelength is decreased by half then the fringe width decreases by half. So teh distance between the consecutive bright spot of dark spot is reduced to half.

b. When the wavelength and the slit width would be tripled then the fringe width is

Thus the fringe width remains same. So the diffraction patternn remains same.

Answer:

b) 16 cm

Magnification, m = v/u

3 = v/u

⇒ v = 3u

Lens formula : 1/v – 1/u = 1/f

1/3u = 1/u = 1/12

-2/3u = 1/12

⇒ u = -8 cm

V = 3 × (-8) = -24

Distance between object and image = u – v = -8 – (-24) = -8 + 24 = 16 cm

Explanation:

Answer: i dont know

Explanation:

E

A horizontal rope is attached from a truck to a 1500-kg car. As the truck tows the car on a horizontal straight road, the rope will break if the tension is greater than 2696 N.

The maximum tension that the rope can handle is 2696 N. To avoid breaking the rope, the tension in the rope when the truck accelerates should not exceed this value. We can use Newton's second law to relate the tension in the rope to the acceleration of the car

Tension = (mass of car) × (acceleration)

Where Tension is the tension in the rope, and the mass of the car is 1500 kg.

Therefore, the maximum acceleration of the truck can be found by dividing the maximum tension by the mass of the car.

Acceleration = Tension / (mass of car)

Acceleration = 2696 N / 1500 kg

Acceleration = 1.8 m/s²

Hence, the maximum possible acceleration of the truck is 1.8 m/s².

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