An electric geyser consumes 2.2 units of electrical energy per hour of its use. It is designed to work on the mains voltage of220V
.
a) What is the ‘power-rating’ of this device?

The velocity of the bus at time \(\(t_2 = 2.15\) s\) is approximately \(\(6.9337\) m/s\).

To find the velocity of the bus at time \(\(t_2 = 2.15\) s\), we can integrate the acceleration function with respect to time to obtain the velocity function.

Let's perform the calculations:

Given:

Acceleration function: \(\(a(t) = \alpha t\)\),

where \(\(\alpha = 1.15\) m/s\(^3\)\)

\(\(t_1 = 1.20\) s\) (initial time)

\(\(v_1 = 5.10\) m/s\) (velocity at \(\(t_1\)\))

\(\(t_2 = 2.15\) s\) (desired time)

Integrating the acceleration function with respect to time will give us the velocity function:

\(\(\int a(t) dt = \int \alpha t dt\)\)

Integrating \(\(\alpha t\)\) with respect to t will yield \(\(\frac{1}{2}\alpha t^2\)\):

\(\(v(t) = \frac{1}{2}\alpha t^2 + C\)\)

To determine the constant of integration \(C\), we can use the initial condition given:

\(\(v(t_1) = \frac{1}{2}\alpha t_1^2 + C\)\)

Substituting \(\(t_1 = 1.20\) s\)  and

\(\(v_1 = 5.10\) m/s\):

\(\(5.10\) m/s = \(\frac{1}{2}(1.15\) m/s\(^3)(1.20\) s\(^2) + C\)\)

Now we can solve for \(C\):

\(\(5.10\) m/s = \(\frac{1}{2}(1.15\) m/s\(^3)(1.44\) s\(^2) + C\)\)

\(\(5.10\) m/s = \(0.828\) m/s\(^3 + C\)\)

\(\(C = 5.10\) m/s - \(0.828\) m/s\(^3\)\)

\(\(C = 4.272\) m/s\)

Now that we have the constant of integration \(C\), we can use it to find the velocity at \(\(t_2\)\):

\(\(v(t_2) = \frac{1}{2}\alpha t_2^2 + C\)\\\(v(t_2) = \frac{1}{2}(1.15\) m/s\(^3)(2.15\) s\(^2) + 4.272\) m/s\)

Calculating:

\(\(v(t_2) = \frac{1}{2}(1.15\) m/s\(^3)(4.6225\) s\(^2) + 4.272\) m/s\\\(v(t_2) = 2.66171875\) m/s + \(4.272\) m/s\\\(v(t_2) \approx 6.9337\) m/s\)

Therefore, the velocity of the bus at time \(\(t_2 = 2.15\) s\) is approximately \(\(6.9337\) m/s\).

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"thanks for the points!!! and have a  (terrible) day"  - puppetasap4 :)

Answer:

"thanks for the points!!! and have a  (terrible) day"  - puppetasap4 :)

Explanation:

Answer:

1m/s

Explanation:

Frequency= 1/10=0.1

Wave length= 10

Velocity= 10*0.1= 1m/s

I hope this helps

The correct option is (d) Externally applied fields don't affect the motion of holes and electrons within the junction zone. The generation currents in a p-n junction, a crucial semiconductor device, are not substantially affected by externally applied electric fields.

This is because externally applied fields do not impact the motion of holes and electrons within the junction zone.

In a p-n junction, the region where p-type and n-type semiconductors meet, the electric field is already present due to the difference in charge carriers. This electric field separates the majority charge carriers, creating a depletion region.

Option (a) suggests that there is already an electric field in the junction zone, which is true. However, this does not explain why externally applied electric fields do not substantially affect the generation currents.

Option (b) states that an external field would inhibit the formation of hole-conduction electron pairs and cancel the increased potential difference. This is not accurate because externally applied electric fields do not directly impact the formation of hole-electron pairs within the junction.

Option (c) suggests that the numbers of conduction holes and electrons generated are small and independent of external fields. However, this is not the main reason why externally applied fields do not substantially affect the generation currents.

Option (d) correctly explains the situation. Externally applied fields do not influence the motion of holes and electrons within the junction zone. The electric field created by the difference in charge carriers dominates the behavior of the charge carriers, and external fields do not significantly alter their motion.

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Answer:

47.515 cm²

In meter square= 4.7515 *10^-3 m²

Explanation:

The radius of a circle = 5.5 cm

In meter= 5.5*0.01

= 0.055 m

the circumference= 34.56 cm

in meters = 34.56 cm*0.01

= 0.3456 meters

the area= 95.03 cm²

in square meters =95.03*(0.01)²

in square meters= 9.503*10^-3 m²

. area under the curve at the right

= Area of semi circle

= Area of circle/2

= 95.03 /2

= 47.515 cm²

In meter square= 4.7515 *10^-3 m²

The dark area (shade) behind an object that obstructs the path of light is called a shadow

We can only see the outline of the object in a shadow. All details about the object cannot be seen. It is always dark in colour despite the colour of light or the colour of the object.

If A rope is vibrating at a high frequency. The length of the rope is 2.40 meters. the wavelength of the wave is 1.60 meters.

Wavelength is the distance between two consecutive points of a wave that are in phase, such as two adjacent crests or troughs. It represents the spatial extent of one complete cycle of the wave.

In this case, since we have one full wave and one-half wave, the total distance covered by the wave is equivalent to one and a half wavelength.

The length of the rope is given as 2.40 meters, which corresponds to one and a half wavelengths. To find the wavelength of the wave, we can divide the total length by the number of wavelengths:

Wavelength = Total length / Number of wavelengths

In this case, the total length is 2.40 meters, and the number of wavelengths is one and a half (1.5):

Wavelength = 2.40 meters / 1.5 = 1.60 meters

Therefore, the wavelength of the wave is 1.60 meters.

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Answer:

the answer is equal to visible light , ultraviolet, x-rays, gamma rayst

The sequence of types of electromagnetic radiation is arranged from that with the least to the most energetic photons will be visible light , ultraviolet, x-rays, gamma rays.

Electromagnetic radiation is a type of energy that is transmitted in incident photons as both magnetic and electric waves.

Electromagnetic radiation has a spectrum with varying wavelengths and frequencies, which gives it different properties.

Electromagnetic radiation is a type of energy that travels in packets of energy called photons as both electrical and magnetic waves.

Wave-particle duality is the name given to this phenomenon. This is a notion that defines how a wave behaves whether it behaves like particles, waves, or both particles and waves.

Hence the sequence of types of electromagnetic radiation is arranged from that with the least to the most energetic photons will be  visible light , ultraviolet, x-rays, gamma rays.

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Answer:

There would be a huge sound like an applause. At 200 decibels (the loudest possible sound ever created on Earth), it could shatter ear drums.

Explanation:

Hope this helps

If everyone on the planet screamed at the same moment, it would be a tremendously loud and chaotic event.

The particular impacts and repercussions, however, would be dependent on a variety of parameters such as the strength of the screams, the length, and the location of persons.

The cumulative cries of billions of individuals would produce a deafening loudness, perhaps reaching levels far beyond what the human ear can handle.

Thus, those in close proximity to people yelling may experience discomfort, confusion, and even bodily pain as a result of the loudness.

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The pressure drop over the 15-m length of the pipe is 0.00108 Pa. This pressure drop is very small and can be neglected for all practical purposes.  

The pressure drop over the 15-m length of the pipe, we can use the formula for pressure drop in a pipe:

Δp = f * L * (ΔT / T)

here Δp is the pressure drop, f is the friction factor, L is the length of the pipe, ΔT is the temperature difference across the pipe, and T is the reference temperature.

Assuming that the temperature difference across the pipe is 10°C, and using a friction factor of 0.018, we can solve for the pressure drop as follows:

Δp = 0.018 * 15 m * (10°C / 20°C)

Δp = 0.00108 Pa

Therefore, the pressure drop over the 15-m length of the pipe is 0.00108 Pa. This pressure drop is very small and can be neglected for all practical purposes.  

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The statement that is true about electromagnets is that these magnet work only when electric current passed through them.

Electromagnets are called temporary magnets because they do not show magnetic properties all the time.

They only show the magnetic properties when the current is passing through them.

As soon as the current in the Electromagnet stops they stops behaving like a magnet.

This property of temporary magnetism of the Electromagnet allows the Electromagnet to be used in devices like magnetic locks, MRI machines, headphones and loudspeakers.

Material like Copper, nickel and Cobalt are used to create electromagnets.

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When the coefficient of kinetic friction is increasing means, the mass of the object is also increasing, so the normal force acting between the contact surfaces will also increase, ultimately increasing the friction.

The ratio obtained by dividing the force required to keep an object moving by the force keeping two sliding surfaces together is known as the coefficient of kinetic friction. The materials used to create the contacting surfaces are particular to this coefficient. The coefficient will be higher on a rougher surface. The ratio of both the friction force to the normal force felt by a body moving on a dry, uneven surface is known as the kinetic friction coefficient or μk.

The force preventing sliding against one another of solid surfaces, fluid layers, and material components is known as friction. There are various kinds of friction, two solid surfaces in touch are opposed to one another's relative lateral motion by dry friction.

Hence, when the coefficient of kinetic friction is increasing means, the mass of the object is also increasing, so the normal force acting between the contact surfaces will also increase, ultimately increasing the friction.

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A detuning of engine crankshaft counterweights is a source of overstress that may be caused by an imbalance in the engine's rotating components.

The crankshaft counterweights are designed to balance out the forces generated by the engine's reciprocating components, such as the pistons and connecting rods. When the counterweights become detuned, it can lead to an imbalance in the rotating assembly, which can cause overstress and lead to premature engine failure.

Detuning can also occur due to wear and tear on the engine components, or as a result of modifications made to the engine that affect the balance of the rotating assembly. Regular maintenance and proper balancing of the engine's rotating components can help prevent detuning and ensure optimal engine performance and longevity.

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The energy due to work done in separating the plates goes into the battery and not into the capacitor. Work done on the charges is not included in the energy stored in the capacitor.

a. The capacitance of a parallel-plate capacitor is given by C = εA/d, where ε is the permittivity of free space, A is the area of each plate, and d is the distance between the plates. Using the given values, we get:

C = \((8.85 × 10^-12 F/m)(0.2 m)^2 / (0.001 m) = 7.08 × 10^-10 F\)

The energy stored in a capacitor is given by \(U = (1/2)CV^2,\) where V is the voltage across the capacitor. Using the given values, we get:

\(U = (1/2)(7.08 × 10^-10 F)(50 V)^2 = 8.85 × 10^-4 J\)

b. The capacitance of the capacitor changes when the distance between the plates is changed. The new capacitance is given by:

\(C' = (8.85 × 10^-12 F/m)(0.2 m)^2 / (0.002 m) = 7.08 × 10^-11 F\)

The voltage across the capacitor remains the same as it is still connected to the 50-V battery. The energy stored in the capacitor now is:

U' = (1/2)(7.08 × 10^-11 F)(50 V)^2 = 1.77 × 10^-5 J

c. When the battery is disconnected, the charge on the plates remains constant. The new capacitance and voltage are given by:

C' = (8.85 × 10^-12 F/m)(0.2 m)^2 / (0.002 m) = 7.08 × 10^-11 F

V' = V(d/2)/(d) = (50 V)(0.001 m) / (0.002 m) = 25 V

The energy stored in the capacitor now is:

\(U' = (1/2)(7.08 × 10^-11 F)(25 V)^2 = 4.42 × 10^-6 J\)

d. The energy stored in the capacitor decreases in (b) because the voltage across the capacitor remains the same, but the capacitance decreases. This means that the charge on the plates decreases, and therefore, the energy stored in the capacitor also decreases.

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Answer:

I think the answer is d; as the object is revolving around the earth in a circular motion due to centripetal force provided by gravity and if you "turned off" the gravity then it would tend to continue away from the planet but in the same direction.

Quick Answer: Science Can’t Figure Out Why Our Bikes Don’t Fall Over When We Ride Th Some physicists thought that the spinning wheels of a bicycle create enough angular momentum

It will take 0.82sec to reach the edge of the table if the system is free to move. If mB=1.0 kg, 99 kg mA must be used if the system's acceleration is to be kept above 1/100 g.

(a) mA = 11 kg

mB = 7 kg

T = mA mB g / (mA+mB) = 77*9.8 / 18 = 41.92 N is the formula for cord tension.

In the case of mA, we have T = mA*a

acceleration of mA, a = T / mA = 41.92 / 11 = 3.81 m/s²

For mB, we have mB*a = mB*g - T a = (mB*g - T) / mB

                                  a = (7*9.8 - 41.92) / 7 = 3.81 m/s²

(b) Initial velocity of mA, u = 0

The distance traveled by mA, s = 1.3 m

We have,

\(s = u t + (1/2) a t^2\\ s = at^2 / 2\\ t = (2s / a)^(1/2) = (2*1.3 / 3.81) (1/2) = 0.82 sec\)

(c) mB = 1 kg

The system's acceleration, a = g/100

The system's acceleration is given by the formula a = mB*g / (mA+mB).

mA = [mB*g / a] - mB = [ 1*9.8*100/9.8 ] - 1 = 99 kg

Acceleration is the rate at which velocity changes with respect to time. It is a vector quantity whose magnitude denotes the amount of change in velocity per unit of time.

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Answer:

t = 1.33 seconds

Explanation:

Given that,

The mass of the ball, m = 5 kg

Initial speed, u = 8 m/s

Final speed, v = 12 m/s

Force, F = 15 N

We need to find the time for long the force is applied. Let the time be t. The force is given by :

\(F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{5\times (12-8)}{15}\\\\=1.33\ s\)

So, the force is applied for 1.33 seconds.

Answer:

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B. Piron > Pwater > Pwax

Wax normally floats in water,it is less dense than water.

Iron is more dense than water, so it sinks.

Answer:

B.    P iron >  P water > P wax

The energy of the characteristic x-ray emitted from a tungsten target 1.36*10³ eV.

The energy of a characteristic x-ray emitted from a tungsten target when an electron drops from an M shell (n=3 state) to a vacancy in the K shell (n=1 state) can be estimated using the equation E =hcZ²Δn²/n². Here Z is the atomic number for tungsten (Z=74) and Δn is the difference between n1, the initial orbital, and n2, the final orbital, (n1 - n2). In this case n1= 3 and n2= 1 giving (3-1=2).

Using the equation E =(hc*74²*2²)/1² = (6.63*10-34*.993*17576)/1= 1.36*10³ eV. Therefore, the estimated energy of the characteristic x-ray emitted from the tungsten target when an electron drops from the M shell (n=3) to a vacancy in the K shell (n=1) is approximately 1.36*10³ eV.

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The correct answer is d) The anode is the cobalt electrode and the cell potential is +0.39 V. In a standard lead-cobalt cell, the anode is the cobalt electrode and the cell potential is +0.39 V.

This is because the cobalt electrode has a higher reduction potential than the lead electrode, meaning that electrons will flow from the cobalt to the lead electrode. The cell potential is the difference in the reduction potentials of the two electrodes, and since the cobalt electrode has a higher reduction potential, the cell potential is positive. The cell potential is measured in volts, and in this case it is +0.39 V.

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The net force on the apple when it is in free fall is 1.0 N, which is equal to its weight.

Option (A) - 0 N and option (B) - 0.1 N are incorrect because they do not take into account the force of gravity acting on the apple. Option (D) - 9.8 N is incorrect because it refers to the force of gravity acting on the apple, rather than the net force.

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\( \huge \boxed { \sf{Answers}}\)

Hope you could understand.

If you have any query, feel free to ask.

The magnitude of the total force of the stone on the man's hand is 1.43N

The rate at which an object's velocity with respect to time changes is referred to as acceleration in mechanics.

The direction of the net force acting on an object determines the direction of its acceleration.

Newton's Second Law's Second Law of Acceleration describes the magnitude of an object's acceleration.

Nis normal force exerted by stone on hand.

According to Newton's third law stone also exerted the same force.

⇒ N-mg= ma

⇒ N=1.3(9.8+1.2)

⇒ N=1.43newton

The magnitude of the total force of the stone on the man's hand is 1.43N

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In an equilibrium system, the sum of all forces is zero but the sum of moments of these forces depends on the location where the moments are calculated. The statement is true.

A body is in a state of equilibrium when there are no changes in its motion. In static equilibrium, the object is stationary, and in dynamic equilibrium, the object is in constant velocity. The state of equilibrium is obtained when the resultant force acting on the object is zero. This implies that the sum of all forces in the body is zero when it is in equilibrium.

The sum of moments of these forces depends on the location where the moments are calculated. This is because when we take a moment, the distance between the point at which the force is applied and the point where we calculate the moment plays a significant role.

As a result, a moment depends on the position of the point at which the force is applied. In conclusion, the statement "In an equilibrium system, the sum of all forces is zero but the sum of moments of these forces depends on the location where the moments are calculated" is true.

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\(\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}\)

\(\\ \sf\longmapsto Pressure=\dfrac{40}{2}\)

\(\\ \sf\longmapsto Pressure=20Pa\)

\( \sf \huge \red {\dag} Given\)

\(\sf \longmapsto \: Force \: = 40 \: Newtons\)

\(\sf \longmapsto \: Area \: = \: 2m \: Square \: ( {2m}^{2} )\)

We know that the formula of Pressure is

\(\sf \longmapsto \: Force/Area\)

Let's start.

\(\sf \longmapsto \: \: P = 40 \: newtons(force)/2(Area)\)

\(\sf \longmapsto Pressure \: 20 \: N/m^2\)

\(\boxed{\sf\red{\boxed{ \sf Pressure ≈ \: 20 \: N/m^2}}}\)

The displacement, in cm, of the spring as a function of m, g, k, and θ is given by the formula,x= mg (1-cosθ) / k.

A block of mass m kg is resting on a frictionless ramp inclined at an angle from the ground. The block is held by a spring with spring constant k n/m.

The displacement, in cm, of the spring as a function of m, g, k, and angle is given by the formula,x= mg (1-cosθ) / k, where x is the displacement of the spring in cm, m is the mass of the block in kg, g is the acceleration due to gravity in m/s², k is the spring constant in N/m, and θ is the angle of the ramp from the ground in degrees.

Therefore, the displacement, in cm, of the spring as a function of m, g, k, and θ is given by the formula,x= mg (1-cosθ) / k.

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First, for this problem, we will need to know the formula for time dilation

Where t is the time in the reference frame on earth

t0 is the time on the ship

v is the velocity relative to lightspeed

and c is light speed in a vacuum

For this problem, we will need to convert .93x10^8 to a fraction of light speed.

.93x10^8 = .31c

Now, we can plug in the numbers we get from the given

t0 = 6 hours since the ship has been in the air for 6 hours

t = 6.31089

For the answer to be valid, we will convert .31089 to minutes by multiplying it by 60 (60 minutes in a hour)

18.654 minutes