Answer:

**Answer: i think you should place it on the red line **

**Explanation:**

**hope this helps **

**and need brainliest**

Ultraviolet light is typically divided into three categories. UV-A, with wavelengths between 400 nm and 320 nm, has been linked with malignant melanomas. UV-B radiation, which is the primary cause of sunburn and other skin cancers, has wavelengths between 320 nm and 280 nm. Finally, the region known as UV-C extends to wavelengths of 100 nm. (a) Find the range of frequencies for UV-B radiation. (b) In which of these three categories does radiation with a frequency of 7.9 * 1014 Hz belong

What causes rain? a.air becomes filled with water vapor b.water vapor condenses on dust particlesc. dust particles can no longer support water droplets

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

Each mass in the figure is 3 kg. Find the magnitude and direction of the net gravitational force on mass A due to the other masses.A. 2.45 × 10–7 N toward BB. 3.75 × 10–7 N toward CC. 2.00 × 10–7 N toward DD. 1.15 × 10–7 N toward D

A hollow sphere of radius 0.25 m is rotating at 13 rad/s about an axis that passes through its center. the mass of the sphere is 3.8 kg. assuming a constant net torque is applied to the sphere, how much work is required to bring the sphere to a stop?

What causes rain? a.air becomes filled with water vapor b.water vapor condenses on dust particlesc. dust particles can no longer support water droplets

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

Each mass in the figure is 3 kg. Find the magnitude and direction of the net gravitational force on mass A due to the other masses.A. 2.45 × 10–7 N toward BB. 3.75 × 10–7 N toward CC. 2.00 × 10–7 N toward DD. 1.15 × 10–7 N toward D

A hollow sphere of radius 0.25 m is rotating at 13 rad/s about an axis that passes through its center. the mass of the sphere is 3.8 kg. assuming a constant net torque is applied to the sphere, how much work is required to bring the sphere to a stop?

Answer:

0.5 lambda(wavelength)

Explanation:

We know that

The first harmonic for both side open ended pipe is

L= 1/2lambda

So L = 0.5*wavelength

1. Which terms (if any) are positive?

2. Which terms (if any) are negative?

3. Which terms (if any) are zero?

b) Determine the energy output by the athlete in SI unit.

c) Determine his metabolic power in SI unit.

d) Another day he performs the same task in 1.2 s.

1. Is the metabolic energy that he expends more, less, or the same?

2. Is his metabolic power more, less, or the same?

**Answer:**

**Explanation:**

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

Positive, negative, and zero terms in the **energy equation**. Calculation of energy output and metabolic power. Comparison of metabolic energy and power for different time durations.

To apply the energy equation to the system, we need to determine whether each term is positive, negative, or zero:

- Positive terms:

- ΔUg - the change in gravitational potential energy is positive as the barbell is lifted vertically from the ground.
- ΔUs - the change in elastic
**potential**energy is positive if there is any stretch or compression in the system.

- ΔK - the change in kinetic energy is negative as the barbell is lifted at a constant speed, so there is no change in velocity.
- ΔEch - the change in chemical potential energy is negative if the athlete is not ingesting any food or drinks during the exercise.

- ΔEth - the change in thermal energy is zero if there is no heat transfer in the system.

To determine the energy output by the athlete, we can calculate the work done on the barbell using the formula W = ΔUg. In this case, the work done is equal to the change in **gravitational **potential energy, which is equal to mgh. Thus, W = 400 N × 2.0 m = 800 J. So the energy output by the athlete is 800 J.

The metabolic power can be calculated using the equation P = W / t, where P is the power, W is the work done, and t is the time taken. Substituting the given values, P = 800 J / 1.6 s = 500 W. Therefore, the **metabolic** power of the athlete is 500 W. If the task is performed in a faster time, the metabolic energy expended will be the same. However, the metabolic power will be greater as the work is done in less time.

#SPJ11

**Answer:**

Gravitational

Electrostatic

magnetic

Frictional

gravitational

electrostatic

magnetic

frictional

hope it helps

pls mark as brainliest

Answer:

0.25( m1m1) , 0.75( m2m2)

Explanation:

Noted the formula for kinetic energy is 1/2(Mass × Velocity).

Therefore the original value of the mass is 0.5, giving half away makes it 0.25 to another mass which is primarily 0.5. This now makes the new mass 0.25+0.5=0.75.

Thank you.

**Answer:**

There is a decrease in modulus of elasticity

**Explanation:**

Young's Modulus of elasticity also known as elastic modulus is the deformation of a body along a particular axis under the action of opposing forces along that axis. at atomic levels, it depends on bond energy or strength.

In cold working processes, plastic deformation a metal occurs below its re-crystallization temperature due to which crystal structure of metal gets distorted and as a result of dislocations fractures also occur resulting in hardening of metal but bonds at atomic levels defining elasticity are temporarily affected.

Thus an increase in cold working results in a decrease in modulus of elasticity.

**Answer :**

(a). The wavelength of electron is 26.22 μm.

(b).The wavelength of car is

**Explanation :**

**Given that,**

Speed = 100 km/hr

Mass of car = 1 ton

**(a). We need to calculate the wavelength of electron**

**Using formula of wavelength**

Put the value into the formula

**(II). We need to calculate the wavelength of car**

**Using formula of wavelength again**

The wavelength of the electron is greater than the dimension of electron and the wavelength of car is less than the dimension of car.

**Therefore, electron is quantum particle and car is classical. **

**Hence, (a). The wavelength of electron is 26.22 μm.**

**(b).The wavelength of car is .**