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mash [69]
3 years ago
14

A particle of mass.25 kg is moving with a speed of 7m/s due north.find it's kinetic energy.​

Physics
1 answer:
Ivenika [448]3 years ago
3 0

Answer:

Explanation:

612J

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1. This heating system maintains room temperature at or near a particular value, known as the .
stiv31 [10]

Answer:

Explanation:

1. This heating system maintains room temperature at or near a particular value, known as the set point.

A temperature setpoint is the level at which the body attempts to maintain its temperature. When the setpoint is raised, the result is a fever.

2. You open the window, and a blast of icy air enters the room. The temperature drops to 17 degrees Celsius, which acts as a STIMULUS to the heating system.

3. The thermostat is a SENSOR that detects the stimulus and triggers a response.

Thermostat is use to turn off or on a switch, when the temperature is high or low

4. The heater turns on, and the temperature in the room INCREASE until it returns to the original setting.

This is the work of the thermostat above, when the temperature of the room is below a certain temperature the thermostat triggered the switch and keep increasing the temperature of the room until normal setting.

5. The response of the heating system reduces the stimulus. This is an example of NEGATIVE feedback.

Body temperature is regulated by negative feedback. The stimulus is when the body temperature exceeds 37 degrees Celsius, the sensors are the nerve cells with endings in the skin and brain, the control is the temperature regulatory center in the brain, and the effector is the sweat glands throughout the body.

6. The way this heating system maintains a stable room temperature is similar to the way an animal's body controls many aspects of its internal environment. The maintenance of a relatively constant internal environment is known as HOMEOSTASIS.

Humans rely on homeostasis to keep their core temperature hovering around 98.6 degrees Fahrenheit, so that their bodies can maintain proper function. Homeostasis is the ability to maintain a relatively stable internal state that persists despite changes in the world outside.

5 0
3 years ago
8. Placing your vehicle between the pilot/escort vehicle and an oversize/overweight vehicle can be dangerous.A. TrueB. False
Sliva [168]

Answer:

A because that's the answer

5 0
3 years ago
How would radiation occur in space?
Marizza181 [45]

Answer:

Space radiation is made up of three kinds of radiation: particles trapped in the Earth's magnetic field; particles shot into space during solar flares (solar particle events); and galactic cosmic rays, which are high-energy protons and heavy ions from outside our solar system.

Explanation:

3 0
2 years ago
A 4.00 kg block is suspended from a spring with k 500 N/m. A 50.0 g bullet is fired into the block from directly below with a sp
Yanka [14]

Answer:

a. A = 0.1656 m

b. % E = 1.219

Explanation:

Given

mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150  m/s , k = 500 N / m

a.

To find the amplitude of the resulting SHM using conserver energy

ΔKe + ΔUg + ΔUs = 0

¹/₂ * m * v²  -  ¹/₂ * k * A² = 0

A = √ mB * vₓ² / k

vₓ = mb * u₁ / mb + mB

vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518

A = √ 4.0 kg * (1.852 m/s)²   /   (500 N / m)

A = 0.1656 m

b.

The percentage of kinetic energy

%E = Es / Ek

Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5

Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N

% E = 13.72 / 1125 = 0.01219 *100

% E = 1.219

6 0
3 years ago
Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it
jek_recluse [69]

Answer:

a)    I= I₀ (cos²θ - cos⁴θ)    b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

         I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

         I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

              θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

          I = I₁ cos² θ'

       

we substitute

         I = (I₀ cos² tea) cos² (θ - 90)

        cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

         I = Io cos² θ sin² θ

        1= cos²θ+ sin²θ

       sin²θ = 1 - cos²θ

        I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

          \frac{dI}{d \theta} = 0

          \frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

            cos θ - 2 cos³ θ = 0

            cos θ ( 1 - 2 cos² θ) = 0  

The zeros of this function are in

           θ = 90º

           1-2cos²θ =0       cos θ = 0.25  θ =  75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

5 0
2 years ago
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