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4 years ago

Can a body have constant speed and still be acclerating ? Give an Example

Physics

1 answer:

Allushta [10]4 years ago

3 0

Hi Pupil Here's Your answer :::

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An object moving with constant speed can be accelerated if direction of motion changes. For example, an object moving with a constant speed in a circular path has an acceleration because its direction of motion changes continuously.

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Hope this Helps . . . . . . . . .

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Which are two characteristics of a proton

Scilla [17]

Positive electric charge and found inside the nucleus

6 0

2 years ago

Read 2 more answers

We intend to measure the open-loop gain (LaTeX: A_{open}A o p e n ) of an actual operational amplifier. The magnitude of LaTeX:

Korvikt [17]

Answer:

voltage divider, R₂ = 1000 R₁

measuring the output in the resistance R₁

Explanation:

Let's analyze the situation, in an op amp in open gain loop, the gain is maximum G = 10⁶ V / V

in this case the signal generator gives a minimum wave of 1 10⁻³ V, after passing through the amplified it becomes 10³ V which saturates the oscilloscope.

To solve this problem we must use a simple voltage divider, for this we use the fact that in a series circuit the voltage is the sum of the voltages of each element.

If we use two resistors whose relationship is

R₂ / R₁ = 10³

R₂ = 1000 R₁

When measuring the output in the resistance R₁ we have the desired divider, with a tolerance range, for the minimum output of the generator (1 10⁻³V) we have a reading of V = 1 V in the oscilloscope, for which we can use voltage up to 10V on the generator

8 0

3 years ago

Jupiter's gravitational field is stronger than Earth's. On Jupiter, an object's__________ will be greater than on Earth. What on

vesna_86 [32]

Answer: Weight

Explanation:The mass of an object is a measure of the quantity of matter in the object. This quantity remains constant under any circumstance.

However, the same cannot be said about the weight of such object.

The weight of the object is very much dependent on the acceleration due to gravity which is a accelerational pull (by convention-- a pull downwards).

This is why an object tends to fall when it is thrown upwards on the earth for instance.

The statements above consequently infer that since the gravitational field of Jupiter is greater than that of the earth, the acceleration due to gravity on Jupiter is greater than that on earth.

And since the weight of an object(W) is a product of its mass and the acceleration due to gravity at that point.

Consequently, the object's weight on Jupiter would be greater than its weight on earth.

Please note; The Mass of the object remains constant everywhere.

3 0

4 years ago

There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is

Mademuasel [1]

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

4 0

3 years ago

Indigenous people sometimes cooked in watertight baskets by placing enough hot rocks into the water to bring it to a boil. What

yaroslaw [1]

Answer:

The rock has a mass of 4.02 kg

Explanation:

Step 1: Data given

Mass of the rock = TO BE DETERMINED

Temperature of the rock = 500 °C

Mass of the water =4.24 kg

⇒ loses 0.044kg as vapor

Initial temperature of the water = 29°C

Final temperature = 100°C

Specific heat of rock = 0.20 kcal/kg °C

Specific heat of water = 1kcal/kg°C

Latent heat of vaporization = 539 kcal/kg

Step 2: formules

Qlost,rock + Qgained,water = 0

Qtotal,water = Qwater +Qvapor

Step 3: Calculate Qvapor

Qvapor = mass of vapor * Latent heat of vapor

Qvapor = 0.044kg * 539 kcal/kg = 23.716 kcal

Step 4: Calculate Qwater

Qwater = mass of water * specific heat * Δtemperature

Qwater = 4.196 kg * 1kcal/kg°C *( 100-29)

Qwater = 297.916 kcal

Step 5: Calculate Qwater,total

Qwater,total = Qwater + Qvapor

Qwater,total = 23.716 kcal + 297.916 = 321.632 kcal

Step 6: Calculate Qrock

Qrock = mass of rock * specific heat rock * Δtemperature

Qrock = mass of rock * 0.20 kcal/kg°C * (100-500)

Qrock = mass of rock * -80 kcal/kg

Step 7: Calculate mass of rock

Qlost,rock + Qgained,water = 0

Qlost,rock = -Qgained,water

mass of rock * -80 kcal/kg = -321.632 kcal

mass of rock = 4.02 kg

The rock has a mass of 4.02 kg

7 0

3 years ago