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DanielleElmas [232]
2 years ago
6

The displacement of a car is a function of time as follows: x(t)=25+3.0t², with x is in meters. Find the average velocity betwee

n t1 = 1.0s and t2 = 4.0s.​
Physics
1 answer:
aniked [119]2 years ago
5 0

Answer: 15m/s

Explanation: <u>Average</u> <u>Velocity</u> is vector describing the total displacement of an object and the time taken to change its position. It is represented as:

v=\frac{\Delta x}{\Delta t}

At t₁ = 1.0s, displacement x₁ is:

x(1)=25+3(1)^{2}

x(1) = 28

At t₂ = 4.0s:

x(4)=25+3(4)^{2}

x(4) = 73

Then, average speed is

v=\frac{73-28}{4-1}

v = 15

The average velocity of a car between t₁ = 1s and t₂ = 4s is 15m/s

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Answer:

Yeah  I think you're right

Explanation:

Because every obj is in motion till acted upon by a force(the branch)

4 0
3 years ago
After fixing a flat tire on a bicycle you give the wheel a spin. Its initial angular speed was 5.45 rad/s and it rotated 14.4 re
Komok [63]

Answer:

(a) α = -0.16 rad/s²

(b) t = 33.2 s

Explanation:

(a)

Applying 3rd equation of motion on the circular motion of the tire:

2αθ = ωf² - ωi²

where,

α = angular acceleration = ?

ωf = final angular velocity = 0 rad/s (tire finally stops)

ωi = initial angular velocity = 5.45 rad/s

θ = Angular Displacement = (14.4 rev)(2π rad/1 rev) = 28.8π rad

Therefore,

2(α)(28.8π rad) = (0 rad/s)² - (5.45 rad/s)²

α = -(29.7 rad²/s²)/(57.6π rad)

<u>α = -0.16 rad/s²</u>

<u>Negative sign shows deceleration</u>

<u></u>

(b)

Now, we apply 1st equation of motion:

ωf = ωi + αt

0 rad/s = 5.45 rad/s + (-0.16 rad/s²)t

t = (5.45 rad/s)/(0.16 rad/s²)

<u>t = 33.2 s</u>

6 0
3 years ago
Why do you think that this type of fossil is called a print fossil?
dem82 [27]

Answer:

Fossils are the remains of plants, animals, fungi, bacteria, and single-celled living things that have been replaced by rock material or impressions of organisms preserved in rock.

Answer ☝ on pic that is non print fossil of dinousor

Answer ☝ on the pic is print fossil of turtle

# CARRY ON LEARNING

# SORRY

4 0
2 years ago
Will mark as BRAINLIEST....... The Displacement x of particle moving in one dimension under the action of constant force is rela
pentagon [3]

Explanation:

It is given that,

The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:

x=4t^3+3t^2-5t+2

Where,

x is in meters and t is in sec

We know that,

Velocity,

v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5

(a) i. t = 2 s

v=12(2)^2+6(2)-5=55\ m/s

At t = 4 s

v=12(4)^2+6(4)-5=211\ m/s

(b) Acceleration,

a=\dfrac{dv}{dt}\\\\a=\dfrac{d(12t^2+6t-5)}{dt}\\\\a=24t+6

Pu t = 3 s in the above equation

So,

a=24(3)+6\\\\a=78\ m/s^2

Hence, this is the required solution.

3 0
3 years ago
Ultrasonic images are obtained from the inside organs of our body. This process uses which property of sound wave?
Temka [501]

This question involves the concepts of echo, ultrasonic images, ultrasonic sound waves.

The process of ultrasonic images uses the "echo" property of the sound waves.

Echo is the property of the sound wave by the virtue of which the sound wave reflects back to the source of the sound after hitting a surface or an object.

Ultrasonic images are obtained from inside organs of our body. This process involves the use of ultrasonic sound waves that have a frequency greater than 20,000 Hz. These sound waves are out of the range of audible sound by the human ear. When these ultrasonic sound waves are sent in form of pulses into the human body by the use of probes, they reflect back from the tissues of different organs to the probe. The probe then records the reflection properties of these sound waves and displays them in form of an image, known as ultrasonic images.

Learn more about echo here:

brainly.com/question/14335186?referrer=searchResults

The attached picture shows the process of ultrasonic imaging.

4 0
2 years ago
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