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Vedmedyk [2.9K]
3 years ago
9

The y-intercept of a position-time graph of an object gives the average velocity of the object.

Physics
1 answer:
finlep [7]3 years ago
4 0

<em>The y-intercept of a position-time graph of an object gives the average velocity of the object is False.</em>

<u>Answer:</u> <em>False. </em>

<u>Explanation:</u>

The initial position of the object can be obtained by calculating the Y intercept of  a position-time graph.  In a position-time graph, the independent  variable is time  and dependent variable is position. Y axis is the position axis and x axis is the time axis.

The equation of the graph is given by y =mx + c. m is the slope of the graph and c is the y intercept. When a particle starts from the origin its y intercept is zero.

The nature of the graph gives us an idea about velocity. When the velocity is positive, the position- time graph has positive slope and when the velocity is positive the graph has negative slope.

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A car of mass 1800 kg can be just be lifted. What is the least force that the electromagnet must use to lift the car? (1 g = 10
Doss [256]

Answer:

<h2>f=a×m</h2>

m=1800kg

1800000g×10N/kg

18000000N force is required to life the car

8 0
3 years ago
A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of
koban [17]

<u>Answer:</u>

 Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

  Speed of truck = 25 m/s north = 25 j m/s

  Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

                          = (1.43 i + 1.00 j) m/s

    Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

    Magnitude of velocity = 26.04 m/s

    Angle from positive horizontal axis = 86.85⁰

 So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.

4 0
3 years ago
A 0.49-kg cord is stretched between two supports,7.3m apart. When one support is struck by a hammer, a transverse wave travels d
Wewaii [24]

Answer:

T= 5.18N

Explanation:

u = mass of chord / length of chord

u = 0.49/ 7.3

u = 0.067 kg/m

Velocity of sound waves (v) =length of chord / time taken for wave to travel

v = 7.3 / 0.83 = 8.795m/s

Tension is calculated below using the formula

T = v² * u

T = (8.795)² x 0.067

T= 5.18N

3 0
3 years ago
A bag of cement has a mass of 62 g. What is the mass of the bag of cement in S.I. units (kg)?
NNADVOKAT [17]

The mass of this bag of cement in S.I. units (kg) is equal to 0.062 kilograms.

<u>Given the following data:</u>

  • Mass of cement = 62 grams.

To calculate the mass of this bag of cement in S.I. units (kg):

<h3>How to convert to S.I. units.</h3>

In Science, kilograms (kg) is the standard unit of measurement or S.I. units of the mass of a physical object. Thus, we would convert the value of the mass of this bag of cement in grams to kilograms (kg) as follows:

<u>Conversion:</u>

1000 grams = 1 kilograms.

62 grams = X kilograms.

Cross-multiplying, we have:

X = \frac{62}{1000}

X = 0.062 kilograms.

Read more on mass here: brainly.com/question/13833323

8 0
2 years ago
The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What
Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

a

 \alpha = 6261 \  rev/minutes^2

b

 \theta  = 613 \ revolutions

Explanation:

From the question we are told that

   The initial  angular speed is w_i =  1120 \ rev/minutes

    The angular speed after t = 13.8 s = \frac{13.8}{60 }  = 0.23 \ minutes  is w_f = 2560 \ rev/minutes

    The time for revolution considered is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  

 Generally the angular acceleration is mathematically represented as

         \alpha = \frac{w_f - w_i }{t}

=>      \alpha = \frac{2560  - 1120 }{0.23}  

=>      \alpha = 6261 \  rev/minutes^2

Generally the number of revolution made is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  is mathematically represented as

           \theta  =  \frac{1}{2}  * (w_i + w_f)*  t

=>      \theta  =  \frac{1}{2}  * (1120+ 2560 )*  0.333

=>      \theta  = 613 \ revolutions

5 0
3 years ago
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