If the velocity of an object changed from 30 m/s to 60 m/s over a period of 10 seconds what would the average acceleration be ?

Answers

Answer 1
3 meters per second squared (3 m/s2)

Related Questions

Shows a car travelling around a bend in the road. The car is travelling at a constant speed. There is a resultant force acting on the car. This resultant force is called the centripetal force. (i) In which direction, A, B, C or D, does the centripetal force act on the car? Tick ( ) one box. A B C D (1) (ii) State the name of the force that provides the centripetal force.

Answers

Answer:

This question is incomplete

Explanation:

This question is incomplete but the missing figure is in the attachment below.

When an object is travelling around a circular path, there is a force that tends to draw that object towards the center of the circular path and keep the object moving in the curved path, that force is called the centripetal force. From this description, it can be deduced that the direction of the centripetal force that acts on the car (in the attachment below) is D.

The name of the force that provides this centripetal force is frictional force. This is the force that prevents the car from slipping off the road; keeping it moving in the curved path.

Which of the following is NOT one of the essential components of an exercise program?

Answers

Answer:

i dont know dude but ask someone who is in your grade lol

Explanation:

help plz The sum of all chemical reactions that take place within an organism is known as ______________. *

Answers

Answer:

Metabolism

Explanation:

metabolism is the sum of all chemical reactions that take place within an organism

Answer:

Metabolism

Explanation:

the fall of a body on the earth's surface cannot be a complete free fall why ?​

Answers

Explanation:

because the boy has larger surface area due to which he offers the larger air resistance which decreases the acceleration so, he will fall towards the earth's surface approximately with constant velocity.

what is the uses of space weather data?​

Answers

Answer:

Space Weather Data

They are used to produce forecasts, such as the monthly sunspot number forecast. Most notably, sunspot numbers represent one of the longest continuous climate records available. The data give scientists the means to make many forecasts and issue space weather advisories and alerts.

Explanation:

A train travels with a speed of 115km/hr. How much time does it take to cover a distance of 470km.​

Answers

It would take approximately 4 hours for the train to cover a distance of 470km.

4 hours 5 minutes 13.04 seconds

A 1500 kg car has an applied forward force of 5000 N and experiences an air resistance of 1250 N. What is the car's acceleration?

Answers

Answer:

[tex]2.33\ m/s^2[/tex]

Explanation:

Net Force

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

Fn = m.a

Where a is the acceleration of the object.  The net force is the sum of the individual vector forces applied to the object.

The m=1500 Kg car has two horizontal forces applied: the forward force of 5000 N that causes the movement and the air resistance force of 1250 N that opposes motion.

The net force is Fn = 5000 N - 1500 N = 3500 N

To find the acceleration, we solve the equation for a:

[tex]\displaystyle a=\frac{Fn}{m}[/tex]

[tex]\displaystyle a=\frac{3500}{1500}[/tex]

[tex]\boxed{a = 2.33\ m/s^2}[/tex]

The car's acceleration is [tex]a = 2.33\ m/s^2[/tex]

Physical values in the real world have two components: magnitude and

Answers

Answer:

Dimension.

Explanation:

- During a certain period, the angular position of a rotating object is given by: = − + , where  is in radian and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the rotating object at = Sec.

Answers

The question is not complete. The complete question is :

During a certain period of time, the angular position of a rotating object is given by [tex]$\theta =2t^2 +10t+5$[/tex], where θ is in radians and t is in seconds.  Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0.00 seconds, (b) at t = 3.00 seconds.

Solution :

Given :

Displacement or angular position of the object, [tex]$\theta =2t^2 +10t+5$[/tex]

∴ Angular speed is   [tex]$\omega = \frac{d \theta}{dt}$[/tex]

                                 ω = 10 + 4t

And angular acceleration is [tex]$\alpha = \frac{d \omega}{dt}$[/tex]

                                              α = 4

a). At time, t = 0.00 seconds :

   Angular displacement is [tex]$\theta =2t^2 +10t+5$[/tex]

                                            [tex]$\theta =2(0)^2 +10(0)+5$[/tex]

                                               = 5 rad

  Angular speed is  ω = 10 + 4t

                                 ω = 10 + 4(0)

                                     = 10 rad/s

Angular acceleration is α = 4 [tex]$rad/s^2$[/tex]

b). At time, t = 3.00 seconds :

   Angular displacement is [tex]$\theta =2t^2 +10t+5$[/tex]

                                            [tex]$\theta =2(3)^2 +10(3)+5$[/tex]

                                               = 53 rad

  Angular speed is  ω = 10 + 4t

                                 ω = 10 + 4(3)

                                     = 22 rad/s

Angular acceleration is α = 4 [tex]$rad/s^2$[/tex]

                                   

                                           

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