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Lerok [7]
3 years ago
6

A body accelerates uniformly from rest at 2ms² for 5 seconds . calculate its average velocity in this time.​

Physics
1 answer:
timama [110]3 years ago
7 0

Answer:

0.4

Explanation:

average velocity =d/t

2/5=0.4

0.4m

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A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the s
egoroff_w [7]

This question is incomplete, the complete question is;

Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.  

1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?

Fwith belt =

2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.

Fwithout belt =

Answer:

1) The Net force on the driver with seat belt is 10.3 KN

2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Explanation:

Given the data in the question;

from the equation of motion, v² = u² + 2as

we solve for a

a = (v² - u²)/2s ----- let this be equation 1

we know that, F = ma ------- let this be equation 2

so from equation 1 and 2

F = m( (v² - u²)/2s )

where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.

1)

Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.

i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m

so we substitute the given values into the equation;

F = 70( ((0)² - (18)²) / 2 × 1.1 )

F = 70 × ( -324 / 2.4 )

F = 70 × -147.2727

F = -10309.09 N

F = -10.3 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwith belt =  10.3 KN

Therefore, Net force of the driver is 10.3 KN

2)

No sit belt,  

m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m

we substitute

F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )

F = 70 × ( -324 / 0.022 )

F = 70 × -14727.2727

F = -1030909.08 N

F = -1030.9 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwithout belt = 1030.9 KN

Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

4 0
3 years ago
In your own words, explain the effects of time dilation.
Natalka [10]
So we want to explain the effects of time dilation. In theory of relativity time dilation is the difference of elapsed time between two events when measured by two observers who are moving relatively to each other. A clock of an observer that is standing still in an inertial frame of reference is going to measure a different time of an event than the clock of an observer that is moving with some velocity with respect to the inertial reference frame that is not moving. In a nutshell, the moving clock is ticking slower than the clock that is standing still.  
7 0
3 years ago
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An interplanetary spacecraft is moving at
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The awnser is. 1728000 kilometers
7 0
3 years ago
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You are hungry and decide to go to your favorite neighborhood fast-food restaurant. You leave your apartment and take the elevat
Viktor [21]

Answer:

A)  R = (200 i ^ + 100 j ^ + 30k ^) m , B)    L = 223.61 m , C)   R = 225.61 m

Explanation:

Part A

This is a vector summing exercise, let's take a Reference System where the z axis corresponds to the height (flights), the x axis is the East - West and the y axis corresponds to the North - South.

Let's write the displacements

Descending from the apartment

10 flights of 3 m each, the total descent is 30 m

                Z = 30 k ^ m

Offset at street level

            L1 = 0.2 i ^ km

            L2 = 0.1 j ^ km

Let's reduce everything to the SI system

          L1 = 0.2 * 1000 = 200 i ^ m

          L2 = 100 j ^ m

The distance traveled is

          R = (200 i ^ + 100 j ^ + 30k ^) m

Part B

The horizontal distance traveled can be found with the Pythagorean theorem for the coordinates in the plane

                L² = x² + y²

                L = √ (200² + 100²)

                L = 223.61 m

Part C

The magnitude of travel, let's use the Pythagorean theorem for the sum

             R² = x² + y² + z²

              R = √ (30² + 200² + 100²)

             R = 225.61 m

7 0
3 years ago
If a 2 cm vector represents speed, and 1 cm equals 5 m/s, how fast is the represented
Ne4ueva [31]

Answer: 10m/s

Explanation: Since a single vector with length of 1cm expresses 5m/s, the vector which has a length doubled from the original vector should have the speed which is also doubled.

6 0
3 years ago
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