All categories

3 years ago

A motorist driving at 25 meters/second decelerates to

Physics

2 answers:

prisoha [69]3 years ago

6 0

Answer:

Kinetic energy = (1/2) (mass) (speed)²

Before slowing down, the car's speed is 25 m/s,

and its kinetic energy is ...

(1/2) (1,500 kg) (25 m/s)²

= (1/2) (1,500 kg) (625 m²/s²)

= 468,750 joules .

After slowing down, the car's speed is 15 m/s,

(1/2) (1,500 kg) (15 m/s)²

= (1/2) (1,500 kg) (225 m²/s²)

= 168,750 joules.

The car lost (468,750 - 168,750) = 300,000 joules

and you heard it from the KING

timama [110]3 years ago

5 0

Kinetic energy = (1/2) (mass) (speed)²

Before slowing down, the car's speed is 25 m/s,
and its kinetic energy is ...

   (1/2) (1,500 kg) (25 m/s)²

      = (1/2) (1,500 kg) (625 m²/s²)

      =    468,750 joules .

After slowing down, the car's speed is 15 m/s,
and its kinetic energy is ...

   (1/2) (1,500 kg) (15 m/s)²

      = (1/2) (1,500 kg) (225 m²/s²)

      = 168,750 joules.

The car lost (468,750 - 168,750) = 300,000 joules of K.E.

The law of Conservation of Energy says:

That 300,000 joules had to go somewhere.

If it's a standard, gas-powered car, then the kinetic energy got
put into the brakes. The energy turned into heat, and the heat
was carried off in the air.

If it's a more modern electric or hybrid car, then the kinetic energy
spun the wheel motors, turning them temporarily into electrical
generators. The generators converted the kinetic energy into
electrical energy, which got put back into the car's batteries, and
could be used again. That's why electric cars use less gas.

You might be interested in

A car is moving at 10 m/s to the right. It accelerates for 10 s after which it is moving at 5 m/s to the left. What was the car'

NNADVOKAT [17]

Answer:

Acceleration, $a=-1.5\ m/s^2$

Explanation:

It is given that,

Initial velocity of the car, u = 10 m/s (in right)

Final velocity of the car, v = -5 m/s (in left)

Time taken, t = 10 s

Let a is the acceleration of the car. It can be calculated using the equation of kinematics. The equation is as :

$v=u+at$

$a=\dfrac{v-u}{t}$

$a=\dfrac{-5-10}{10}$

$a=-1.5\ m/s^2$

So, the acceleration of the car is $-1.5\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

Which of these steps would most likely be part of a lab procedure?

Snezhnost [94]

Answer:

These are a part of lab procedures:

1. Write a hypothesis to answer a question.

2. Write a title at the top of a completed lab report.

3. Record the time to complete a chemical reaction.

These are NOT a part of lab procedures:

1. Create a question on the cause of a chemical reaction.

8 0

3 years ago

An object is placed on a surface. A student tries to apply various combinations of forces on the object. Which pair of forces wi

AVprozaik [17]

Answer:

See Explanation

Explanation:

The question is incomplete, as there are no diagrams or options to provide more information to the question.

The general explanation is as follows:

For the object not to move

(1): The forces acting on the object must opposite each other. i.e. if force A acts at the right (or positive direction), force B will act at the left (or negative direction).

(2) The two forces must be equal.

So, for instance:

If the pair of forces are 5N and 5N in opposite directions, the object wil not move.

However, if one of the forces is greater, the object will move towards the direction of the greater force.

7 0

3 years ago

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 44.1m/s^2 . The acce

ollegr [7]

The acceleration and distance is related to the following expression:
y=v0*t + a*t^2/2 ; v0=0
y=44.1*100/2 = 2205m
hence, the speed will be
v=0 + a*t = 441m/s
from that height it will just be subjected to the gravitational acceleration
0=v_acc^2 -2g*y_free
y_free = v_acc^2/2g = 9922.5m
<span>y_max = y_acc+y_free = 441+9922.5 =10363.5m</span>

7 0

3 years ago

Light propagate faster through medium “a” than medium “b”

dangina [55]

1) Medium "b" has more optical density

2) Light must hit the interface between the two mediums perpendicularly

Explanation:

Refraction occurs when light propagates from a medium into a second medium.

The optical density of a medium is given by its index of refraction, which is defined as:

$n=\frac{c}{v}$

where

c is the speed of light in a vacuum

v is the speed of light in a medium

Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.

In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.

We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

$n_1, n_2$ are the index of refraction of the two mediums

$\theta_1, \theta_2$ are the angle of incidence and of refraction (the angle that light makes with the normal to the surface in medium 1 and medium 2)