Help me please!

You apply the brakes of your car abruptly and your book starts sliding off the front seat. Three observers sitting in the car ex

krek1111 [17]

Answer:

All the observers are correct.

Explanation:

This is simply a problem of reference frames from which the motion of the book is being viewed by the various observers.

From their various reference frames, they are all correct.

Observer A must be in the inertial reference frame.

<em>Observers who can explain the behavior of the book and the car by using the relationship between the sum of the forces and changing velocity are said to be observers in inertial reference frames.</em>

This is clearly shown by what observer A noticed. There was a relative motion between the book and the car as she pointed out, making her to be in an inertial reference frame.

<em>Similarly, observers in inertial reference frames can also explain the changes in velocity of objects by considering the forces exerted on them by other objects.</em>

This is shown by observer B as he is able to notice how the force of the car affects the velocity of the book.

Observer C is actually in a non-inertial reference frame, as newtons law of force motion relationship are no longer observed. This occurs in the non inertial reference frame.

7 0

2 years ago

Plz i need help for the 5 problems. plz show the work!!!

Artemon [7]

Answer:

1. 3 $m/s^{2}$

2. 1.5 $m/s^{2}$

3. 3 seconds

4. 0 $m/s^{2}$

5. 2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

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

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = $\frac {30-0}{10}=3 m/s^{2}$

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

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

Substituting u=10m/s, v=22m/s and t=8 seconds

a = $\frac {22-10}{8}=1.5 m/s^{2}$

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

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

Substituting u=0m/s since at rest, v=15m/s and a=5 $\frac {m}{s^{2}}$

= $\frac {15-0}{5}=3s$

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

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

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = $\frac {20-20}{10}=0 m/s^{2}$

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

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

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 $\frac {m}{s^{2}}$

= $\frac {0-9}{-4.1}=2.2s$

8 0

2 years ago

How would you write the number 6,500,000,000 in scientific notation?

photoshop1234 [79]

Scientific form = 6.5 x 109.

8 0

3 years ago

Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags m

stepladder [879]

Answer: $F_{2}=\frac{3}{4}F_{1}$

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

In this case we have two situations:

1) Two bags with masses $4M$ and $4M$ mutually exerting a gravitational attraction $F_{1}$ on each other:

$F_{1}=G\frac{(4M)(4M)}{r^2}$ (1)

$F_{1}=G\frac{16M^2}{r^2}$ (2)

$F_{1}=16\frac{GM^2}{r^2}$ (3)

2) Two bags with masses $2M$ and $6M$ mutually exerting a gravitational attraction $F_{2}$ on each other (assuming the distance between both bags is the same as situation 1):

$F_{2}=G\frac{(2M)(6M)}{r^2}$ (4)

$F_{2}=G\frac{12M^2}{r^2}$ (5)

$F_{2}=12\frac{GM^2}{r^2}$ (6)

Now, if we isolate $\frac{GM^2}{r^2}$ from (3):

$\frac{F_{1}}{16}=\frac{GM^2}{r^2}$ (7)

Substituting $\frac{GM^2}{r^2}$ found in (7) in (6):

$F_{2}=12(\frac{F_{1}}{16})$ (8)

$F_{2}=\frac{12}{16}F_{1}$ (9)

Simplifying, we finally get the expression for $F_{2}$ in terms of $F_{1}$ :

$F_{2}=\frac{3}{4}F_{1}$

5 0

3 years ago

What will police do with witnesses at the scene?

NNADVOKAT [17]

Answer:

Speaking to witnesses is a job they do. If anyone stops after seeing the accident occur, the officers who respond to the accident take their personal information, they get their contact information, and they add their statements to their report.

Explanation:

4 0

2 years ago

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