All categories

4 years ago

Inertia law what happens when You are standing on the bus and the bus stops abruptly:

Physics

2 answers:

lord [1]4 years ago

6 0

Answer:

You will fall towards the front of the bus i.e towards its front seat

Explanation:

When the bus is moving, you are supposed to be in that motion. Your body is experiencing a forward force as the bus is moving forward So when bus stops suddenly your inertia does not allow you to stay in that standing position it will let you move forrward to maintain your inertia so you will feel a forward push. (If you were sitting on a seat with seatbelt on you that seatbelt will make you come back when car or bus stops.)

I am Lyosha [343]4 years ago

4 0

Answer:

You will fly forward in the bus until you hit something.

Explanation:

While standing there on the bus, you are traveling at the same speed as the bus. If the bus suddenly stops, you will still be traveling at the same speed you started with. That is until you hit something hard enough or big enough to stop you.

You might be interested in

In places such as hospital operating rooms or factories for electronic circuit boards, electric sparks must be avoided. A person

FrozenT [24]

Time taken by the rubber-soled shoes to reduce a person's potential from 3.00×10³ V to 100V is 3.91s

The initial potential difference $\Delta V_o$ is related to the final potential difference $\Delta V$ by

$\Delta V=\Delta V_{o} e^{-t / \tau}$

Our target is to find t, so we rearrange equation (1) for t to be

$\Delta V=\Delta V_{o} e^{-t / RC}\\ \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) = \frac{-t}{RC}\\ t= -RC \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) \tag{2}$

The body has a capacitance 150 pF while the foot has a capacitance 80 pF and both are in parallel connection. So, the equivalent capacitance is

$C= 150 \mathrm{~pF} + 80 \mathrm{~pF} = 230 \mathrm{~pF}$

The shoes of the rubber-soled has resistance $R=5000 \mathrm{~M\Omega}$ . Plug the values for $\Delta V, \Delta V_o$ and C into equation (2) to get the time t.

$t&= -RC \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) \\ &= -( 5000 \times 10^{6} \mathrm{~\Omega})(230 \times 10^{-12} \mathrm{~F}) \ln \left ( \frac{100\mathrm{~V}}{3000 \mathrm{~V}} \right ) \\ &= {3.91 \mathrm{~s}}$

Learn more about capacitance here:

brainly.com/question/12644355

#SPJ4

3 0

2 years ago

The product side of a chemical reaction is shown. → 7Ti2(SO4)3 Which number represents a coefficient?

solniwko [45]

7 is the answer.................................

3 0

3 years ago

Read 2 more answers

What happened to the speed of light if it travels from air into glass?

vovikov84 [41]

Answer:

it slows down

Explanation:

the answer

8 0

3 years ago

The following data were collected during a short race between two friends. Velocity (m/s) 0 0.5 1 1.5 2 2 4 6 2 0 Time (s) 0 2 4

scoundrel [369]

The characteristics of the kinematics allow to find the results for the questions about the movement of the body are:

a) we have four sections;

0 to 8 s The body is accelerating.

8 to 10 s The body goes at a constant speed, the acceleration is zero.

10 to 14 Body accelerating.

14 to 18 Body slowing down.

b) The acceleration is the first 8 s is: a = 0.25 m / s²

c) The maximum acceleration is: a = 1 m / s²

d) The displacement is: i) d₁ = 8m, ii) $d_{total}$ = 16 m

e) maximum speed is: v = 6 m / s

Kinematics studies the movement of bodies by finding relationships between the position, speed and acceleration of bodies.

v = v₀ + a t

y = v₀ t + ½ a t²

where v and v₀ is the current and initial velocity, respectively, a is the acceleration and t is time.

In many circumstances graphs are made for their analysis, in a graph of speed versus time when we have a horizontal line the speed is constant, the acceleration is zero and in the case of a slope there is an acceleration, we have two cases:

Positive slope the body is accelerating and the speed is increasing.

Negative slope the body is stopping, the speed decreases.

Let's answer the different questions about the system.

a) in the attached we have a graph of the velocity versus time, each section corresponds to a change in the slope of the graph, we have four sections;

0 to 8 s The body is accelerating.

8 to 10 s The body goes at a constant speed, the acceleration is zero.

10 to 14 Body accelerating.

14 to 18 Body slowing down.

b) The acceleration is the first 8 s

v = v₀ + a t

$a = \frac{v-v_o}{\Delta t}$

$a = \frac{2-0}{8-0}$

a = 0.25 m / s²

c) The maximum acceleration is when the slope is maximum.

$a = \frac{6-2}{ 14-10}$

a = 1 m / s²

Therefore the acceleration is maximum in the section between 10 and 14 s

d) The total displacement is the sum of the displacements of each section.

$d_{total } = d_1 +d_2 + d_3 +d_4$

We look for every displacement.

d₁ = v₀ + ½ a₁ Δt²

d₁ = 0 + ½ 0.25 8²

d₁ = 8 m

In the second section the velocity is constant

d₂ = v₂ Δt₂

d₂ = 2 (10-8)

d₂ = 4 m

The third section.

d₃ = v₀ + ½ a t²

d₃ = 2 + ½ 1 (14-10) ²

d₃ = 10 m

The distance of the fourth section.

we look for acceleration

a₄ = $\frac{v-v_o}{\Delta t}$

a₄ = $\frac{0-6}{18-14}$

a₄ = -1.5 m / s²

d₄ = 6 + ½ (-1.5) (1814) ²

d₄ = -6 m

The total displacement is;

$d_{total}$ = 8 + 4 + 10 -6

$d_{total}$ = 16 m

e) The maximum speed is the highest point in the graph of speed versus time that in the attachment we can see corresponds to

v = 6 m / s

In conclusion using the characteristics of kinematics we can find the results for the questions about the motion of bodies are:

a) we have four sections;

0 to 8 s The body is accelerating.

8 to 10 s The body goes at a constant speed, the acceleration is zero.

10 to 14 Body accelerating.

14 to 18 Body slowing down.

b) The acceleration is the first 8 s is: a = 0.25 m / s²

c) The maximum acceleration is: a = 1 m / s²

d) The displacement is: i) d₁ = 8m, ii) $d_{total}$ = 16 m

e) maximum speed is: v = 6 m / s

Learn more about kinematics here: brainly.com/question/24783036

3 0

2 years ago

Assuming the same current is running through two separate coils, why is it easier to thrust a magnet into a wire coil with one l

s2008m [1.1K]

The magnetic field strength in a coil is directly proportional to the number of turns, or loops, in the coil.
Therefore, when there are four loops instead of one, the magnetic field strength has increased four times, making it harder to push the magnet in.

6 0

3 years ago

Read 2 more answers