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3 years ago

What happens to the pressure in a tire if air is slowly leaking out of the tire? explain your answer?

1 answer:

3 0

It's pretty simple. When air is leaking out of a tire, like a tiny hole or something, the pressure in the tire decreases, because without air in the tire, there is no pressure.

You might be interested in

How long will the ball be in the air if the cliff is 120 m tall and the ball falls to the base of the cliff?

Readme [11.4K]

Answer:

Explanation:

4.95s≈5s

Use equation :

h=G*t²/2

h=120m ----hight of cliff

G=9.81m/s²

t=?

-----------------------

h=G*t²/2

120m=9.81m/s²*t²/2

240m=9.81m/s²*t²

t²=240m/ 9.81m/s²

t²=24.46s

t=√24.46s²

t=4.95 s≈5s

3 0

3 years ago

Investigar la función de los espejos esféricos y como funciona, en el microscopio.

NikAS [45]

Answer:

yea

Explanation:

4 0

2 years ago

6. Which Thanksgiving dish contains the most dietary fiber ? (Corn) (Mashed Potatoes) (Sweet Potatoes) (Pecan Pie)

pashok25 [27]

Answer:

The correct option is;

Sweet Potatoes

Explanation:

The dietary fiber in Corn = 2.7 g per 100 g

The dietary fiber in Mashed Potatoes = 1.3 g per 100 g

The dietary fiber in Sweet Potatoes = 3 g per 100 g

The dietary fiber in Pecan Pie = 2.1 g per 100 g

Therefore, the Thanksgiving dish that contains the most dietary fiber per 100 g is Sweet Potatoes

8 0

2 years ago

bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. what is the veloc

USPshnik [31]

Answer:

The final velocity of the car A is -1.053 m/s.

Explanation:

For an elastic collision both the kinetic energy and the momentum of the system are conserved.

Let us call

$m_A$ = mass of car A;

$v_{A1}$ = the initial velocity of car A;

$v_{A2}$ = the final velocity of car A;

and

$m_B$ = mass of car B;

$v_{B1}$ = the initial velocity of car B;

$v_{B2}$ = the final velocity of car B.

Then, the law of conservation of momentum demands that

$m_Av_{A1}+m_Bv_{B1} =m_Av_{A2}+m_Bv_{B2}$

And the conservation of kinetic energy says that

$\dfrac{1}{2} m_Av_{A1}^2+\dfrac{1}{2}m_Bv_{B1}^2=\dfrac{1}{2}m_Av_{A2}^2+\dfrac{1}{2}m_Bv_{B2}^2$

These two equations are solved for final velocities $v_{A2}$ and $v_{B2}$ to give

$v_{A2} =\frac{m_A-m_B}{m_A+m_B} v_{A1}+\frac{2m_B}{m_A+m_B} v_{B1}$

$v_{B2} =\frac{2m_A}{m_A+m_B} v_{A1}+\frac{m_B-m_A}{m_A+m_B} v_{B1}$

by putting in the numerical values of the variables we get

$v_{A2} =\frac{281-209}{281+209} (2.82)+\frac{2*209}{281+209} (-1.72)$

$\boxed{v_{A2} = -1.05m/s}$

and

$v_{B2} =\frac{2*281}{281+209} (2.82)+\frac{209-281}{281+209} (-1.72)$

$\boxed{v_{B2} = 3.49m/s}$

Thus, the final velocity of the car A is -1.053 m/s and of car B is 3.49 m/s.

4 0

3 years ago

A dust particle with mass of 5.4×10−2 g and a charge of 2.3×10−6 C is in a region of space where the potential is given by V(x)=

user100 [1]

Answer:

1.6 m/s^2

Explanation:

Hello!

To calculate the acceleration we must know the electric field. The electric field and the potential are related by:

$E = -\frac{dV}{dx} =- 2.6(\frac{V}{m^{2}})x + 8.1(\frac{V}{m^{3}})x^{2}$

If the particle starts at 2.3m, the electric field is:

E = 36.869 V/m = 36.869 N/C

So, the force on the particle is:

F = q E = 2.3×10^−6 C * 36.869 N/C = 8.48 x 10^-5 N

And its acceleration is :

a = F/m = 8.48 x 10^-5 N / 5.4×10−5 kg = 1.57 m/s^2

Rounded to two significant figures:

1.6 m/s^2

6 0

3 years ago