1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
levacccp [35]
3 years ago
15

Due to the friction between a tire and the road surface, the temperature of the air inside a tire increases as one drives. If th

e tire pressure is 37 psi when the car is in the driveway at 25° C, What is the pressure of the tire when the temperature increases to 50° C?
Physics
1 answer:
inysia [295]3 years ago
5 0

Answer:

40.10 atm .

Explanation:

Here volume of tire is assumed to be nearly constant .

Modified gas law ,

P₁ / T₁ = P₂ / T₂

37 / ( 273 + 25 ) = P₂ / ( 273 + 50)

37 / 298 = P₂ / 323

P₂ = 37 x 323 / 298

= 40.10 atm .

You might be interested in
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas
Pavel [41]

Complete question is:

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.

Answer:

(V_A) = 31.32 m/s

Explanation:

We are given;

car's mass, m = 1200 kg

h_A = 100 m

h_B = 150 m

v_B = 0 m/s

From law of conservation of energy,

the distance from point A to B is;

h = 150m - 100 m = 50 m

From Newton's equations of motion;

v² = u² + 2gh

Thus;

(V_B)² = (V_A)² + (-2gh)

(negative next to g because it's going against gravity)

Thus;

(V_B)² = (V_A)² - (2gh)

Plugging in the relevant values;

0² = (V_A)² - 2(9.81 × 50)

(V_A) = √981

(V_A) = 31.32 m/s

3 0
3 years ago
4) A satellite, mass m, is in circular orbit (radius r) around the earth, which has mass ME and radius Re. The value of r is lar
defon
<h2>Answers:</h2>

(a) The kinetic energy of a body is that energy it possesses due to its movement and is defined as:

K=\frac{1}{2}m{V}{2}     (1)

Where m is the mass of the body and V its velocity.

In this specific case of the satellite, its kinetic energy K_m taking into account its mass m is:

K_{m}=\frac{1}{2}m{V}^{2}     (2)

On the other hand, the velocity of a satellite describing a circular orbit is constant and defined by the following expression:

V=\sqrt{G\frac{ME}{r}}     (3)

Where G is the gravity constant, ME the mass of the Earth and r the radius of the orbit <u>(measured from the center of the Earth to the satellite). </u>

Now, if we substitute the value of V from equation (3) on equation (2), we will have the final expression of the kinetic energy of this satellite:

K_{m}=\frac{1}{2}m{\sqrt{G\frac{ME}{r}}}^{2}     (4)

Finally:

K_{m}=\frac{1}{2}Gm\frac{ME}{r}     (5)  >>>>This is the kinetic energy of the satellite

(b) According to Kepler’s 2nd Law applied in the case of a circular orbit, its Period T is defined as:

T=2\pi\sqrt{\frac{r^{3}}{\mu}}     (6)

Where \mu is a constant and is equal to GME. So, this equation in these terms is written as:

T=2\pi\sqrt{\frac{r^{3}}{GME}}     (7)

As we can see, <u>the Period of the orbit does not depend on the mass of the satellite </u>m, it depends on the mass of the greater body (the Earth in this case) ME, the radius of the orbit and the gravity constant.

(c) The gravitational force described by the law of gravity is a central force and therefore is <u>a conservative force</u>. This means:

1. The work performed by a gravitational force to move a body from a position A to a position B <u>only depends on these positions and not on the path followed to get from A to B. </u>

2. When the path that the body follows between A and B is a c<u>losed path or cycle</u> (as this case with a <u>circular orbit</u>), <u>the gravitational work is null or zero</u>.

<h2>This is because the gravity force that maintains an object in circular motion is a centripetal force, that is, <u>it always acts perpendicular to the movement</u>. </h2>

Then, in the case of the satellite orbiting the Earth in a circular orbit, its movement will always be perpendicular to the gravity force that attracts it to the planet, at each point of its path.

(d)  The total Mechanical Energy E of a body is the sum of its Kinetic Energy K and its Potential Energy P:

E=K+P     (8)

But in this specific case of the circular orbit, its kinetic energy will be expresses as calculated in the first answer (equation 5):

K_{m}=\frac{1}{2}Gm\frac{ME}{r}     (5)

And its potential energy due to the Earth gravitational field as:

P_{m}=-G\frac{mME}{r}     (9)

This energy is negative by definition.

So, the total mechanical energy of the orbit, also called the Orbital Energy is:

E=\frac{1}{2}Gm\frac{ME}{r}+(- G\frac{mME}{r})      (10)

Solving equation (10) we finally have the Orbital Energy:

E=-\frac{1}{2}mME\frac{G}{r}     (11)

At this point, it is necessary to clarify that a satellite (or any other celestial body) orbiting another massive body, can describe one of these types of orbits depending on its Orbital Total Mechanical Energy E:

-When E=0:

We are talking about an <u>open orbit</u> in which the satellite escapes from the attraction of the planet's gravitational field. The shape of its trajectory is a parabola, fulfilling the following condition:

K_{m}=-P_{m}

Such is the case of some comets in the solar system.

-When E>0:

We are also talking about <u>open orbits</u>, which are hyperbolic, being K_{m}>P_{m}

<h2>-When E: >>>><u>This case</u></h2>

We are talking about <u>closed orbits</u>, that is, the satellite will always be "linked" to the gravitational field of the planet and will describe an orbit that periodically repeats with a shape determined by the relationship between its kinetic and potential energy, as follows:

-Elliptical orbit: Although E is constant, K_m and P_m are changing along the trajectory .

-Circular orbit: When at all times both the kinetic energy K_m and the potential P_m remain constant, resulting in a total mechanical energy E as the one obtained in this exercise. This means that the speed is constant too and <u>is the explanation of why this Energy has a negative sign. </u>

3 0
4 years ago
The rate of chage of displacement is called __________​
faltersainse [42]

Answer:

Velocity is the rate of change of displacement.

8 0
3 years ago
Read 2 more answers
HELP ME PLEASE
Stels [109]
The answer is B. the germinal stage
7 0
3 years ago
Read 2 more answers
Pls help i’m in a test and don’t know what to do
Tresset [83]

Answer:

I think its C, even with the typo.

6 0
3 years ago
Other questions:
  • One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
    9·1 answer
  • Why is it expensive to bring electricity into urban areas?
    15·2 answers
  • A fire truck has a searchlight with a resistance of 60 (ohm) which is placed across a 24-V battery. What is the current in this
    15·1 answer
  • What is the acceleration of an object with mass of 42.6 kg when an unbalanced force of 112 N is applied to it
    5·2 answers
  • An object is placed 11.0 cm in front of a concave mirror whose focal length is 24.0 cm. The object is 2.60 cm tall. What is the
    10·1 answer
  • If I move 15ft foward, 15 ft backwards, 15 ft to the right, 15ft to the left where am I?
    7·2 answers
  • Why do we see only one side of the moon from earth?
    12·2 answers
  • How much kinetic energy does a person of mass 50kg running at 3m/s have? Provide
    13·1 answer
  • Based on Archimedes' principle, the greatest buoyant force an object can experience in water is determined by which quantity?
    12·1 answer
  • What type of image can be larger or smaller than the object?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!