Ask question

Ask question

All categories

2 months ago

10

PROBLEMA DI FISICA:il volume di un attizzatoio da camino in ferro, a temperatura ambiente (20°C), è 12,2 cm cubi. A contatto con

le braci i riscalda e passa al volume di 12,4 cm cubi. ?calcola temperatura delle braci?

1 answer:

Maru [3.3K]2 months ago

4 0

Responder:

20.3 ° C

Explicación:

According to Charles's Law: when the pressure on a sample of dry gas remains constant, temperature and volume will be directly proportional.

Step one:

given data

Temperature T1 = 20 ° C

Temperature T2 =?

Volume V1 = 12.2 cm ^ 3

Volume V2 = 12.4 cm ^ 3

Apply the relation between temperature and volume

$\frac {V_{1}}{T_{1}}=\frac {V_{2}}{T_{2}}$

substituting we have

$\frac {12.2}{20}=\frac {12.4}{T_{2}}\\\\$

Cross-multiplying we find

$T_2=\frac{20*12.4}{12.2} \\\\T_2=\frac{248}{12.2}\\\\T_2=20.3$

Temperature of the embers 20.3°C

You might be interested in

A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa

kicyunya [3294]

The result is 15.67 seconds. Using the first equation of motion, we have: Final Velocity = Initial Velocity + (Acceleration * Time). This translates mathematically to v = u + at. Now, plugging in our values: v = 3, u = 50, and a = -4 (indicating negative acceleration). We find that 3 = 50 + (-4 * t) leads to -47/-4 = t, giving us a time of 15.67 seconds.

6 0

1 month ago

Identify the situations that have an unbalanced force. Check all that apply. A baseball speeds up as it falls through the air. A

kicyunya [3294]

<span>As a baseball descends through the air, it accelerates.
Indeed. Unbalanced forces act on the balloon.
The balloon accelerates, indicating that the gravitational force
exceeds the air resistance acting upwards.

A soccer ball remains still on the ground.
No. Since the ball isn't accelerating, it implies that the forces acting on it
are in balance.
The force of gravity pulling the ball downwards is equal to the force
exerted by the ground upwards.

An ice skater moves straight at a steady speed.
No. The skater's speed and direction remain unchanged, hence he is not
accelerating. This indicates that the forces acting on him are balanced.

A bumper car collides with another and veers at an angle.
Yes. The car's trajectory changed due to the collision.
This constitutes acceleration, confirming that the forces acting on it are unbalanced,
at least at the moment of collision.

When released, a balloon swiftly moves across the room.
Yes. Initially, the balloon was stationary. But as the nozzle was
opened, it began darting around the room. Therefore, its speed altered.
Additionally, as it zooms about the room, its direction changes as well.
This indicates significant acceleration, which means the forces on it
are not balanced.</span>

5 0

1 month ago

Read 2 more answers

a truss is made by hinging two uniform, 150-N rafters as shown in Fig. 5-21. They rest on an essentially frictionless floor and

Softa [3030]

<span>Let L and R represent the left and right vertical reactions on the floor
let T signify the tension in the rope
the total force must equal zero

R + L - 500 - 150 - 150 = 0

R + L = 800

given that the triangle formed is isosceles, the legs must be equal
R = L = 400 N

by applying the torque summation

L [ 4] - T[ 3 - (2 *3/5)] = 0
400 [ 4] - T[ 1.8] = 0
T = 1600/1.8
T = 889 N </span>

7 0

1 month ago

. A magnetic field has a magnitude of 0.078 T and is uniform over a circular surface whose radius is 0.10 m. The field is orient

Keith_Richards [3271]

Response:

The magnetic flux across the surface amounts to $2.22 \times 10^{-3}$ Wb

Clarification:

Provided:

Strength of magnetic field $B = 0.078$ T

Circle's radius $r = 0.10$ m

Angle of incidence between the field and the surface normal $\theta =$ 25°

Utilizing the flux formula,

$\phi = B.A$

$\phi = BA\cos \theta$

Where $\theta =$ the angle represents the orientation of the magnetic field line relative to the surface normal, $A =$ and the area pertains to the circular surface.

$A = \pi r^{2}$

$A = 3.14 \times (0.10) ^{2}$

$A = 0.0314$ $m^{2}$

The calculation for magnetic flux is expressed as

$\phi = 0.078 \times 0.0314 \times \cos 25$

$\phi = 2.22 \times 10^{-3}$ Wb

Thus, the magnetic flux through the surface calculates to $2.22 \times 10^{-3}$ Wb

6 0

1 month ago

What fraction is this of the satellite's weight at the surface of the earth? take the free-fall acceleration at the surface of t

kicyunya [3294]

I have included the additional details regarding the question.
It is important to remember that weight represents the strength of the gravitational pull on an object.
The weight at the surface can be expressed as:
$F_g=mg$
Here, we utilize the standard g value. However, g varies with elevation. The reason this variation is rarely observed is due to the fact that the height must be considerable relative to Earth's radius for it to make a notable impact.
According to Newton's law of gravitation, we have:
$F_g=G\frac{m_em}{r^2}$
In this scenario, r refers to the distance between the centers of mass of the interacting objects. While standing on the Earth's surface, r is equivalent to Earth's radius:
$F_g'=G\frac{m_em}{r_e^2}\\ ma=G\frac{m_em}{r_e^2}\\ a=G\frac{m_e}{r_e^2}$
This acceleration is what we denote as g.
When at a certain height, the expression is:
$F_g=G\frac{m_em}{(r_e+h)^2}\\ ma=G\frac{m_em}{(r_e+h)^2}\\ a=G\frac{m_e}{(r_e+h)^2}\\$
Let’s call this new acceleration g'. Dividing it by g gives us:
$\frac{g'}{g}=\frac{r_e^2}{(r_e+h)^2}\\ g'=g\cdot\frac{r_e^2}{(r_e+h)^2}$
Thus, the weight can be expressed as:
$F_g'=mg'=mg\cdot\frac{r_e^2}{(r_e+h)^2}$
By substituting the values, we find:
$F_g'=2.02N$
For Earth, we obtain:
$F_g=22050N$
Consequently, the resulting ratio is:
$\frac{F_g'}{Fg}=\frac{2.02}{22050}=0.000091=0.0091\%$

3 0

29 days ago