Find the momentum of a 25 kg object traveling at a speed of 4 m/s

Two objects that are not initially in thermal equilibrium are placed in close contact. After a while, the temperature of the cod

Dima020 [189]

Answer:

If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops, then the specific heats of both objects will be equal.

Explanation:

If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops when the two objects of same mass are brought into contact, then their specific heat capacity is equal.

We can prove this by the equation of heat for the two bodies:

According to given condition,

$\Delta T_1=\Delta T_2$

$\frac{Q_1}{m_1.c_1} = \frac{Q_2}{m_2.c_2}$

when there is no heat loss from the system of two bodies then $Q_1=Q_2$

$\frac{1}{m.c_1} =\frac{1}{m.c_2}$

$\Rightarrow c_1=c_2$

Thermal conductivity is ultimately affects the rate of heat transfer, however the bodies will attain their final temperature based upon their mass and their specific heat capacities.

The temperature of the colder object will rise twice as much as the temperature of the hotter object only in two cases:

when the specific heat of the colder object is half the specific heat of the hotter object while mass is equal for both.

OR

the mass of colder object is half the mass of the hotter object while their specific heat is same.

3 0

3 years ago

At a particular instant, a hot air balloon is 100 m in the air and descending at a constant speed of 2.0 m/s. at this exact inst

rewona [7]

Answer:

86.4 m horizontal from landing spot

Explanation:

Find out how long before the ball hits the ground

vertical speed of ball = -2 m/s gravity = - 9.81 m/s^2

find time to hit ground from 100 m

( height will be zero when it hits the ground)

0 = 100 - 2 t - 1/2 ( 9.81) t^2

use Quadratic Formula to find t = 4.32 seconds

horizontal speed of ball = 20 m/s

in 4.32 seconds it will travel horizontally 20 m/s * 4.32 s = 86.4 m

3 0

2 years ago

Un satélite geoestacionario se encuentra a una distancia de 120.000 km sobre la superficie de Júpiter. Determine: a. El periodo

Lisa [10]

Answer:

a) a geostationary satellite is that it is always at the same point with respect to the planet,

b) f = 2.7777 10⁻⁵ Hz

c) d) w = 1.745 10⁻⁴ rad / s

Explanation:

a) The definition of a geostationary satellite is that it is always at the same point with respect to the planet, that is, its period of revolutions is the same as the period of the planet

T = 10 h (3600 s / 1h) = 3.6 104 s

b) the period the frequency are related

T = 1 / f

f = 1 / T

f = 1 / 3.6 104

f = 2.7777 10⁻⁵ Hz

c) the distance traveled by the satellite in 1 day

The distance traveled is equal to the length of the circumference

d = 2pi (R + r)

d = 2pi (69 911 103 + 120 106)

d = 1193.24 m

d) the angular velocity is the angle traveled between the time used.

.w = 2pi /t

w = 2pi / 3.6 10⁴

w = 1.745 10⁻⁴ rad / s

how fast is

v = w r

v = 1.75 10-4 (69.911 106 + 120 106)

v = 190017 m / s

5 0

3 years ago

How do you determine the specific heat of a substance?

svetlana [45]

Answer:

I think it's A

Explanation:

3 0

3 years ago

A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit

vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

$\omega =\sqrt{\dfrac{K}{m}}$

The maximum speed v given as

$v=\omega A$

A=Amplitude

$v=\sqrt{\dfrac{K}{m}}\times A$

$0.32=\sqrt{\dfrac{13.5}{0.75}}\times A$

A=0.075 m

A= 0.75 cm

The speed at distance x

$v=\omega \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}$

v= 0.21 m/s

5 0

3 years ago