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Alenkasestr [34]
3 years ago
9

PLEASE HELP THIS TEST IS DUE IN TEN MINS

Physics
2 answers:
Sveta_85 [38]3 years ago
4 0

Answer:

A I hope you get it !!!!!!!

nekit [7.7K]3 years ago
3 0

Answer:

A. A statement of how the volume of a gas is related to its temperature.

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A 5 kg block moves with a constant speed of 10 ms to the right on a smooth surface where frictional forces are considered to be
nirvana33 [79]

Answer:

Work done, W = 19.6 J

Explanation:

It is given that,

Mass of the block, m = 5 kg

Speed of the block, v = 10 m/s

The coefficient of kinetic friction between the block and the rough section is 0.2

Distance covered by the block, d = 2 m

As the block passes through the rough part, some of the energy gets lost and this energy is equal to the work done by the kinetic energy.

W=\mu_kmgd

W=0.2\times 5\times 9.8\times 2

W = 19.6 J

So, the change in the kinetic energy of the block as it passes through the rough section is 19.6 J. Hence, this is the required solution.

5 0
3 years ago
Read 2 more answers
A 600 g model rocket is on a cart that is rolling to the right at a speed of 2.5 m/s. The rocket engine, when it is fired, exert
Sergio039 [100]

Answer: 8.43 m

Explanation:

Given

Mass of the rocket, m = 600 g = 0.6 kg

Speed of riling, v = 2.5 m/s

F = mg, where

F = weight of the rocket in N

m = mass of the rocket in kg

g = acceleration due to gravity in m/s²

F = 0.6 * 9.8

F = 5.88 N

Since the the vertical thrust on the rocket is 8 N and 5.88 N of it will be used to counter the rockets weight. Thus, the remaining 2.12 N is what is available to provide acceleration.

Upward acceleration of the rocket is..... F = ma, a = F / m

a = 2.12 / 0.6

a = 3.53 m/s²

using equation of motion, we will calculate how long it will take to rise 20 m in the air

S = ut + 1/2at², where u = 0

20 = 1/2 * 3.53 * t²

t² = 4. / 3.53

t² = 11.33

t = √11.33 = 3.37 s

The distance then is

v = Δx / t, such that

Δx = v * t

Δx = 2.5 * 3.37

Δx = 8.43 m

4 0
3 years ago
Read 2 more answers
If a train travel from Addis Ababa to Dire Dawa at a constant velocity of 400Km/hr and the to Djibouti at a constant velocity of
madreJ [45]

The value of V is 640 km/hr.

<h3>What is Average velocity?</h3>

Average velocity is the change in position or displacement (∆x) divided by the time intervals (∆t) in which the displacement occurs.

To calculate the value of V from the question, we use the formula below.

  • V' = (V+U)/2............ Equation 1

Making V the subject of the equation

  • V = 2V'-U.............. Equation 2

From the question,

⇒ Given:

  • V' = 520 km/hr
  • U = 400 km/hr

Substitute these values into equation 2

  • V = (520×2)-400
  • V = 640 km/hr

Hence, the value of V is 640 km/hr.

Learn more about average velocity here: brainly.com/question/4931057

#SPJ1

4 0
2 years ago
Jupiter has a magnetic field that is a. much stronger than that of Earth and greatly extended. b. very strong and localized clos
irina [24]

Answer:

a. much stronger than that of Earth and greatly extended.

Explanation:

Jupiter is a gas giant. It mainly consists of hydrogen and helium and also may have rocky core. Jupiter has the fastest rotational period among the planets of the solar system. One day on Jupiter lasts for 11 hours. This rotation causes the hydrogen in the atmosphere to rotate at great speed.

The rotation of the hydrogen generates the magnetic field of Jupiter. It is almost 14 times greater than the Earth's magnetic field in terms of strength and the second largest in terms of distance after the sun.

7 0
3 years ago
A point charge q1 is at the center of a sphere of radius 20 cm. Another point charge q2 = 5 nC is located at a distance r = 30 c
valina [46]

Answer:

q_1 =7.08*10^{-9}C.

Explanation:

Gauss's Law says that the electric flux \Phi_E through a closed surface is directly proportional to the charge Q_{enc} inside it. More precisely,

$\Phi_E=\oint_S E\cdot dA = \dfrac{Q_{enc}}{\epsilon_0}. $

This means what is outside this closed surface S does not contribute to the flux through it because field lines that go in must come out, <em>resulting a zero flux from an external charge. </em>

In our context, this means the charge q_2 which is outside the sphere will have zero flux through the surface; therefore, Gauss's law will only be concerned with charge q_1 which is inside the sphere; Hence,

$\Phi_E=\oint_S E\cdot dA = \dfrac{q_1}{\epsilon_0} = 800 N\cdot m^2/C. $

Solving for q_1 gives

$ q_1= (800 N\cdot m^2/C)\epsilon_0, $

$ q_1= (800 N\cdot m^2/C)*(8.85*10^{-12}C^2/N\cdot m^2) $

\boxed{q_1 =7.08*10^{-9}C. }

which is the charge inside the sphere.

5 0
3 years ago
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