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Daniel [21]
3 years ago
15

What is the kinetic energy of a 14kg object traveling at 10m/s

Physics
1 answer:
Sauron [17]3 years ago
5 0

Answer:

Explanation:

kinetic energy=1/2*mass*velocity^2

=1/2*14kg*10^2

=7*100

=700 joule

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A steel, rigid container is filled with an ideal gas. The gas inside is heated such that its temperature, as measured in kelvin,
nikitadnepr [17]

Answer:

If the gas inside a rigid steel vessel (constant volume) is heated in such a way that its temperature, measured in Kelvin degrees, doubles then <u><em>the internal pressure will increase by a factor of 2</em></u>.

Explanation:

Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T} =k

where V = volume, T = temperature, K = Constant

This law indicates that the ratio between pressure and temperature is constant.

This law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of shocks against the walls increases, that is, the pressure increases. That is, <em>the gas pressure is directly proportional to its temperature.</em> This must be fulfilled since the relationship must remain constant

In short, when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature decreases, gas pressure decreases.

It is desired to study two different states, an initial state and an final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. When the temperature varies to a new T2 value, then the pressure will change to P2, and the following will be true:

\frac{P1}{T1} =\frac{P2}{T2}

Given the above, it is possible to say that <u><em>if the gas inside a rigid steel vessel (constant volume) is heated in such a way that its temperature, measured in Kelvin degrees, doubles then the internal pressure will increase by a factor of 2 (doubles too) </em></u>

8 0
3 years ago
If a water wave completes one cycle in 2 seconds, what is the period of the wave?
Anika [276]
The definition of the period of a wave is the time that it takes for the wave to complete one cycle. As stated in the problem, the certain water wave completes one cycle in 2 seconds. From this information alone, we can conclude that the period of the wave is indeed, 2 seconds. 
8 0
3 years ago
Read 2 more answers
A 39.7 n object is in free fall. what is the magnitude of the net force which acts on the object? answer in units of n.
Vika [28.1K]
If there is no air resistance, the value would still be the same. So therefore, the answer is 39.7 N.
Supposing that there is an air resistance of 19.8 N.With a 19.8 N air resistance: ∑F = 39.7 N - 19.8 N = 19.9 N would be the net force. The important point here is that the force of gravity equivalent the weight of the object.
6 0
3 years ago
An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 2.14 mT. If the speed of the
JulsSmile [24]

Answer:

(a) 3.9cm

(b) 1.66 x 10⁻⁸s

Explanation:

Since the electron is moving in a circular path, the centripetal acceleration needed to keep it from slipping off is provided by the magnetic force. This force (F), according to Newton's second law of motion is given by,

F = m x a          --------------(i)

Where;

m = mass of the particle

a = acceleration of the mass

The centripetal acceleration is given by;

a = v² / r          [v = linear velocity of particle, r = radius of circular path]

<em>Therefore, equation (i) becomes;</em>

F = m v²/ r             --------------------(ii)

The magnitude of the magnetic force on a moving charge in a magnetic field as stated by Lorentz's law is given by;

F = qvBsinθ          -------------(iii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = angle between the velocity and the magnetic field

<em>Combine equations (ii) and (iii) as follows;</em>

m (v² / r) = qvBsinθ         [divide both side by v]

m v / r = qBsinθ              [make r subject of the formula]

r = (m v) / (qBsinθ)              ---------(iv)

(a) From the question;

v = 1.48 x 10⁷m/s

B = 2.14mT = 2.14 x 10⁻³T

θ = 90°          [since the direction of velocity is perpendicular to magnetic field]

m = mass of electron = 9.11 x 10⁻³¹kg

q = charge of electron = 1.6 x 10⁻¹⁹C

Substitute these values into equation (iv) as follows;

r = (9.11 x 10⁻³¹ x 1.48 x 10⁷) / (1.6 x 10⁻¹⁹ x 2.14 x 10⁻³ sin 90°)

r = 3.9 x 10⁻²m

r = 3.9cm

Therefore, the radius of the circular path is 3.9cm

(b) The time interval required to complete one revolution is the period (T) of the motion of the electron and it is given by

T = d / v          --------------(*)

Where;

d = distance traveled in the circular path in one complete turn = 2πr

v = velocity of the motion = 1.48 x 10⁷m/s

d = 2 π (3.9 x 10⁻²)            [Take π = 22/7 = 3.142]

d = 2(3.142)(3.9 x 10⁻²) = 0.245m

Substitute the values of d and v into equation (*) as follows;

T = 0.245 / 1.48 x 10⁷

T = 0.166 x 10⁻⁷s

T = 1.66 x 10⁻⁸s

Therefore, the time interval is 1.66 x 10⁻⁸s

6 0
3 years ago
NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supp
IRINA_888 [86]

Complete question :

NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supply storage area of the lunar outpost where gravity is 1.63m/s/s can only support 1 x 10 over 5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost?

Answer:

601000 N

Explanation:

Given that :

Acceleration due to gravity at lunar outpost = 1.6m/s²

Supported Weight of supplies = 1 * 10^5 N

Acceleration due to gravity on the earth surface = 9.8m/s²

Maximum weight of supplies as measured on EARTH :

Ratio of earth gravity to lunar post gravity:

(Earth gravity / Lunar post gravity) ;

(9.8 / 1.63) = 6.01

Hence, maximum weight of supplies as measured on EARTH should be :

6.01 * (1 × 10^5)

6.01 × 10^5

= 601000 N

3 0
3 years ago
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