What does it mean to say that two systems are in thermal equilibrium

(a) The initial speed of the rock as it left the astronaut's hand is 19.35 m/s.

(b) The speed of the satellite is 50.24 m/s.

g = GM/R²

where;

g = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(63200)²

g = 0.13 m/s²

v² = u² - 2gh

where;

at maximum height, v = 0

u² = 2gh

u = √2gh

u = √(2 x 0.13 x 1,440)

u = 19.35 m/s

v = √GM/r

r = radius of planet Glob + radius of the satellite

r = 63200 m + 145,000 m = 208,200 m

v = √[(6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(208,200)]

v = 50.24 m/s

Thus, the initial speed of the rock as it left the astronaut's hand is 19.35 m/s.

The speed of the satellite is 50.24 m/s.

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Answer:

Create space

Watch your breath

Watch your thoughts

Be gentle with yourself

Affirm what you want

Explanation:

(A) The spring constant of the spring is 4.94 N.

(B) The speed of the cart after pushing it is 0.47 m/s.

(C) The force applied to the cart is 0.75 N.

ω = √k/m

where;

0.5 rev/s = 0.5(2π) rad/s = π rad/s = 3.142 rad/s

ω² = k/m

k = mω²

k = 0.5 x (3.142)²

k = 4.94 N/m

E = ¹/₂kA²

where;

A is amplitude

E = ¹/₂(4.94)(0.15)²

E = 0.056 J

E = ¹/₂mv²

2E = mv²

v² = 2E/m

v² = (2 x 0.056)/(0.5)

v² = 0.224

v = √0.224

v = 0.47 m/s

E = ¹/₂FA

2E = FA

F = 2E/A

F = (2 x 0.056)/(0.15)

F = 0.75 N

Thus, the spring constant of the spring is 4.94 N. The speed of the cart after pushing it is 0.47 m/s. The force applied to the cart is 0.75 N.

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(a) The weight of the drop is 1.54 x 10⁻²⁵ N,

(b) The the charge on the drop, in terms of e is 5 x 10⁻¹²e and

(c) The excess electrons is 5 x 10⁻¹² electron.

The weight of the drop is calculated as follows;

Volume of the drop; V = ⁴/₃πr³

V = ⁴/₃π(1.64 x 10⁻⁶)³ = 1.845 x 10⁻¹⁷ m³

mass = density x volume

mass =  0.851 g/cm³ x  1.845 x 10⁻²³ cm³ = 1.572 x 10⁻²³ g =  1.57 x 10⁻²⁶ kg.

Weight = 1.57 x 10⁻²⁶ kg x 9.8 m/s² = 1.54 x 10⁻²⁵ N.

q = F/E

q = (1.54 x 10⁻²⁵ N)/(1.92 x 10⁵)

q = 8.01 x 10⁻³¹ C

1.6 x 10⁻¹⁹ C = 1e

8.01 x 10⁻³¹ C = ?

= 5 x 10⁻¹²e

1.6 x 10⁻¹⁹ C ------- 1 electron

8.01 x 10⁻³¹ C ------- ?

= 5 x 10⁻¹² electron

Thus, the weight of the drop is 1.54 x 10⁻²⁵ N, the the charge on the drop, in terms of e is 5 x 10⁻¹²e and the excess electrons is 5 x 10⁻¹² electron.

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The highest point of the wheel is the position of the wheel when its potential energy is greatest.

The position of the wheel when its potential energy is greatest when it is at the highest point because potential energy depends on the height of an object. If the object is at more height then it has more potential energy and vice versa.

So we can conclude that the highest point of the wheel is the position of the wheel when its potential energy is greatest.

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C. The opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.

The law of increasing opportunity cost is an economic principle that describes how opportunity costs increase as resources are applied.

As the student gives up his sleep or night rest in the place of his exam preparation, we say that the opportunity cost is the sleep or rest.

Thus, the opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.

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Answer:

See below

Explanation:

I will assume the force is in a DOWNWARD direction ( I believe it makes no answer difference)

Horizontal component is then 100 cos 36° =80.9 N

F = ma

80.9 = 25 kg  *a

a = 3.24 m/s^2

Answer:

See image

Explanation:

Plato

The solution for the questions below is mathematically given as

(A)

Generally, the equation for the flux is  mathematically given as

[tex]\Phi _{B}=B.A[/tex]

[tex]\Phi _{B}=B.A\\\\\Phi _{B}=B(40*30*10^{-4}) ,[/tex]

So, the magnetic flux through the coil is constant.

From faradays law,

[tex]\varepsilon =-\frac{\mathrm{d} \Phi _{B}}{\mathrm{d} t}[/tex]

---(1) for the induced emf. Since magnetic flux is constant, LHS. of (1) =0

induced emf=0

(B)

Let x be the length of the coil's magnetic field area.

[tex]then, \Phi _{B}=B.A=B(40*x*10^{-4}) \\\\\frac{\mathrm{d}\Phi _{B} }{\mathrm{d} t}=40\\B*10^{-4}*\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]

induced emf=0.0132V

(C)

In conclusion, Therefore, there is no variation in the magnetic flux across the coil when magnetic flux=0.

induced emf=0

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(a) The maximum height reached by Jane is 1.8 m.

(b) The length of the vine will affect the time of her motion, which will impact on speed and maximum height attained.

apply the principle of conservation of energy;

P.E = K.E

mgh = ¹/₂mv²

h = v²/2g

where;

h = (6²)/(2 x 9.8)

h = 1.84 m

Assuming Jane to be in simple harmonic motion, the time of motion is calculated as;

T =  2π√(L/g)

where;

Thus, the length of the vine will affect the time of her motion, which will impact on speed and maximum height attained.

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Since upward forces must be equal to the downward forces, the tension in the cable is 225.3 N

The two conditions are;

The given parameters are;

Tsinθ = mg

Tsin 66 = 21 x 9.8

T = 205.8 / 0.914

T = 225.3 N

Therefore, the tension in the cable is 225.3 N

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Answer:

No

Explanation:

Plastic bags are not magnetic materials, only magnetic materials (such as iron) can be attracted by the magnet.

Hope this helps.

Hi there!

We can use Biot-Savart's Law for a moving particle:
[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }[/tex]

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }[/tex]

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }[/tex]

[tex]B = \boxed{7.07 *10^{-10} T}[/tex]

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

c)

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

[tex]B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}[/tex]

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

[tex]\boxed{B = 0 T}[/tex]

Answer:

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The work done is given by 742.5 J while the coefficient of kinetic friction between the block and the surface is 0.46.

The work done is given by the use of the formula;

W = F * x

Where;

F = force applied

x = distance covered

W = 150 N *  4.95 m = 742.5 J

Now;

The coefficient of kinetic friction is given by;

μ = F/mg

μ = 150/ 33 * 9.8

μ = 0.46

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The magnitude of the cannons velocity after the ball is fired will be 2.4m/s.

To find the answer, we have to know more about the law of conservation of linear momentum.

                      [tex]M_B=8kg\\M_C=1*10^3 kg\\P_f=2.4*10^3 kgm/s.\\[/tex]

                   [tex]P_f=M_CV_C\\V_C=\frac{P_f}{M_C}=\frac{2.4*10^3}{1*10^3} =2.4m/s \\west[/tex]

Thus, we can conclude that, the magnitude of the cannons velocity after the ball is fired is 2.4m/s.

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After the ball is shot, the cannon will move at a speed of 2.4 m/s.

We must learn more about the rule of conservation of linear momentum in order to locate the solution.

How can I determine the cannon's post-ball velocity magnitude?

                          [tex]m=8kg\\M=1000kg\\P_f=2.4*10^3kgm/s[/tex]

                               [tex]P_f=MV\\V=\frac{P_f}{M} =2.4m/s[/tex]

Thus, we can infer that the cannon's maximum speed once the ball is fired is 2.4 m/s.

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The Aster's final position from her initial position is 64 m

The angle is 300° and She has to head in West north direction to return to her initial position

The displacement is the distance travelled in a specific direction. Displacement is a vector quantity.

Given that Aster walks first 70 m in the direction 37° north of east, and then walks 82 m in the direction 20° south of east, and finally walks 28 m in the direction 30° west of north.

This can be solved by using bearing method. Cosine formula will be the best to solve for the distance D.

[tex]D^{2}[/tex] = [tex]70^{2}[/tex] + [tex]82^{2}[/tex] - 70 x 82 x Cos (37 + 20)

[tex]D^{2}[/tex] = 4900 + 6724 - 5740Cos57

[tex]D^{2}[/tex] = 11624 - 3126.23

[tex]D^{2}[/tex] = 8497.8

D = [tex]\sqrt{8497.8}[/tex]

D = 92.2 m

a) The Aster's final position from her initial position is 92.2 - 28 = 64 m

The angle = 270° + 30° = 300°

b) She has to head in West north direction to return to her initial position

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The spring constant, k, in newtons per meter is 1,955.9 N/m.

h = v²sin²θ/2g

v = √[(2gh)/sin²θ]

v = √[(2 x 9.8 x 4.4)/ (sin 39)²]

v = 14.76 m/s

E = K.E + P.E

E = ¹/₂m(v cosθ)²  +  mgh

E =  ¹/₂(1.1)(14.76 cos39)²  +  (1.1 x 9.8 x 4.4)

E = 119.8 J

E = ¹/₂kx²

k = 2E/x²

k = (2 x 119.8)/(0.35²)

k = 1,955.9 N/m

Thus, the spring constant, k, in newtons per meter is 1,955.9 N/m.

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(a) The proton’s potential energy change is 3.6 x 10⁻¹⁸ J.

(b) The potential difference between the negative plate and a point midway between the plates is 11.25 V.

(c) The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.

U = qΔV

where;

U = q(Ed)

U = (1.6 x 10⁻¹⁹)(1500 x 1.5 x 10⁻²)

U = 3.6 x 10⁻¹⁸ J

ΔV = E(0.5d)

ΔV = 0.5Ed

ΔV = 0.5 (1500)(1.5 x 10⁻²)

ΔV = 11.25 V

U = ¹/₂mv²

U = mv²

v² = 2U/m

where;

v² = (2 x 3.6 x 10⁻¹⁸) / ( 1.67 x 10⁻²⁷)

v² = 4.311 x 10⁹

v = √(4.311 x 10⁹)

v = 6.57 x 10⁴ m/s

Thus, the proton’s potential energy change is 3.6 x 10⁻¹⁸ J.

The potential difference between the negative plate and a point midway between the plates is 11.25 V.

The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.

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Answer:

Explanation:

the value is -8 cm

The thermal energy that is generated due to friction is 344J.

Now we know that the total mechanical energy in the system is constant. The loss in energy is given by the loss in energy.

Thus, the kinetic energy is given as;

KE = 0.5 * mv^2 =0.5 * 15.0-kg * (1.10 m/s)^2 = 9.1 J

PE = mgh = 15.0-kg * 9.8 m/s^2 *  2.40 m = 352.8 J

The thermal energy is; 352.8 J - 9.1 J = 344J

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Answer:

Explanation:

The angular speed of the rotor is 200 rad/s.

The torque needed to be transmitted by the engine is 180 Nm.

The power of the rotor required to transmit energy to apply a torque τ to rotate a motor with angular speed ω,

P=τω

=180×200W

=36kW

An astronaut's spacesuit is considerably more than just a pair of garments they put on during spacewalks. An entire spacesuit is actually a single-person spacecraft. The Extravehicular Mobility Unit, or EMU, is the official name for the spacesuit used on the International Space Station and Space Shuttle. "Extravehicular" refers to something that is outside of a vehicle or spaceship. To be "mobile" implies to be able to move around while wearing an astronaut suit. The astronaut is shielded by the spacesuit from the perils of being outdoors in space.

The human body cannot survive in space's hostile atmosphere without protection. A spacesuit must shield the astronaut from the vacuum of space, the drastic temperature changes in space, and, if at all possible, it must lessen the astronaut's exposure to radiation. Therefore, the suit's primary function is to act as a pressure vessel. It must keep an air environment close to the skin constant, deliver a constant supply of clean air to the lungs, and expel stale, carbon dioxide-rich air. Every 90 minutes or so, an astronaut in low Earth orbit (LEO) experiences day and night. They can fast chill to -250 F during the night and reach 250 F while the sun is shining on them (-156 C). The body's normal temperature of 98.6 F must be maintained by the suit (37 C).

To do all this, the current NASA suit has 14 layers. The liquid ventilation and cooling garment is the first three layers. It resembles a body-hugging spandex garment that has tubes inside that carry cool water across the body to dissipate extra heat. Air below the next layer cannot escape since it is a pressure vessel made of nylon coated with urethane. The following layer resembles a tent since it is constructed of Dacron. Its goal is to exert pressure on the pressure garment so that it keeps its form and doesn't expand. Neoprene coasted nylon ripstop is the following layer. It is quite resilient to tears. Seven layers of mylar film and foil blanket are used to cover it. These restrict warmth transfer into or out of the suit by acting like a thermos. Due to its several layers, it also provides defense against very small micrometeroids, which drain energy as they pierce and break each layer. The top layer is made of orthofabric, a goretex, nomex, and kevlar mixture. It is very impervious to tearing and thermally reflecting to help regulate temperature. The suits are made by ILC Dover. Incorporating materials that minimize radiation penetration through the suit is something they are always working on.

Answer:

D. Impulse

Explanation: Hope this helps

The speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.

The speed of the roller coaster at any position is calculated by applying the principle of conservation of energy.

K.E = P.E

¹/₂mv² = mgh

v = √2gh

where;

v(2) = √[(2 x 9.8)(39 - 0)]

v(2) = 28 m/s

v(3) = √[(2 x 9.8)(39 -  26)]

v(3) = 16 m/s

v(4) = √[(2 x 9.8)(39 - 14)]

v(4) = 22 m/s

Thus, the speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.

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There is a 446N force in the horizontal massless cable CD. The vertical component of the pivot A's force on the aluminum pole, which is 328.3N. The force that the pivot A applies to the aluminum pole has a horizontal component of 446N.

We need to be aware of the force in order to discover the solution.

where, H=3.8m, l=7.6m, m=10kg, M=23.5 kg and alpha= 37 degree,

                      [tex]F_V-mg-Mg=0\\F_v=328.2N\\[/tex]

                  [tex]F_H=T=446N[/tex]

In light of this, we may say that the tension in the horizontal massless cable CD and the horizontal component of the force applied by the pivot A to the aluminum pole are identical and each exert 466N of force, whereas the vertical component of that force is 328.3N.

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A. The magnitude of the spring force (in N) acting upon the object is 15.9 N

B. The magnitude of the object's acceleration (in m/s²) is 30.58 m/s²

C. The direction of the acceleration vector points toward the equilibrium position (i.e., to the left in the figure).

F = Ke

F = 106 × 0.15

F = 15.9 N

F = ma

Divide both sides by m

a = F / m

a = 15.9 / 0.52

a = 30.58 m/s²

Considering the diagram, we can see that the spring was pulled away from the equilibrium point.

Thus, when the spring is released, it will move toward the equilibrium point. This is also true about the acceleration.

Therefore, we can conclude that the direction of the acceleration vector is towards the equilibrium point.

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The maximum height h of the curved path is 7.38 m.

Apply the following kinematic equation;

v² = u² - 2gh

where;

(u² - v²)/2g = h

(39² - 37.1²)/(2 x 9.8) = h

7.38 m = h

Thus, the maximum height h of the curved path is 7.38 m.

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Answer:

Approximately [tex]25\; {\rm N}[/tex] assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].

Explanation:

This mass is in a circular motion of radius [tex]r[/tex]. Hence, when the velocity of the mass is [tex]v[/tex], the acceleration of this mass should be [tex](v^{2} / r)[/tex]. The net force on this mass should be [tex](\text{net force}) = (m\, v^{2}) / r[/tex] towards the center of the circle.

When this [tex]m = 0.80\; {\rm kg}[/tex] mass is at the top of this circle, both gravitational pull and the force of the string (tension) point downwards. Hence, the net force on this mass would be:

[tex](\text{net force}) = (\text{weight}) + (\text{tension})[/tex].

Thus:

[tex]\begin{aligned} (\text{tension}) &= (\text{net force}) -(\text{weight})\\ &= \frac{m\, v^{2}}{r} - m\, g \\ &= m\, \left(\frac{v^{2}}{r} - g\right) \\ &= 0.80\; {\rm kg}\times \left(\frac{(9.0\; {\rm m\cdot s^{-1}})^{2}}{2.0\; {\rm m}} - 9.81\; {\rm m\cdot s^{-2}}\right) \\ &\approx 25\; {\rm kg \cdot m\cdot s^{-2}} \\ &= 25\; {\rm N}\end{aligned}[/tex].