Which of the following categories of motion is mutually exclusive with each of the others? A. Translational motion B. Rectilinear motion C. Rotational motion D. Curvilinear motion

Answer:

Explanation:

ASSUMING that block = sled AND that the rope is parallel to the slope.

The force acting parallel due to the weight is

13.6(9.81)sin35.5 = 77.475 N

The maximum friction force is

(0.45)13.6(9.81)cos35.5 = 48.877 N

If rope tension is T

77.475 - 48.877 < T < 77.475 + 48.877

            28.6 N < T < 126 N

28.6 N will occur if the block is on the verge of sliding downhill

126 N will occur if the block is on the verge of sliding uphill

Could be any value between them.

Answer:

freezing point and melting point

stronger, posturestronger, posture

hope that helped

Answer:

It is to produce sunlight in the forest plants

the question is about tyres of a race car, which are made of rubber and will be in contact with a race track, which is generally made from asphalt, the static coefficient of friction is in the range of (0.5–0.8), in dry conditions (Source: Friction and Friction Coefficients ).

Explanation:

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Answer:

Hi I hope this is correct!

Explanation:

You can use this formula to solve this question T = mv^2/R

m = 0.3 kg , v = 6.0 m/s , R = 1.5 m

T = (0.3 kg)(6.0 m/s)^2 / 1.5 m  

  = 7.2 Newtons

Hope this helps! Best of luck <3

Answer:

independent variable

Explanation:

Answer:

x = 41.28 m

Explanation:

This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.

Let's start by using trigonometry to find the initial velocity

         cos 25 = v₀ₓ / v₀

         sin 25 = Iv_{oy} / v₀

         v₀ₓ = v₀ cos 25

         v_{oy} = v₀ sin 25

         v₀ₓ = 22 cos 25 = 19.94 m / s

         v_{oy} = 22 sin 25 = 0.0192 m / s

let's use movement on the vertical axis

         y = y₀ + v_{oy} t - ½ g t²

when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m

         0 = 21 + 0.0192 t - ½ 9.81 t²

         4.905 t² - 0.0192 t - 21 = 0

         t² - 0.003914 t - 4.2813 =0

we solve the quadratic equation

        t = [tex]\frac{ 0.003914\ \pm \sqrt{0.003914^2 + 4 \ 4.2813 } }{2}[/tex]

        t = [tex]\frac{0.003914 \ \pm 4.13828}{2}[/tex]

        t₁ = 2.07 s

        t₂ = -2.067 s

since time must be a positive scalar quantity, the correct result is

        t = 2.07 s

now we can look up the distance traveled

         x = v₀ₓ t

         x = 19.94  2.07

         x = 41.28 m

Answer:

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Explanation:

ysggsghxuxgscsixigdvgsibxhdhshshjf

Complete Question

Partial tides __________.

Question 7 options:  

a. represent various components of local tides that are resolved mathematically  

b. are added together to predict the height and timing of astronomical tides  

c. consist of 4 components due to the influence of celestial bodies  

d. consist of up to 60 components due to astronomical and non-astronomical factors  

e. All of the above except c are correct.

Answer:

Option E

Explanation:

Generally

Partial tides represent various components of local tides that are resolved mathematically

Partial tides  are added together to predict the height and timing of astronomical tides

Partial tides consist of up to 60 components due to astronomical and non-astronomical factors

But Partial tides do not consist of 4 components due to the influence of celestial bodies

Therefore

All of the above except c are correct.

Option E

Answer:

Explanation:

The air enters their lungs at the same pressure as the water at that depth.

If they hold their breath as they rise to atmospheric pressure, the expanding volume of air (due to decreasing pressure) trapped in their lungs will hyperextend the alveoli in their lungs, likely tearing blood lines and risking death by drowning in their own blood.

Answer:

opposite direction

Explanation:

An electric field is defined as a physical field which surrounds the electrically charged particles that exerts force on the other particles on the field.

Now when an electron or a negatively charged particle enters a uniform electric field, the electric forces acts on the negatively charged particles and it forces the particle to move in the direction which is opposite to the direction of the field. In an uniform electric field, the field lines are parallel.

Answer:

Explanation:

negatively charged particle is placed inside uniform electric field

The force on the charge due to the electric field is

F = q E

when the charge is negative so the force on the charge is opposite to the direction of electric field.

The electric field is opposite to the force.

Answer:

The answer is "[tex]0.3336\ m^3[/tex]"

Explanation:

Using the Promideal gas law:

[tex]P_A=P_B\\\\P_A(V_A-\eta_A b)= \eta_A RT......(1)\\\\P_B V_B=\eta_B \bar{R}T........(2)\\\\From (1) \zeta (2)\\\\[/tex]  

[tex]\frac{\eta_A}{V_A-\eta_A b}=\frac{\eta B}{V B}\\\\ \frac{V A- \eta_A b}{V B}=\frac{\eta A}{\eta B }\\\\ \frac{V A-b}{V B}=\frac{1}{2}\\\\V A+V B=1\\\\V B =1- V A\\\\\frac{V A-b}{1-V A}=\frac{1}{2}\\\\2V A-2b=1-V A\\\\3 V A=1+2b\\\\V A=\frac{1+2b}{3}\\\\[/tex]

      [tex]=\frac{1+2(4\times 10^{-4})}{3}\\\\=0.3336\ m^3[/tex]

The equilibrium value of Va is 0.3336 m³.

The equilibrium value of Va is determine by applying ideal gas law as shown below;

Pressure of gas A = Pressure of gas B

Pa = Pb

Pa(Va - nab) = naRT----(1)

PbVb = nbRT -----(2)

Solve equation (1) and (2)

[tex]\frac{P_b}{RT} = \frac{n_b}{V_b} \\\\\frac{P_b}{P_a(V_a- n_ab)/n_a} = \frac{n_b}{V_b}\\\\\frac{n_a}{V_a - n_ab} = \frac{n_b}{V_b} \\\\\frac{V_a - n_ab}{V_b} = \frac{n_a}{n_b} \\\\\frac{V_a - b}{V_b} = \frac{1}{2}[/tex]

Va + Vb = 1

Vb = 1 - Va

[tex]\frac{V_a - b}{1 - V_a} = \frac{1}{2}[/tex]

2Va - 2b = 1 - Va

3Va = 1 + 2b

[tex]V_ a = \frac{1 + 2b}{3} \\\\V_a = \frac{1 + (2 \times 4\times 10^{-4})}{3} \\\\V_a = 0.3336 \ m^3[/tex]

Thus, the equilibrium value of Va is 0.3336 m³.

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Answer:

When a stone is going around a circular path, the instantaneous velocity of stone is acting as tangent to the circle. When the string breaks, the centripetal force stops to act. Due to inertia, the stone continues to move along the tangent to circular path. So, the stone flies off tangentially to the circular path

Answer:

when the string's rotation is broken, there will be no centripetal force to keep the stone stationary. Thus, the stone will flung away when the rotation is stopped

Answer:

the weight of the wire + 25kg

Explanation:

Answer:

2

Explanation:

pulling force because of it force

Answer:

5.9 cm

Explanation:

f: frequency of oscillation

frequency of oscillationk: spring constant

frequency of oscillationk: spring constantm: the mass

[tex]f = \frac{1}{2\pi} \sqrt{ \frac{k}{m} } [/tex]

in this problem we know,

F= 1.4 Hz

m= 0.26 kg

By re-arranging the formula we get

[tex]k = {(2\pi \: f )}^{2} m = {(2\pi(1.4hz))}^{2} 0.26kg = 20.1 \frac{n}{m} [/tex]

The restoring force of the spring is:

F= kx

where

F= 1.2 N

k= 20.1 N/m

x: the displacement of the block

[tex]x = \frac{f}{k} = \frac{1.2 \: n}{20.1 \frac{n}{m} } = 0.059m \: = 5.9 \: cm[/tex]

Answer:

Semiconductors are not normal materials. They have special properties which conductors/metals cannot exhibit. The main reason for the behavior of semiconductors is that they have paired charge carriers-the electron-hole pair. This is not available in metals.

Answer:

amplification

Explanation:

reflection can happen

some amount of lighr get absorbed

something gets refracted

but amplification cant

Answer:

W = F • ∆x

so for work to be done, a force and displacement has to be in the same direction. (Ex: a box is being pushed forward and it's also moving forward.)

Answer:

Part A - 4.084 mJ

Part B - 0.908 mJ

Part C - 8.168 mJ

Explanation:

Part A: How much energy is stored in the capacitor network as shown in (Figure 1)?

Since capacitors C₂ and C₃ are in series, their equivalent capacitance is C',

1/C' = 1/C₂ + 1/C₃      (Since C₁ = C₂ = C₃ = C)

1/C' = 1/C + 1/C

1/C' = 2/C

C' = C/2

Since C' is in parallel with C₁, the equivalent capacitance for the circuit is C" = C₁ + C' = C + C/2 = 3C/2

C" = 3C/2

The energy stored in the circuit, W = 1/2C"V² where C" = equivalent capacitance = 3C/2 and V = voltage = 15.0 V

W = 1/2C"V²

W = 1/2(3C/2)V²

W = 3CV²/4

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = 3CV²/4

W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/4

W = 3 × 24.2 × 10⁻⁶ F × 225 V²/4

W = 16335/4 × 10⁻⁶ FV²

W = 4083.75 × 10⁻⁶ J

W = 4.08375 × 10⁻³ J

W = 4.08375 mJ

W ≅ 4.084 mJ

Part B: How much energy would be stored in the capacitor network if the capacitors were all in series?

If the capacitors are connected in series, their equivalent resistance is C'

and 1/C' = 1/C₁ + 1/C₂ + 1/C₃

Since C₁ = C₂ = C₃ = C

1/C' = 1/C + 1/C + 1/C

1/C' = 3/C

C' = C/3

The energy stored in the circuit, W = 1/2C'V² where C' = equivalent capacitance = C/3 and V = voltage = 15.0 V

W = 1/2C'V²

W = 1/2(C/3)V²

W = CV²/6

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = CV²/6

W = 24.2 × 10⁻⁶ F (15.0 V)²/6

W = 24.2 × 10⁻⁶ F × 225 V²/6

W = 5445/6 × 10⁻⁶ FV²

W = 907.5 × 10⁻⁶ J

W = 0.9075 × 10⁻³ J

W = 0.9075 mJ

W ≅ 0.908 mJ

Part C: How much energy would be stored in the capacitor network if the capacitors were all in parallel?

If the capacitors are connected in parallel, their equivalent resistance is C'

and C' = C₁ + C₂ + C₃

Since C₁ = C₂ = C₃ = C

C' = C + C + C

C' = 3C

The energy stored in the capacitor network, W = 1/2C'V² where C' = equivalent capacitance = 3C and V = voltage = 15.0 V

W = 1/2C'V²

W = 1/2(3C)V²

W = 3CV²/2

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = 3CV²/2

W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/2

W = 3 × 24.2 × 10⁻⁶ F × 225 V²/2

W = 16335/2 × 10⁻⁶ FV²

W = 8167.5 × 10⁻⁶ J

W = 8.1675 × 10⁻³ J

W = 8.1675 mJ

W ≅ 8.168 mJ

I assume the curve is flat and not banked. A car making a turn on the curve has 3 forces acting on it:

• its weight, mg, pulling it downward

• the normal force from contact with the road, n, pushing upward

• static friction, f = µn, directed toward the center of the curve (where µ is the coefficient of static friction)

By Newton's second law, the net forces on the car in either the vertical or horizontal directions are

∑ F (vertical) = n - mg = 0

∑ F (horizontal) = f = ma

where a is the car's centripetal acceleration, given by

a = v ²/r

and where v is the maximum speed you want to find and r = 25 m.

From the first equation, we have n = mg, and so f = µmg. Then in the second equation, we have

µmg = mv ²/r   ==>   v ² = µgr   ==>   v = √(µgr )

So the maximum speed at which the car can make the turn without sliding off the road is

v = √(0.80 (9.80 m/s²) (25 m)) = 14 m/s

Answer:

Explanation:

Since energy is conserved:

2

mu  

2

​

=  

2

mv  

2

​

+mgh

⇒u  

2

=v  

2

+2gh

⇒(3)  

2

=v  

2

+2(9.8)(0.5−0.5cos60)

⇒v=2m/s

Acceleration of the simple pendulum is 2.62 m/s².

When a point mass is suspended from a fixed support by a light, non-extensible string, the instrument is said to be a simple pendulum.

Here,

Let the mass of the bob be m. The simple pendulum is attached to the fixed support with a string having length l. The pendulum makes an angle of 15° with the vertical from the equilibrium point.

Let T be the tension acting on the string.

As, the bob passes through the angle,

The weight of the bob becomes equal to the vertical component of the tension.

mg = T cos15°

Also, the horizontal component of the tension,

T sin15° = ma

By solving these two equations, we get that,

Acceleration of the simple pendulum,

a = g tan15°

a = 9.8 x 0.267

a = 2.62 m/s²

Hence,

Acceleration of the simple pendulum is 2.62 m/s².

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Answer:

The energy stored is 1.4 x 10^-9 J.

Explanation:

Side of square, L = 10 cm = 0.1 m

Distance, d = 2 mm = 0.002 m

Electric field, E = 4000 V/m

The energy stored in the capacitor is

[tex]U = 0.5 C V^2[/tex]

The capacitance is given by

[tex]C = \frac{\varepsilon o A}{d}\\\\So \\\\U = 0.5\frac{\varepsilon o A}{d}\times E^2 d^2\\\\U = 0.5\times 8.85\times 10^{-12}\times 0.1\times 0.1\times 4000\times 4000\times 0.002\\\\U = 1.4\times10^{-9} J[/tex]

The percentage by which the water density increased is 4.1[tex]\mathbf{\overline 6}[/tex] %

The known values are;

The increase in pressure per 10 meter increase in depth = 1.0 atm

The depth of the deepest ocean = 12 km = 12,000 m

The compressibility of the ocean = 5.0 × 10⁻⁵ 1/atm

The unknown

The percentage the density of water increased in the deepest ocean

Strategy;

Find the pressure at the deepest point of the deepest ocean and apply the compressibility

We have;

[tex]\mathbf{Compressibility = \dfrac{1}{V} \times \dfrac{\partial V}{\partial p}}[/tex]

The change in pressure, [tex]\partial p[/tex] = (12,000 m/(10 m)) × 1.0 atm = 1,200 atm

Therefore, we have for one cubic meter of water

[tex]\mathbf{5.0 \times 10^{-5} \ atm^{-1} = \dfrac{1}{1 \, m^3} \times \dfrac{\partial V}{1,200 \, atm}}[/tex]

Therefore;

[tex]\mathbf{\partial}[/tex]V = 5.0 × 10⁻⁵ atm⁻¹ × 1 m³ × 1,200 atm = 0.06 m³

The new volume = V - [tex]\mathbf{\partial}[/tex]V

∴ The new volume = 1 m³ - 0.06 m³ = 0.94 m³

The initial density = mass/(1 m³)

The new density = mass/(0.96 m³)

The percentage increase in density, [tex]\partial[/tex]ρ%, is given as follows;

[tex]\mathbf{\partial p \% = \dfrac{ \dfrac{Mass}{0.96 \ m^3} - \dfrac{Mass}{1 \ m^3} }{ \dfrac{Mass}{1 \ m^3}} \times 100 = \dfrac{25}{6} \% = 4.1 \overline 6 \%}[/tex]

∴  [tex]\mathbf{\partial}[/tex]ρ% =  4.1[tex]\mathbf {\overline 6}[/tex] %

The percentage by which the water density increased, [tex]\partial[/tex]ρ% = 4.1[tex]\mathbf{\overline 6}[/tex] %

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Group of answer choices.

a) the capacity to learn from experience, solve problems, and to adapt to new situations.

b) how behavior changes as a result of experience.

c) the factors directing behavior toward a goal.

d) the ability to generate novel

Answer:

a) the capacity to understand the world, think rationally, and use resources effectively.

Explanation:

Psychology can be defined as the scientific study of both the consciousness and unconsciousness of the human mind such as feelings, emotions and thoughts, so as to understand how it functions and affect human behaviors in contextual terms.

This ultimately implies that, psychology focuses on studying behaviors and the mind that controls it.

In this scenario, Ashley who is a psychology major, stated that she's interested in the study of intelligence.

Intelligence can be defined as a measure of the ability of an individual to think, learn, proffer solutions to day-to-day life problems and effectively make informed decisions.

In other words, Ashley is interested in the capacity of humans to understand the world, think rationally, and use resources effectively to produce goods and services that meet the unending requirements, needs or wants of the people (consumers or end users) living around the world.

Explanation:

N2 + H2 --> NH3

balance them:

N2 + 3 H2 --> 2 NH3

so if 6 moles of N2 react, 12 moles of NH3 will form.

(you have to look at the big number in front, in this case its N2 and 2 NH3, therefore the amount of N2 will produce double the amount of NH3 )

Answer:

Resistance of the armature coil = 6.61 ohms

Back emf developed at normal speed = 98.62 V (Approx.)

Current drawn by the motor at one-third normal speed = 12.73 A

Explanation:

Given:

Potential difference V = 117 V

Current = 17.7 A

Motor drawn current = 2.78 A

Find:

Resistance of the armature coil

Back emf developed at normal speed

Current drawn by the motor at one-third normal speed

Computation:

A] Resistance of the armature coil R = V/ I

Resistance of the armature coil = 117 / 17.7

Resistance of the armature coil = 6.61 ohms

B] Back emf developed at normal speed  = V- IR

Back emf developed at normal speed = 117 V - (2.78 A)(6.61 ohms)

Back emf developed at normal speed = 117 V - 18.37

Back emf developed at normal speed = 98.62 V (Approx.)

C] Current drawn by the motor at one-third normal speed = 17.7 A - (98.62/3)/(6.61 ohms)

Current drawn by the motor at one-third normal speed = 17.7 - 4.97

Current drawn by the motor at one-third normal speed = 12.73 A

Answer:

C. inertia

Explanation:

inertia describes an object’s resistance to change in motion (or to get in motion due to a lack of motion), and momentum describes how much motion it has.

both are connected, as inertia depends on the object's momentum, but the answer here is inertia.

Answer: [tex]5.77\ rps[/tex]

Explanation:

Given

Initial angular velocity is [tex]\omega_i=8\ rps[/tex]

rate of reduction [tex]\alpha=0.9 rev/s^2[/tex]

after 17 revolution i.e. [tex]\theta =17\ rev[/tex]

using [tex]\Rightarrow \omega_f^2-\omega_i^2=2\alpha\theta[/tex]

Insert the values

[tex]\Rightarrow \omega_f^2=8^2-2\times (0.9)\times17\\\Rightarrow \omega_f^2=33.4\\\Rightarrow \omega_f=5.77\ rps[/tex]