It is said that a gas fills all the space available to it. Why then doesn't the atmosphere go off into space?

Answer:

c. Only the linear acceleration is zero.

Explanation:

The linear acceleration is defined as the rate of change of linear velocity. Since the bicycle is moving in the same direction, with the same speed, without speeding up or slowing down. Therefore, there will be no change in linear velocity and as a result, linear acceleration will be zero.

The angular acceleration is the rate of change of angular velocity. Since the angular velocity is changing its direction constantly. Therefore, it has a certain component of acceleration at all times called centripetal acceleration.

Therefore, the correct option is:

c. Only the linear acceleration is zero.

Use the work-energy theorem to find the velocity of the block when it's released by the spring. The work done by the spring on the block as it's restored to equilibrium is

W = 1/2 kx ²

where k is the spring constant and x is the compression of the spring. So

W = 1/2 (1400 N/m) (0.170 m)² = 20.23 J

This is equal to the block's change in kinetic energy ∆K,

W = ∆K

and since it starts from rest, the initial K is zero, leaving us with

W = 1/2 mv ²

where m is the mass of the block and v is its speed, so that

20.23 J = 1/2 (0.200 kg) v ²

==>   v ≈ 14.2 m/s

The block slides at this speed across the frictionless surface until it hits the incline which introduces friction.

First, you need to find the length of the incline. It forms a 45° angle, and the underlying 45°-45°-90° triangle has a hypotenuse of length √2 (2.0 m) ≈ 2.83 m.

Next, you need to find the total work done on the block as it slides up the incline. Use Newton's second law to examine the forces acting on the block during this phase:

• the net force acting on the block in the direction perpendicular to the incline is

∑ F = n - mg cos(45°) = 0

where n = mg cos(45°) ≈ 1.39 N is the magnitude of the normal force and mg cos(45°) ≈ 1.39 N is the perpendicular component of the block's weight;

• the net force acting on the block parallel to the surface is

∑ F = -f - mg sin(45°) = ma

where f = µn = 0.210n ≈ 0.291 N is the magnitude of kinetic friction, mg sin(45°) ≈ 1.39 N is the parallel component of the weight, and a is the acceleration of the block.

Only the parallel forces do work on the block, and this work is negative because friction and weight oppose the block's sliding up the incline. The total work done on the block is then

W = (-0.291 N - 1.39 N) (2.83 m) ≈ -4.74 J

Use the work-energy theorem again to find the block's new speed v at the top of the incline:

W = ∆K

==>   -4.74 J = 1/2 (0.200 kg) v ² - 1/2 (0.200 kg) (14.2 m/s)²

==>   v ≈ 12.4 m/s

And now this becomes a projectile problem. The block travels a horizontal distance x after being launched at an angle of 45° with initial speed 12.4 m/s after time t according to

x = (12.4 m/s) cos(45°) t

Its height y from the 2.0 m-high surface at time t is given by

y = (12.4 m/s) sin(45°) t - 1/2 gt ²

The block lands on the surface when y = 0, which occurs after t ≈ 1.79 s, at which point the block has covered a distance d ≈ 15.7 m.

The block sail through the air at the distance of "15.8 m"

Given:

Spring constant,

Mass,

Block's coefficient,

By using Work energy theorem, we get

→ [tex]W_{spring}+W_g+W_f = KE_f-KE_i[/tex]

By substituting the values, we get

→ [tex]\frac{1}{2}\times 1400\times (0.17)^2- (0.2\times 9.8\times 2)-(0.21\times 0.2\times \frac{9.8}{\sqrt{2} }\times \sqrt{2}\times 2 )= \frac{1}{2}\times 0.2\times V_f^2[/tex]

here,

[tex]V_f = 12.44 \ m/s[/tex]

→ [tex]d = \frac{V_f^2 Sin 2 \Theta}{g}[/tex]

     [tex]= \frac{(12.44)^2 Sin 90^{\circ}}{9.8}[/tex]

     [tex]= 15.8 \ m[/tex]

Thus the answer above is right.  

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Answer:

The time of flight is [tex]T = \frac{2 u sin A}{g}[/tex].

Explanation:

Let the initial velocity is u and the angle of projection is A.

Use first equation of motion for vertical motion

Let the time to reach the maximum height is t.

[tex]v = u - gt\\\\0 = u sin A - gt \\\\t = \frac{ u sin A}{g}[/tex]

Total time of flight is

T =2  t

[tex]T = \frac{2 u sin A}{g}[/tex]

Answer:

Rahul's weight

Explanation:

In fact, the force Rahul exerts on Earth corresponds to the force of gravity. But Rahul's weight is, in fact, the force of gravity exerted by the Earth on Rahul, and these two forces correspond to the action-reaction pair of Newton's third law, which states that the two forces are equal.

Using formulas, Rahul's weight is equal to

W=mg

where m is Rahul mass and g is the gravitational acceleration (g=9.81 m/s^2).

Answer:

wavelength=75.0

speed of sound(v)=350 .0m/s

frequency(f)=?

we know,

v=f*wavelengh

350.0 =f*750

f. =350/75

=4.667

pls mark me brainlest

Answer:

Explanation:

Of course because it's Newton's Law that if body A exerts force on body B, then body B will exert equal but opposite force on body A.

HAPPY LEARNING:)

The first charged object is exerting a force on the second charged object. The second charged object necessarily exerting a force on the first is correct.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Newton's third law states that for every action (force) in nature there is an equal and opposite reaction, from that we can understand when force is exerted on first charged object is exerting a force on the second charged object. The second charged object necessarily exerting a force on the first.

The first charged object is exerting a force on the second charged object. The second charged object necessarily exerting a force on the first is correct.

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Answer:

120 beats per minute.

Explanation:

If I take your pulse and observe 80 heartbeats in a minute. Then the period of your heartbeat is 0.8 s and frequency is 1.3Hz.

A pulse is the term used in medicine to describe the tactile arterial palpation of the cardiac cycle (heartbeat) by skilled fingertips. Any location where an artery can be compressed close to the surface of the body, such as the carotid artery in the neck, the radial artery in the wrist, the femoral artery in the groyne, the popliteal artery behind the knee, the posterior tibial artery near the ankle joint, and on the foot, can be used to palpate the pulse (dorsalis pedis artery). Heart rate may be determined by monitoring pulse, or the number of arterial pulses per minute. Auscultation, which is the process of counting the heartbeats while listening to the heart using a stethoscope, is another way to determine the heart rate. Typically, three fingers are used to gauge the radial pulse.

Given,

heart beat = 80 beats/min = 1.3 beats/s

Frequency is nothing but how much beats is heart having in one second and that is 1.3 beats/s. Hence frequency of heart is 1.3Hz.

The Period is reciprocal of frequency,

T = 1/f = 0.8 s

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Answer:

concave mirror use in headlights and torches

Explanation:

Concave mirrors are used in headlights and torches. The shaving mirrors are also concave in nature since these mirrors can produce enlarged clear images. Doctors use concave mirrors as head mirrors to have a clearer view of eyes, noses, and ears. The dental mirrors used by dentists are also concave.

Explanation:

the answer is in the above image

(a) The maximum voltage V is  190 Volts.

(b) The rms voltage V is 95√2 Volts.

(c) The rms current in Amperes is  7.9 A.

(d) The peak current Amperes is  11.18 A.

(e)The current when t = 0.0045 s is 7.26 A.

The current is the stream of charges which flow inside the conductors when connected across the end of voltage.

Given is an AC source is connected across a 17.0 Ω resistor, the output voltage is given by Δv = (190 V)sin(50πt).

(a) From the given voltage equation, maximum voltage V is 190 Volts.

(b) rms voltage =Vmax/√2

Put the values, we get

Vrms = 190/√2 = 95√2 Volts

(c) rms current = Vrms/Resistance

Put the values, we get

Irms =  95√2 /17

Irms = 7.9 Amperes.

(d) peak current  =√2 Irms

Substitute the values, we get

Peak current = 7.9 √2 = 11.18 A

(e) The current when t = 0.0045 s is written as

I =  (190 V)sin(50πt)/R

Substitute the values, we have

I =  (190 )sin(50π x0.0045)/17

I = 7.26 A.

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Answer:

Explanation:

This is an interference exercise, which the case of constructive interference is described by the expression

          d sin θ  = m λ

in this case they indicate that we are in the ninth order (m = 9).

To be able to observe the pattern, the dispersion angle must be less than 90º

we substitute

           sin 90 = 1

           d = m  lang

let's calculate

           d = 9 λ

           d = 9 380 10⁻⁰

            d = 3.42 10⁻⁶  

           d2 = 9 760 10⁻⁹

           d2 = 6.84 10₋⁶

Answer:

air  f = 4.6527 10¹⁴ Hz

water f = 3.4914 10¹⁴ Hz

glass f = 3.1027 10¹⁴ Hz

Explanation:

The refractive index of a material is given by

         n = c / v

where c is the speed of light in a vacuum c = 3 108 m / s and v is the speed of light in the material medium.

the speed of the wave is

         v = λ f

we substitute

         c / n = λ f

          f = [tex]\frac{c}{n \ \lambda}[/tex]

The refractive indices are

air 1,00029

water 1.3330

glass 1.5

let's calculate the frequencies

vaccum

         f = 3 10⁸ / 1 644.6 10⁻⁹

         f = 4.6540 10¹⁴ Hz

air

         f = 3 10⁸ / 1,00029 644.6 10⁻⁹

         f = 4.6527 10¹⁴ Hz

Water

         f = 3 10⁸ / 1.333 644.6 10⁻⁹

         f = 3.4914 10¹⁴ Hz

glass

        f = 3 10 ^ 8 / 1.5 644.6 10⁻⁹

        f = 3.1027 10¹⁴ Hz

Answer:

the answer is b. reflective

Answer:

12+2=24+30+2=66

Explanation:

Answer:

I = 15A

Explanation:

V = I*R

120V = I*8.0ohms

I = 120V/8.0ohms

I = 15A

Answer:

I=15A

Explanation:

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Give them the brain list.

Answer:

Explanation:

1.00kg×20m/s×cos40=15.3

Answer:

Gravitational potential energy is mass of the object times the gravitational constant times the height of the object:

U = mgh (I will use 10 for the gravitational constant but you can use 9.8 or 9.81 or something even more accurate)

U = 280

The gravitational potential of the object is 280 joules

Answer:

Mass of car's engine = 342.85 kg (Approx.)

Explanation:

Given values:

Force applied by car = 1,200 N

Acceleration of car' engine = 3.5 m/s²

Find:

Mass of car's engine

Computation:

⇒ Mass = Force / Acceleration

⇒ Mass of car's engine = Force applied by car / Acceleration of car

⇒ Mass of car's engine = 1,200 / 3.5

⇒ Mass of car's engine = 342.85 kg (Approx.)

Answer:

[tex]F=5(240t^2i+72tj)\ N[/tex]

Explanation:

Given that,

The mass of the object, m = 5 kg

The position vector is, [tex]r=20t^4i+12t^3j[/tex]

Velocity, [tex]v=\dfrac{dr}{dt}=80t^3i+36t^2j[/tex]

Acceleration, [tex]a=\dfrac{dv}{dt}=240t^2i+72tj[/tex]

Newton's second law of motion is given as follows:

F = ma

Put all the values,

[tex]F=5(240t^2i+72tj)\ N[/tex]

Hence, this is the required solution.

Answer:

171.5J

Explanation:

K=1/2 *m *U²

K=1/2 *7 *7²

K=171.5 J

Answer:

We use a spring of large spring constant.

Explanation:

The spring constant is defined as the force applied on the spring per unit extension or compression in length.

F = k x

where, F is the force, x is the extension, k is the spring constant.

Its unit is N/m.

To get the comfortable ride, we use the spring of large spring constant, so that the spring gets stiffer and we get comfort.

Answer:

30 kilometers is a reasonable measurement

Answer:

a) True

Explanation:

A program-specific message provided to an individual or group with the intention of raising awareness of a health condition, motivating behavior change, removing perceived barriers to participating in a health habit, or something else relating to the program's aims and objectives. The most effective intervention messages are usually theory-based and culturally adapted.

Answer:

m v = M V  conservation of momentum after  throwing camera

V = m * v / M =   .928 * 12 / 60.9 = .183 m/s

t = S / V = 36.5 / .183 = 199 sec = 3.32 min

Answer: Hazmat products are allowed in your FC are:

Explanation:

Hazmat products consist of flammable, corrosive and harmful substances which are actually very hazardous to human health and environment.  

Hazardous material allowed in FC are as follows.

So, hazmat allowed products are GPS unit (lithium batteries) and subwoofer (magnetized materials).

Thus, we can conclude that hazmat products are allowed in your FC are:

This is the solution of the problem. The problem is calculated from a closed system energy balance equation

Answer:

The water usually rushes back too enthusiastically, causing a splash – and the bigger the rock, the bigger the splash. The splash then creates even more ripples that tend to move away from where the rock went into the water.

Answer:

a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm

Explanation:

For isothermal expansion PV = constant

So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,

So, P₁V₁ = P₂V₂

P₂ = P₁V₁/V₂

Since V₂/V₁ = 0.19,

P₂ = P₁V₁/V₂

P₂ = 1 atm (1/0.19)  

P₂ = 5.26 atm

For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas

So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,

So, P₂V₂ⁿ = P₃V₃ⁿ

P₃ = P₂V₂ⁿ/V₃ⁿ

P₃ = P₂(V₂/V₃)ⁿ

Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19

1/0.19,

P₃ = P₂(V₂/V₃)ⁿ

P₃ = 5.26 atm (0.19)⁽⁵/³⁾

P₃ = 5.26 atm × 0.0628

P₃ = 0.33 atm

Using the ideal gas equation

P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion  P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)

P₃V₃/T₃ = P₄V₄/T₄

T₃ = P₃V₃T₄/P₄V₄    

T₃ = (P₃/P₄)(V₃/V₄)T₂

Since V₃ = V₄ = V₁ and P₄ = P₁

V₃/V₄ = 1 and P₃/P₄ = P₃/P₁

T₃ = (P₃/P₁)(V₃/V₄)T₂

T₃ = (0.33 atm/1 atm)(1)273 K  

T₃ = 90.1 K

So,

a. The highest temperature attained by the gas is T₁ = 273 K

b. The lowest temperature attained by the gas = T₃ = 90.1 K

c. The highest pressure attained by the gas is P₂ = 5.26 atm

d. The lowest pressure attained by the gas is P₃ = 0.33 atm

Answer: hello some data related to your question is missing attached below is the missing data and diagram related to the solution

answer:

P = 141.21 N

Explanation:

Given data:

Mass of crate = 50 kg

coefficient  of static friction ( μ ) = 0.25

Calculate minimum horizontal force ( P ) that holds the crate from sliding

∑fx = 0

     = P + Fcos θ - N*sinθ = 0

     = P + 0.25N cos 30° - Nsin30°  = 0

∴ P = 0.2835 N = 0

P - 0.2853 N = 0 ------- ( 1 )

∑fy = 0

     - 50g + Ncosθ + Fsinθ

     - 50*9.81 + Ncos30° + 0.25Nsin30°

∴ N = 494.942 N ----- ( 2 )

input 2 into 1

P - 0.2853 ( 494.942 ) = 0

P = 141.21 N