Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a molecule at the top of the container (assuming the potential energy is zero at the bottom) with the average kinetic energy of the molecules. Is it reasonable to neglect the potential energy?

Answer:

1000 m upwards

Explanation:

Displacement Formula: Average Velocity = Displacement/Total Time

Simply plug in our known variables and solve:

100 m/s = x m/10 seconds

100 m/s(10 s) = x m

m = 1000

The question is incomplete. Here is the complete question.

A uniform electric field of 2kN/C points in the +x-direction.

(a) What is the change in potential energy of a +2.00nC test charge, [tex]U_{electric,b} - U_{electric,a}[/tex] as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?

(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?

(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?

Answer: (a) ΔU = 3.2×[tex]10^{-6}[/tex] J

(b) KE = 2×[tex]10^{-6}[/tex] J

Explanation: Potential Energy (U) is the amount of work done due to its position or condition and its unit is Joule (J). Kinetic Energy (KE) is the ability to do work by virtue of velocity and the unit is also (J). Mechanical Energy is the sum of Potential and Kinetic Energies of a system.

(a) Related to electricity, Potential Energy can be calculated as:

ΔU = Eqd

where E is the electric field (in N/C);

q is the charge (in C);

d is the distance between plaques (in m);

For a at x = - 30cm and b at x = 50 cm:

E = 2×[tex]10^{3}[/tex] N/C

q = 2×[tex]10^{-9}[/tex] C

d = 50 - (-30) = 80×[tex]10^{-2}[/tex] = 8×[tex]10^{-1}[/tex]m

ΔU = [tex]U_{electric,b} - U_{electric,a}[/tex] = Eqd

[tex]U_{electric,b} - U_{electric,a}[/tex] = 2×[tex]10^{3}[/tex] .  2×[tex]10^{-9}[/tex] . 8×[tex]10^{-1}[/tex]

ΔU = 3.2×[tex]10^{-6}[/tex] J

(b) Mechanical Energy is constant, so:

[tex]KE_{i} + U_{i} = KE_{f} + U_{f}[/tex]

Since the initial position is zero and there is no initial kinetic energy:

[tex]KE_{f} = - U{f}[/tex]

[tex]KE_{f} =[/tex] - (2×[tex]10^{3}[/tex]. 2×[tex]10^{-9}[/tex] . 5×[tex]10^{-1}[/tex])

[tex]KE_{f} = - 2.10^{-6}[/tex] J

(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.

The complete question is;

In an amusement park ride called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.

Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?

Answer:

F_top = 385.36 N

Explanation:

We are given;

mass;m = 52 kg

Time;t = 4.3 s

Diameter;d = 16m

So,Radius;r = 16/2 = 8m

The formula for the centrifugal force is given as;

F_c = mω²R

Where;

R = radius

Angular velocity;ω = 2πf

f = frequency = 1/t = 1/4.3 Hz

F_c = 53 × (2π × 1/4.3)² × 8 = 905.29 N.

The force at top would be;

F_top = F_c - mg

F_top = 905.29 - (9.81 × 53) N

F_top = 385.36 N

The force at the top of ride will be "385.36 N".

According to the question,

Rider's mass, m = 52 kg

Time, t = 4.3 s

Diameter, d = 16 m

Radius, r = [tex]\frac{16}{2}[/tex] = 8 m

Frequency, f = [tex]\frac{1}{t}[/tex] = [tex]\frac{1}{4.3}[/tex] Hz

We know the formula,

Centrifugal force,  [tex]F_c[/tex] = mω²R

or,

Angular velocity, ω = 2πf

By substituting the values in the above formula,

[tex]F_c = 53(2\pi \times (\frac{1}{4.3})^2\times 8 )[/tex]

    [tex]= 905.29[/tex] N

hence,

The top force will be:

→ [tex]F_{top} = F_c[/tex] - mg

By substituting the values,

          [tex]= 905.29-(9.81\times 53)[/tex]

          [tex]= 385.36[/tex] N

Thus the above response is correct.  

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Answer:

The perceived frequency is higher than the actual emitted sound frequency. that means that the received sound waves are of shorter wavelength.

Explanation:

When the source of a sound wave is moving toward the observer, the perceived frequency of the wave changes in relation to the observer producing a change in pitch. The effect is called Doppler effect in honor of the physicist who formulated the physical explanation.

In the case of the sound source approaching the observer, the perceived frequency is higher than that actually emitted by the sound source.

Explanation:

It is given that, the radius of a ball is [tex](5.2\pm 0.2)\ cm[/tex].

We need to find the percentage  error in the volume of the ball. The volume of a sphere is : [tex]V=\dfrac{4}{3}\pi r^2[/tex]

The percentage  error is given by :

[tex]\dfrac{\Delta V}{V}=3\dfrac{\Delta r}{r}\times 100[/tex]

We have, [tex]\Delta r=0.2[/tex] and r = 5.2

So,

[tex]\dfrac{\Delta V}{V}=3\times \dfrac{0.2}{5.2}\times 100\\\\\%=11\%[/tex]

So, the percentage  error in the volume of the ball is 11%

Answer:

Explanation:

Ohm's law can be expressed as;

                    V = IR.

For the given question, two cases would be considered.

1. If the circuit is a series circuit.

For a series circuit, the potential difference across each bulb in the circuit varies. The current has the same value at any point in the circuit.

When the second bulb is added, less current would flow through the bulbs. The current reduces, same as the potential difference across each bulb. Thus the brightness of the first bulb decreases because the current is shared between the two bulbs.

2. If the circuit is a parallel circuit.

In a parallel circuit, the potential difference across its ends is the same. But the current is shared between the bulbs. The potential difference is the same, while the current varies. Therefore, the second light bulb would produce light without affecting the brightness of the first light bulb in the circuit.

Complete question:

An ideal measuring device is one that does not alter the very measurement it is meant to take. Therefore, in comparison with the resistance being measured, the resistances of an ideal ammeter and an ideal voltmeter must be, respectively: Select the best answer from the choices provided.

a) very small; very small

b) very large; very small

c) very small; very large

d) very large; very large

Answer:

c) very small; very large

Explanation:

Ammeters can be said to be a device which measures the flow of electric current through a conductor. An ideal ammeter is said to have zero internal resistance. This is because there will be little or no voltage drop as electric current flows through it.

Therefore the resistance of an ideal ammeter must be very small.

A voltmeter can be said to be a device that measures the difference in potential difference between two points in a given circuit. The internal resistance of a voltmeter is said to be infinite, which means it could be very large. This means no current will flow through the voltmeter and the measured voltage will have little or no error.

Therefore the resistance of an ideal voltmeter must be very large.

Answer:

Explanation:

Distance of charges placed on y axis from given point on x axis

= √ (3² + 4² )

= 5 m

Potential at point A due to given charge

= Q / 4πε₀ R

Potential due to each of charges

= 4 x 10⁻⁶ x 9 x 10⁹ / 5  [ 1/ 4πε₀ = 9 x 10⁹ ]

= 7.2 x 10³ V

Potential due to both the charges

= 2 x 7.2 x 10³

= 14.4 kV .

Answer:

0.0241 m

Explanation:

mass of the hockey player m1 = 90 kg

mass of puck m2 = 0.150 kg

puck velocity v1= 45 m/s

distance traveled by puck to reach the goal =15.0 m.

now accoding to momentum conservation law

90×45+0.15×v2 = 0 [ since, If both are initially at rest and if the ice is frictionless,]

therefore, v2= -0.0725 m/s.

Now time taken by the puck to reach the goal

t= 15/45 = 1/3 sec.

therefore, how far does the player recoil in the time

=0.0725×1/3= 0.0241 m.

the distance travelled by the player( recoil ) in the time the puck reach the goal is 0.025m.

Given the data in the question

Since they were both at rest initially

Using conservation of liner momentum:

[tex]mu + m_1u_1 = mv+ m_1v_1[/tex]

Now, Since they were both at rest initially

[tex]0 = mv+ m_1v_1[/tex]

We substitute in our values to find the velocity of the player after the hit ( recoil velocity )

[tex]0 =[ 0.150kg * 45.0m/s ] + [ 90.0kg * v_1 ]\\\\0 = 6.75kg.m/s + [ 90.0kg * v_1 ]\\\\90.0kg * v_1 = -6.75kg.m/s \\\\v_1 = -\frac{6.75kg.m/s}{90.0kg} \\\\v_1 =- 0.075m/s[/tex]

{ The negative sign shows that the velocity of both the player and the puck are in opposite direction }

Hence, recoil velocity of the player is 0.075m/s

Now, we determine the time taken for the puck to trach the goal using the relation between distance, velocity and time .

Time = Distance / Velocity

We substitute our values into the expression

[tex]t = \frac{s}{v} \\\\t = \frac{15.0m}{45m/s} \\\\t = 0.3333s[/tex]

Hence, the time taken for the puck to reach the goal is 0.3333 seconds.

Next, we determine the distance travelled by the player( recoil ) in the time the puck reach the goal using the relation between distance, velocity and time .

Time = Distance / Velocity

We substitute in our values

[tex]t = \frac{s}{v}\\\\0.3333s = \frac{s}{0.075m/s} \\\\s = 0.3333s * 0.075m/s\\\\s = 0.025m[/tex]

Therefore, the distance travelled by the player( recoil ) in the time the puck reach the goal is 0.025m.

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Answer:

[tex]4.9\ m/sec^2[/tex]

Explanation:

The computation of acceleration of the particle down the slope is shown below:-

data provided in the question

Particle of mass = 1.3 kg i,e sliding down

Inclined = 30 to the horizontal

based on the above information

Force is given by

[tex]N = mg\ cos \theta[/tex] ............ 1

and sliding force is given by

[tex]F = mg\ sin\alpha[/tex]

[tex]a = g(sin\ 30^{\circ})[/tex]

[tex]= 9.8\times \frac{1}{2} m/sec^2[/tex]

= [tex]= 4.9\ m/sec^2[/tex]

Hence, the acceleration of the particle down the slope is 4.9 m/sec^2

Explanation:

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Answer:

The maximum potential difference across capacitor is 20 V

Explanation:

Given;

maximum emf of the ac circuit, [tex]V_{max}[/tex] = 25 V

maximum potential difference across resistor, [tex]V_R[/tex]  = 15 V

maximum potential difference across capacitor, [tex]V_C[/tex] = ?

Determine the maximum potential difference across capacitor using the equation below;

[tex]V_{max} = \sqrt{V_R^2 + V_C^2} \\\\25^2 = V_R^2 + V_C^2\\\\V_C^2 = 25^2 - V_R^2\\\\V_C^2 = 25^2 - 15^2\\\\V_C^2 = 400\\\\V_C = \sqrt{400} \\\\V_C = 20 \ V[/tex]

Therefore, the maximum potential difference across capacitor is 20 V

Answer:

The terminal velocity at sea level is 7.99 m/s

The terminal velocity at an altitude of 5000 m is 10.298 m/s

Explanation:

mass of sphere m  = 10 kg

radius of sphere r = 0.5 m

air density at sea level p = 1.22 kg/m^3

drag coefficient Cd = 0.8

terminal velocity = ?

Area of the sphere A = [tex]4\pi r^{2}[/tex] = 4 x 3.142 x [tex]0.5^{2}[/tex] = 3.142 m^2

terminal velocity is gotten from the relationship

[tex]Vt = \sqrt{\frac{2mg}{pACd} }[/tex]

where g = acceleration due to gravity = 9.81 m/s^2

imputing values into the equation

[tex]Vt = \sqrt{\frac{2*10*9.81}{1.22*3.142*0.8} }[/tex] = 7.99 m/s

If  at an altitude of  5000 m where air density = 0.736 kg/m^3, then we replace value of air density in the relationship as 0.736 kg/m^3

[tex]Vt = \sqrt{\frac{2mg}{pACd} }[/tex]

[tex]Vt = \sqrt{\frac{2*10*9.81}{0.736*3.142*0.8} }[/tex] = 10.298 m/s

The value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.

Given data:

The mass of sphere is, m = 10 kg.

The radius of sphere is, r = 0.5 m.

The density of air is, [tex]\rho = 1.22 \;\rm kg/m^{3}[/tex].

The drag coefficient of object is, [tex]C_{d}=0.8[/tex].

The altitude is, h = 5000 m.

The density of air at altitude is, [tex]\rho' =0.736 \;\rm kg/m^{3}[/tex].

The mathematical expression for the terminal velocity of an object is,

[tex]v_{t}=\sqrt\dfrac{2mg}{\rho \times A \times C_{d}}[/tex]

here,

g is the gravitational acceleration.

A is the area of sphere.

Solving as,

[tex]v_{t}=\sqrt{\dfrac{2 \times 10 \times 9.8}{1.22 \times (4 \pi r^{2}) \times C_{d}}}\\\\\\v_{t}=\sqrt{\dfrac{2 \times 10 \times 9.8}{1.22 \times (4 \pi \times 0.5^{2}) \times 0.8}}\\\\\\\v_{t}=7.99 \;\rm m/s[/tex]

Now, the terminal velocity at the altitude of 5000 m is given as,

[tex]v_{t}'=\sqrt\dfrac{2mg}{\rho' \times A \times C_{d}}[/tex]

Solving as,

[tex]v_{t}'=\sqrt{\dfrac{2 \times 10 \times 9.8}{0.736 \times (4 \pi \times 0.5^{2}) \times 0.8}}\\\\\\v_{t}'=10.30 \;\rm m/s[/tex]

Thus, we can conclude that the value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.

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Answer:

   M = 1433.5 kg

Explanation:

This exercise is solved using the Archimedean principle, which states that the hydrostatic thrust is equal to the weight of the desalinated liquid,

              B = ρ g V

with the weight of the truck it is in equilibrium with the push, we use Newton's equilibrium condition

           Σ F = 0

           B-W = 0

           B = W

       body weight

           W = M g

the volume is

           V = l to h

           rho_liquid g (l to h) = M g

           M = rho_liquid l a h

we calculate

            M = 1000 4.7 6.10 0.05

           M = 1433.5 kg

Answer:

Option A. 48°C

Explanation:

The following data were obtained from the question:

Mass (m) = 0.3 Kg

Initial temperature (T1) = 20°C

Heat (Q) added = 35 KJ

Specific heat capacity (C) = 4.18 KJ/Kg°C

Final temperature (T2) =..?

The final temperature of water can be obtained as follow:

Q = MC(T2 – T1)

35 = 0.3 x 4.18 (T2 – 20)

35 = 1.254 (T2 – 20)

Clear the bracket

35 = 1.254T2 – 25.08

Collect like terms

1254T2 = 35 + 25.08

1.254T2 = 60.08

Divide both side by the coefficient of T2 i.e 1.254

T2 = 60.08/1.254

T2 = 47.9 ≈ 48°C

Therefore, the final temperature of the water is 48°C.

Answer:

May be egg shaped

Has no new stars being formed.

Has almost no gas or dust between stars.

Explanation:

Elliptical galaxy is the collection of many stars which are bounded together gravitationally, which is smooth and ellipsoidal and shape and the appearance is featureless.

Elliptical galaxy is ovoid or spherical masses of stars.

It is found in galaxy clusters and compact galaxies.

It has no gas or dust between stars which result in low rates of star formation.

It is formed When two spirals collide, they lose their familiar shape, morphing into the less-structured elliptical galaxies.

Elliptical galaxy is made of old stars and have no gas and dust.

An example is elliptical galaxy m60 which shines brightly and is egg shaped.

The distance the second child needs to sit from the center in order to balance the seasaw is 1.76 m.

The can be defined as the total length between two points.

To calculate the distance the second child needs to sit from the center, we use the formula below.

Note: For the seesaw to be balanced, sum of clockwise moment is equal to sum of anti clockwise moment acting on it.

Formula:

 Make D the subject of the equation

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the distance of the second child from the center is 1.76 m.

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Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

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

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\[/tex]

Now, determine the amplitude of oscillation, A;

[tex]E = \frac{1}{2} kA^2[/tex]

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

[tex]E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm[/tex]

Therefore, the amplitude of the oscillation is 2.82 cm

Answer:

18 km

Explanation:

Convert km/h to m/s:

120 km/h × (1000 m/km) × (1 h / 3600 s) = 33.3 m/s

The time it takes the bomb to travel the 2000 meters is:

2000 m / (33.3 m/s) = 60 s

So it takes 60 seconds for the bomb to fall.  The distance it fell is therefore:

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (0 m/s) (60 s) + ½ (10 m/s²) (60 s)²

Δy = 18,000 m

Δy = 18 km

Answer: b. Ultraviolet Light

Explanation:

Electromagnetic wave is defined as the wave which is associated with both electrical and magnetic component associated with them. They can travel in vacuum as well and travel with the speed of light.

The electromagnetic radiations consist of radio waves, microwaves, infrared ,Visible , ultraviolet, X rays and gamma rays arranged in order of increasing frequency and decreasing wavelengths.Thus ultraviolet light has more energy than visible light as energy and frequency are directly proportional.

Ultraviolet Light is more energetic than visible light, and have potential to damage skin cells and lead to skin cancer.

Answer:

it attracts

Explanation:

since in a magnetic body there are two poles

(north and south poles)if the first pole was repeled when brought near the North Pole therefore the other end is going to attarct because the first end was also a North Pole while the second end will be a south pole

Answer:

Milton Hershey

Steve Jobs

Simon Cowell

Thomas Edison

Explanation:

Answer:

8.94*10^22 kg

Explanation:

Given that

Mass of Lo, M = ?

Radius of Lo, r = 1.82*10^6 m

Acceleration on Lo, g = 1.80 m/s²

Gravitational constant, G = 6.67*10^-11

Using the formula

g = GM/r²

Solution is attached below

Answer is 8.94*10^22 kg

Answer:

363m.s-1

Explanation:

Answer:

It would be 450 or 640. My final answer would be about 450

Explanation: Because it would't be to high if it was shot Voy = 100

btw i think i know what i know what i am talking about.

The answer would be about 450 m.

The top isn't the standard for a cliff to be reckoned as a cliff as such. Any steep rock face particularly at the edge of the sea can be specified as a cliff.

A 'clifftop' just refers to any pinnacle of a cliff. A 'plateau' is any flat extended geologic floor. An 'overhang' is a part of a structure or formation that protrudes from the primary frame and rests such that it is 'overhanging' the ground (striking above it).

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Answer:

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

Explanation:

In order to calculate the magnitude and direction of the magnetic field, you take into account the following equation for the magnetic force on the proton:

[tex]\vec{F_B}=q\vec{v}\ X\ \vec{B}[/tex]       (1)

v: speed of the proton = 9.9*10^5 m/s

q: charge of the proton = 1.6*10^-19C

B: magnetic field = ?

FB: magnetic force on the proton = 1.6*10^-13N

When the proton travels in the positive y direction (^j), you have that the proton experiences a force in the positive z direction (+^k). To obtain this direction of the magnetic force on the proton, it is necessary that the magnetic field points in the negative x direction, in fact, you have:

^j X (-^i) = -(-^k)=^k

To obtain the magnitude of the magnetic field you use:

[tex]F_B=qvBsin90\°=qvB\\\\B=\frac{F_B}{qv}=\frac{1.6*10^{-13}N}{(1.6*10^{-19}C)(9.9*10^5m/s)}\\\\B=1.01T[/tex]

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

Answer:

Density = 10,933.93 kg/m^3

the density of the material is 10,933.93 kg/m^3

Explanation:

Density is the mass per unit volume

Density = mass/volume = m/V

Volume of a cylinder V = πr^2 h

Given;

Height h = 48.5mm = 0.0485 m

Radius r = diameter/2 = 49mm÷2 = 24.5mm = 0.0245m

Substituting the values;

Volume V = π×(0.0245^2)×0.0485

V = 0.000091458438030 m^3

V = 0.000091458 m^3

The mass is given as;

Mass = 1 kg

So, the density can be calculated as;

Density = 1/0.000091458

Density = 10933.92825785 kg/m^3

Density = 10,933.93 kg/m^3

the density of the material is 10,933.93 kg/m^3

Answer:

74kW

Explanation:

See attached file

Answer:

α(0) = 0 rad/s²

α(5) = 15 rad/s²

Explanation:

The angular velocity of the flywheel is given as follows:

w(t) = A + B t²

where, A and B are constants.

Now, for the angular acceleration, we must take derivative of angular velocity with respect to time:

Angular Acceleration = α (t) = dw/dt

α(t) = (d/dt)(A + B t²)

α(t) = 2 B t

where,

B = 1.5

AT t = 0 s

α(0) = 2(1.5)(0)

α(0) = 0 rad/s²

AT t = 5 s

α(5) = 2(1.5)(5)

α(5) = 15 rad/s²