The average speed of a nitrogen molecule in air is about 6.70×102 m/s, and its mass is 4.68×10-26 kg.
(a) If it takes 3.00×10-13 s for a nitrogen molecule to hit a wall and rebound with the same speed but
moving in the opposite direction, what is the average acceleration of the molecule during this time
interval?
(b) What average force does the molecule exert on the wall?
(c) What is the total force exerted
on the wall if it is struck by 1015 such molecules all at once?

Answers

Answer 1

Answer:

a)   a = 3.06 10¹⁵ m / s , b)    F= 1.43  10⁻¹⁰ N, c)    F_total = 14.32 10⁻²⁶ N

Explanation:

This exercise will average solve using the moment relationship.

a ) let's use the relationship between momentum and momentum

          I = ∫ F dt = Δp

          F t = m [tex]v_{f}[/tex] - m v₀

          F = m (v_{f} -v₀o) / t

 in the exercise indicates that the speed module is the same, but in the opposite direction

          F = m (-2v) / t

if we use Newton's second law

          F = m a

we substitute

            - 2 mv / t = m a

            a = - 2 v / t

let's calculate

            a = - 2 4.59 10²/3 10⁻¹³

            a = 3.06 10¹⁵ m / s

b)      F= m a

        F= 4.68 10⁻²⁶ 3.06 10¹⁵

        F= 1.43  10⁻¹⁰ N

c) if we hit the wall for 1015 each exerts a force F

            F_total = n F

            F_total = n m a

            F_total = 10¹⁵  4.68 10⁻²⁶ 3.06 10¹⁵

            F_total = 14.32 10⁻²⁶ N


Related Questions


A projectile is fired at time t = 0.0 s from point o at the edge of a cliff, with initial velocity components of Vox = 30 m/s and Voy = 100 m/s. The projectile rises, and then falls into the sea
at point P. The time of flight of the projectile is 25 s. Assume air resistance is negligible.
t
What is the height of the cliff?
560 m
450 m
780 m
400 m
640 m

Answers

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.

What peak is considered a cliff?

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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A 90.0-kg ice hockey player hits a 0.150-kg puck, giving the puck a velocity of 45.0 m/s. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 15.0 m away

Answers

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

Mass of the player; [tex]m_1 = 90.0kg[/tex]Mass of puck; [tex]m = 0.150kg[/tex]

Since they were both at rest initially

Initial velocity of player; [tex]u_1 = 0[/tex]Initial velocity of puck; [tex]u = 0[/tex]Velocity of player after the hit; [tex]v_1 = \ ?[/tex]Velocity of puck after the hit; [tex]v = 45.0m/s[/tex]Distance to the goal; [tex]s = 15.0m[/tex]

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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Four identical charges particles of charge 1Uc, 2Uc,
3Uc and 4Uc
are placed at x = lm, x=2m,
x=3m and
x=5m. The electric field intensity
at origin is?

Answers

Answer:

17.94 kN/C is the electric field intensity at the origin due to the charges.

Explanation:

From the question, we are told that

The distance of 1 μC from origin = 1 m

The distance of 2 μC from origin = 2 m

The distance of 3 μC from origin = 3 m

The distance of 4 μC from origin = 5 m

Therefore, for us to find the electric field intensity, we'll solve below:

The formula for Electric field intensity = ( k * q ) / ( r * r )

where , r is distance ,

k = 9 * 10^9 ,

and , q is charge .

now ,

electric field intensity at the origin = [ k * 10^(-6) / 1 * 1 ] +[ k * 2 * 10^(-6) / 2 * 2 ] + [ k * 3 * 10^(-6) / 3 * 3 ] + [ k * 4 * 10^(-6) / 5 * 5 ]

=> electric field intensity at the origin = k * 10^(-6) [ 1 + 1/2 + 1/3 + 4/25 ] N/C

=> electric field intensity at the origin = 9 * 10^9 * 10^(-6) * 1.99 N/C

=> electric field intensity at the origin = 17.94 kN/C

Final naïve case: If the highest-pitch string on the piano is made of spring steel (density = 7800 kg/m3) with a diameter of 1/32" (= 0.794 mm), what will the linear density of such a string be (in kg/m)?

Answers

Answer:

The linear density is  [tex]K = 3.863 *10^{-3 } \ kg/m[/tex]

Explanation:

From the question we are told that

     The density of  steel is  [tex]\rho = 7800 \ kg/m^3[/tex]

      The diameter of the string is  [tex]d = 0.794 \ mm = 7.94 *10^{-4} \ m[/tex]

       The  radius of  the string is  evaluated as  [tex]r = \frac{D}{2} = \frac{7.94 *10^{-4}}{2} = 3.97*10^{-4} \ m[/tex]

The volume of the string is  mathematically evaluated as

       [tex]V = \pi * r ^2 * L[/tex]

Now assuming that the length of the string is  L = 2 m  

      So  

         [tex]V = 3.142 * (3.97 *10^{-4})^2 * (2)[/tex]

        [tex]V = 9.9041 *10^{-7} \ m^3[/tex]

Then the mass of the string would be  

       [tex]m = \rho * V[/tex]

substituting value  

       [tex]m = 7800*9.904 14 *10^{-7}[/tex]

      [tex]m = 7.73*10^{-3} \ kg[/tex]

Looking at the question we see that the unit of the linear density is  [tex]\frac{kg}{m}[/tex]

Hence the linear density is evaluated as

        [tex]K = \frac{m}{L}[/tex]

substituting value  

        [tex]K = \frac{7.73 *10^{-3}}{2}[/tex]

        [tex]K = 3.863 *10^{-3 } \ kg/m[/tex]

 

Which of the following is not considered a behavior?
A. eating
B. anxiety
C. sleeping
D. crying

Answers

━━━━━━━☆☆━━━━━━━

▹ Answer

B. Anxiety

▹ Step-by-Step Explanation

Anxiety isn't a behavior since it's a feeling. Behavior and feeling are different things therefore, anxiety is the correct answer.

Hope this helps!

- CloutAnswers ❁

Brainliest is greatly appreciated!

━━━━━━━☆☆━━━━━━━

Eating, sleeping, and crying all are considered as behaviors. However, anxiety cannot be considered as a behavior because it is a feeling. Thus, the correct option is B.

What is Anxiety?

Anxiety is an intense feeling of excessive, and persistent worry and the fear about everyday situations. This includes fast heart rate, rapid breathing, sweating, and feeling tired constantly may occur.

Behavior is the range of actions and mannerisms which are made by individuals, organisms, systems or the artificial entities in some environment. These systems can include other systems or organisms as well as the inanimate physical environment. Behaviors include eating, sleeping, and crying. Anxiety is not a behavior, it is a feeling.

Therefore, the correct option is B.

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Identical 4.0-uC (microCoulomb) charges are placed on the y axis at y = +/-4.0 m. (plus or minus 4.0 m) Point A is on the x axis at x = +3.0 m. Determine the electric potential of point A (relative to zero at the origin).

Answers

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 .

I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block and it oscillates with a period of 0.13 s. What is the amplitude of oscillation

Answers

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

A sphere has a mass of 8200 kg and a radius of 90 cm and floats with nearly zero friction on a thin layer of pressurized water. Suppose you pushes on the sphere tangent to its surface with a steady force of F = 75N and that the pressured water provides a frictionless support.

Required:
How long will it take her to rotate the sphere one time, starting from rest?

Answers

Answer:

Time is 22seconds

Explanation:

See attached file

a particle of mass 1.3kg is sliding down a frictionless slope inclined at 30 to the horizontal. the acceleration of the particle down the slope is

Answers

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

Water is flowing at a rate of 0.25 m3/s, and it is assumed that head loss in the system is 2V2/2g from the reservoir to the gage, where V is the velocity in the 30 cm pipe. What power must the pump supply

Answers

Answer:

74kW

Explanation:

See attached file

Which of the following controls the normal breathing system ​

Answers

What are the options?

An RC series circuit is connected to an ac generator with a maximum emf of 25 V. If the maximum potential difference across the resistor is 15 V, then the maximum potential difference across the capacitor is

Answers

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

6. The radius of a ball is (5.2 + 0.2) cm. The percentage
error in the volume of the ball is;​

Answers

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%

A bomb is dropped from a bomber traveling at the speed of 120 km / h, destroying a military objective located at a distance of 2000 m. How high was the plane traveling?

Answers

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

Angular velocity in the z direction of a flywheel is w(t)=A + Bt2 The numerical values of the constants are A=2.75 and B=1.50. What is the angular acceleration α(t) when t=0s and t=5.00s?

Answers

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

Use Coulomb’s law to derive the dimension for the permittivity of free space.



Answers

Answer: M^-1 L^-3T^4A^2

Explanation:

From coloumb's law

K = q1q2 / (F × r^2)

Where;

q1, q2 = charges

k = constant (permittivity of free space)

r = distance

Charge (q) = current(A) × time(T) = TA

THEREFORE,

q1q2 = (TA) × (TA) = (TA)^2

Velocity = Distance(L) / time(T) = L/T

Acceleration = change in Velocity(L/T) / time (T)

Therefore, acceleration = LT^-2

Force(F) = Mass(M) × acceleration (LT^-2)

Force(F) = MLT^-2

Distance(r^2) = L^2

From ; K = q1q2 / (F × r^2)

K = (TA)^2 / (MLT^-2) (L^2)

K = T^2A^2M^-1L^-1T^2 L^-2

COLLEXTING LIKE TERMS

T^2+2 A^2 M^-1 L^-1-2

M^-1 L^-3T^4A^2

What is the final temperature of 0.3kg of water initially at 20 Celsius after 35 KJ Of heat is added?

Answers

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.

Identify the term used to describe the ability of a liquid to flow against gravity up a narrow tube.

Answers

Answer:

This would be capillary action.

Explanation:

The physics behind it is gravity adhesion. The forces that attract between dissimilar molecules or atoms, in our case the contact area between the particles of the liquid and the particles forming the tube.

How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 20.0 cm in diameter to produce an electric field of 1450 N/C just outside the surface of the sphere

Answers

Answer:

1.007 × 10^(10) electron

Explanation:

We are given;

Electric Field;E = 1450 N/C

Diameter;d = 20 cm = 0.2 m

So, Radius: r = 0.2/2 = 0.1 m

Formula for Electric field just outside the sphere is given by the formula;

E = kq/r²

Where;

E is the magnitude of the electric field. q is the magnitude of the point charge r is distance from the point charge

k is a constant with a value of 9 x 10^(9) N.m²/C²

Making q the subject, we have;

q = Er²/k

Thus,

q = 1450 × 0.1²/(9 × 10^(9))

q = 1.61 × 10^(-9) C

Now, total charge q is also given by the formula;

q = Ne

Where;

e is charge on electron which is 1.6 × 10^(-19)

N is number of excess electrons

Making N the formula, we have;

N = q/e

N = (1.61 × 10^(-9))/(1.6 × 10^(-19))

N = 1.007 × 10^(10) electron

The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 400 g. The vibrating section of the string is 1.90 m long.What tension is needed to tune this string properly?Express your answer with the appropriate units.

Answers

Answer:

1456 N

Explanation:

Given that

Frequency of the piano, f = 27.5 Hz

Entire length of the string, l = 2 m

Mass of the piano, m = 400 g

Length of the vibrating section of the string, L = 1.9 m

Tension needed, T = ?

The formula for the tension is represented as

T = 4mL²f²/ l, where

T = tension

m = mass

L = length of vibrating part

F = frequency

l = length of the whole part

If we substitute and apply the values we have Fri. The question, we would have

T = (4 * 0.4 * 1.9² * 27.5²) / 2

T = 4368.1 / 2

T = 1456 N

Thus, we could conclude that the tension needed to tune the string properly is 1456 N

A proton moving in the positive x direction with a speed of 9.9 105 m/s experiences zero magnetic force. When it moves in the positive y direction it experiences a force of 1.6 10-13 N that points in the positive z direction. Determine the magnitude and direction of the magnetic field.

Answers

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

(5 pt) You tie a cord to a pail of water, and your swing the pail in a vertical circular 0.700 m. What is the minimum speed must you give the pail at the highest point of the circle if no water is to spill from it

Answers

Answer:

The minimum speed required  is 2.62m/s

Explanation:

The value of  gravitational acceleration = g = 9.81 m/s^2

Radius of the vertical circle = R = 0.7 m

Given the mass of the pail of water = m

The speed at the highest point of the circle = V

The centripetal force will be needed must be more than the weight of the pail of water in order to not spill water.

Below is the calculation:

[tex]\frac{mV^{2}}{R} = mg[/tex]

[tex]V = \sqrt{gR}[/tex]

[tex]V = \sqrt{9.81 \times 0.7}[/tex]

[tex]V = 2.62 m/s[/tex]

What is the change in potential energy of a 2.00 nC test charge, Uelectric, b - Uelectric, a, as it is moved from point a at x

Answers

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.

Two children sit on different sides of a seesaw. The first child of mass 27 kg sits 1.5 m from the center. How far must the second child of mass 23 kg sit from the center on the opposite side in order to balance the seesaw

Answers

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

What is distance?

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:

mgd = MgD......... Equation 1

 Make D the subject of the equation

D = md/M ............ Equation 2

Where:

D = Distance of the second child from the centerm = Mass of the first childd = Distance of the first child from the centerM = Mass of the second child

From the question,

Given:

m = 27 kgd = 1.5 mM = 23 kg

Substitute these values into equation 1

D = (27×1.5)/23D = 1.76 m

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

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In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the diver, in rad/s, during this time interval.

Answers

Answer:

The  angular speed is [tex]w = 5.89 \ rad/s[/tex]

Explanation:

From the question we are told that

    The time taken is  [tex]t = 1.6 s[/tex]

    The number of somersaults  is n  =  1.5

The total angular displacement during the somersault is mathematically represented as

         [tex]\theta = n * 2 * \pi[/tex]

substituting values

        [tex]\theta = 1.5 * 2 * 3.142[/tex]

       [tex]\theta = 9.426 \ rad[/tex]

 The angular speed is mathematically represented as

         [tex]w = \frac{\theta }{t}[/tex]

substituting values

         [tex]w = \frac{9.426}{1.6}[/tex]

          [tex]w = 5.89 \ rad/s[/tex]

     

Find the ratio of the gravitational force between two planets if the masses of both planets are quadrupled but the distance between them stays the same.

Answers

Answer:

The ratio of the new force over the original force is 16

Explanation:

Recall the formula for the gravitational force between two masses M1 and M2 separated a distance D:

[tex]F_G=G\,\frac{M_1\,\,M_2}{D^2}[/tex]

So now, if the masses M1 and M2 are quadrupled and the distance stays the same, the new force becomes:

[tex]F'_G=G\,\frac{4M_1\,\,4M_2}{D^2}=G\,\frac{16\,\,M_1\,\,M_2}{D^2}=16\,\,G\,\frac{M_1\,\,M_2}{D^2}= 16\,\,F_G[/tex]

which is 16 times the original force.

So the ratio of the new force over the original force is 16

The ratio of the gravitational force between two planets if the masses of both planets are quadrupled but the distance between them stays the same is 16:1.

What does Newton's law of gravitation state?

Newton's law of gravitation states that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them.

The formula for Newton's law of gravitation is:

[tex]F = G \frac{m_1m_2}{r^{2} }[/tex]

where,

F is the gravitational force.G is the gravitational constant.m₁ and m₂ are the masses of both objects.r is the distance between the objects.

The initial force between the planets is:

[tex]F_1 = G \frac{m_1m_2}{r^{2} }[/tex]

The force between the planets if the masses of both planets are quadrupled but the distance between them stays the same is:

[tex]F_2 = G \frac{4m_14m_2}{r^{2} } = 16 G \frac{m_1m_2}{r^{2} }[/tex]

The ratio of F₂ to F₁ is:

[tex]\frac{F_2}{F_1} =\frac{16 G \frac{m_1m_2}{r^{2} }}{G \frac{m_1m_2}{r^{2} }} = \frac{16}{1}[/tex]

The ratio of the gravitational force between two planets if the masses of both planets are quadrupled but the distance between them stays the same is 16:1.

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How much heat had to be put in during the process to increase the internal energy of the gas by 2.00×104 JJ ?

Answers

Complete question:

The pv diagram in the figure below (see attached file) shows a process abc involving 0.920 of an ideal gas.

How much heat had to be put in during the process to increase the internal energy of the gas by 2.00×10⁴ J ?

Answer:

The amount of heat required to increase the internal energy of the gas is 2,000 J

Explanation:

Work done in gas is given as;

W = ΔPV

The pressure of the gas at "a" = 2 x 10⁵ Pa

The pressure of the gas at "b" = 5 x 10⁵ Pa

The volume of the gas at "a" = 0.01 m³

The volume of the gas at "b" = 0.07 m³

The work done = (5 x 10⁵ Pa - 2 x 10⁵ Pa) x (0.07 m³ - 0.01 m³)

The work done = 3 x 10⁵ Pa x 0.06 m³

The work done = 18000 J

The work done = 1.8 x 10⁴ J

Determine, the amount of heat required to increase the internal energy of the gas by 2.00×10⁴ J.

1.8 x 10⁴ J + H = 2.00 x 10⁴ J

H = 2.00 x 10⁴ J - 1.8 x 10⁴ J

H = 0.2 x 10⁴ J

H = 2,000 J

Therefore,  the amount of heat required to increase the internal energy of the gas by 2.00×10⁴J is 2,000 J

When can an electric field be measured at any point from the force on a stationary test charge at that point?

Answers

Answer:

Electric field be measured at any point from the force on a stationary test charge at that point, when the force on the test charge is generated only by the coulomb field (E) of static charges.

Explanation:

Total force on a test charge is given as;

F = qE + q(V x B)

F = q(E + V x B)

Where;

F is the total force on the test charge

q is the test charge

E is the electric field

B is the magnetic field

V is the velocity of the test charge

F = q(E + V x B)

for a stationary test charge, V = 0

The equation above reduces to;

F = qE

Thus, electric field be measured at any point from the force on a stationary test charge at that point, when the force on the test charge is generated only by the coulomb field (E) of static charges.

A standard 1 kilogram weight is a cylinder 48.5 mm in height and 49.0 mm in diameter. What is the density of the material? kg/m3

Answers

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

A particle accelerator fires a proton into a region with a magnetic field that points in the +x-direction (a) If the proton is moving in the ty-direction, what is the direction of the magnetic force on the proton? +x-direction -x-direction +y-direction -y-direction +z-direction -z-direction zero force What is the formula used to find the vector magnetic force? What is the right-hand rule for a cross prod (b) If the proton is moving in the -y-direction, what is the direction of the magnetic force on the proton? +x-direction -x-direction +y-direction -y-direction +z-direction -z-direction zero force (c) If the proton is moving in the x-direction, what is the direction of the magnetic force on the proton?

Answers

Answer:

a) -z direction, b) +z direction, c) F=0  

Explanation:

The magnetic force is given by the expression

        F = q v x B

the bold indicate vectors, this equation can be separated in its module

        F = a v B sin θ

and where θ is the angles between the speed and the magnetic field.

The direction of the force can be found with the right-hand rule. For a positive charge, the thumb goes in the direction of speed, the fingers extended in the direction of the magnetic field and the palm points in the direction of the force, if the charge is negative the force is in the opposite direction.

a) Let's apply this to our case

the proton is positively charged

moves in the direction of + x

The magnetized field goes in the direction of y

therefore applying the right hand rule the force must be in the direction of the negative part of the z-axis (-z)

The right-hand rule is used to find this address.

b)  in this case it indicates that the proton moves in the recode of -y

again we apply the right hand rule and the force is in the direction of + z

c)   The proton moves in the x direction

In this case the force is zero because the angle between the field and the speed is zero and the sine is zero, therefore the force is zero

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