wo charged spheres are 1.5 m apart and are exerting an electrostatic force (Fo) on each other. If the charge on each sphere decreases by a factor of 9, determine (in terms of Fo) how much electrostatic force each sphere will exert on the other.

Answer:

Weight on Venus = 443.75 N

Explanation:

Weight of a body is the product of mass and acceleration due to gravity.

So we have

       Weight =  Mass x Acceleration due to gravity

        W = mg

        Mass, m = 50 kg

        Acceleration due to gravity, g = 8.875 m/s²

        W = 50 x 8.875 = 443.75 N

Weight on Venus = 443.75 N

Answer:

443.75 N

Explanation:

Weight is the force with which a planet can attract anybody towards its centre.

Weight = mass of body × acceleration due to gravity on that planet

Weight = 50 × 8.875 = 443.75 N

Answer:

The mass of the planet is [tex]2.7\times10^{27}\ kg[/tex].

Explanation:

Given that,

Semi major axis [tex]a= 4.3\times10^{8}[/tex]

Orbital period T=1.516 days

Using Kepler's third law

[tex]T^2=\dfrac{4\pi^2}{GM}a^3[/tex]

[tex]M=\dfrac{4\pi^2}{GT^2}a^3[/tex]

Where, T = days

G = gravitational constant

a = semi major axis

Put the value into the formula

[tex]M=\dfrac{4\times(3.14)^2}{6.67\times10^{-11}(1.516\times24\times60\times60)^2}(4.3\times10^{8})^3[/tex]

[tex]M=2.7\times10^{27}\ kg[/tex]

Hence, The mass of the planet is [tex]2.7\times10^{27}\ kg[/tex].

Answer:

164.33 N

Explanation:

Given:

The mass of the ball, m = 5.57 kg

Angle made by the sag = 19.4°

it is required to find the tension [tex]F_T[/tex] in the string

Note: Refer the attached figure

The tension in the string will be caused by the weight of the wall

thus,

Weight of the ball, W = 5.57 kg × 9.8 m/s² = 54.586 N

Now the resolving the tension [tex]F_T[/tex] in the string into components as shown in the figure attached, the weight of the ball will be balanced by the sin component of the tension

thus,

W = [tex]F_T[/tex]sinΘ

or

54.586 N = [tex]F_T[/tex] sin 19.4°

or

[tex]F_T[/tex] = 164.33 N

Hence, the tension in the string will be 164.33 N

Answer:

Option D is the correct answer.

Explanation:

Since value of angular acceleration is constant, the body has only centripetal acceleration.

Centripetal acceleration

               [tex]a=\frac{v^2}{r}=\frac{(r\omega )^2}{r}=r\omega ^2[/tex]

We have radius = 7.112 cm = 0.07112 m

Frequency, f = 1975 rpm = 32.92 rps

Angular frequency, ω = 2πf = 2 x π x 32.92 = 206.82 rad/s

Substituting in centripetal acceleration equation,

              [tex]a=r\omega ^2=0.07112\times 206.82^2=3042.17m/s^2[/tex]

Option D is the correct answer.                

Answer:

The horizontal distance traveled by the ball before it hits the ground is 48.85 meters.

Explanation:

It is given that,

Speed of golf ball, v = 25 m/s

Angle above horizontal or angle of projection, θ = 65°

We need to find the distance travelled by the ball before it hots the ground or in other words we need to find the range. It is given by R.

[tex]R=\dfrac{v^2\ sin2\theta}{g}[/tex]

[tex]R=\dfrac{(25\ m/s)^2\ sin2(65)}{9.8\ m/s^2}[/tex]

R = 48.85 m

So, the distance travelled by the ball before it hots the ground is 48.85 meters. Hence, this is the required solution.

Answer:

The magnetic field strength of the proton is 0.026 Tesla.

Explanation:

It is given that,

Speed of the proton, [tex]v=0.25\times 10^7\ m/s[/tex]

The radius of circular path, r = 0.975 m

It is moving perpendicular to a magnetic field such that the magnetic force is balancing the centripetal force.

[tex]qvB\ sin90=\dfrac{mv^2}{r}[/tex]

[tex]B=\dfrac{mv}{qr}[/tex]

q = charge on proton

[tex]B=\dfrac{1.67\times 10^{-27}\ kg\times 0.25\times 10^7\ m/s}{1.6\times 10^{-19}\ C\times 0.975\ m}[/tex]

B = 0.026 Tesla

So, the magnetic field strength of the proton is 0.026 Tesla.

Answer:

Option (a), (b) and (c)

Explanation:

The resistance of a conductor depends on the length of the conductor, area of crossection of the conductor and the nature of the conductor.

The formula for the resistance is given by

R = ρ x l / A

Where, ρ is the resistivity of the conductor, l be the length of the conductor and A be the area of crossection of the conductor.

So, It depends on the length, area and the type of material.

Resistance of a material is determined by its length, cross-sectional area, and the type of material; voltage and current do not determine resistance but are influenced by it.

The Quantities Determining the Resistance of a Material

The resistance of a piece of material is determined by several key factors, specifically:

It is important to note that the voltage across the material and the current flowing through the material are not factors that determine resistance. These two quantities are actually influenced by the resistance according to Ohm's law, which states that Voltage (V) equals Current (I) times Resistance (R), or V = IR.

The average induced current in the loop is 0.218 A.

The emf induced in the loop is determined by applying Faraday's law as shown below;

emf = dФ/dt

emf = BA/t

where;

A is the area

A = πr² = πd²/4

A = π x (0.025)²/4

A = 4.908 x 10⁻³ m²

emf = (2 x 4.908 x 10⁻³)/(0.45)

emf = 2.18 x  10⁻³ V

The average induced current in the loop is calculated as follows;

I = emf/R

I = 2.18 x  10⁻³/0.01

I = 0.218 A

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Answer:

Worker's change in gravitational potential energy = 641.69 kJ

Explanation:

Potential energy = Mass x Acceleration due to gravity x Height

PE = mgh

Mass, m = 79 kg

Acceleration due to gravity, g = 9.81 m/s²

Height, h = 828 m

Potential energy, PE = 79 x 9.81 x 828 = 641691.72 J = 641.69 kJ

Worker's change in gravitational potential energy = 641.69 kJ

Answer:

a)  The velocity of the ball upon leaving the foot = 600 m/s

b)  Time to reach the goal posts 40.0 m away = 0.07 seconds

c)  The kick won't e going inside goal post, it is higher by 10.34m.

Explanation:

a) Rate of change of momentum = Force

   [tex]\frac{\texttt{Final momentum - Initial momentum}}{\texttt{Time}}=\texttt{Force}\\\\\frac{0.040v-0.040\times 0}{0.010}=2400\\\\v=600m/s[/tex]

  The velocity of the ball upon leaving the foot = 600 m/s

b) Horizontal velocity = 600 cos20 = 563.82 m/s

   Horizontal displacement = 40 m

   Time

            [tex]t=\frac{40}{563.82}=0.07s[/tex]

   Time to reach the goal posts 40.0 m away = 0.07 seconds

c) Vertical velocity = 600 sin20 = 205.21 m/s

    Time to reach the goal posts 40.0 m away = 0.07 seconds

    Acceleration = -9.81m/s²

    Substituting in s = ut + 0.5at²

             s = 205.21 x 0.07 - 0.5 x 9.81 x 0.07²= 14.34 m

    Height of ball = 14.34 m

    Height of post = 4 m

    Difference in height = 14.34 - 4 = 10.34 m

    The kick won't e going inside goal post, it is higher by 10.34m.

I have seen this question before and the correct answer would be B

Hope this helped!!

Answer:

The correct answer would be B

Explanation:

To calculate the total resistance of three resistors connected in parallel, use the formula 1/R = 1/R1 + 1/R2 + 1/R3. The maximum error in the calculated value of R can be estimated by multiplying the sum of the errors in each resistance by the calculated value of R.

To find the total resistance of three resistors connected in parallel, we use the formula 1/R = 1/R1 + 1/R2 + 1/R3. Given the resistances R1 = 100 Ω, R2 = 25 Ω, and R3 = 10 Ω, we can substitute these values into the formula to calculate the total resistance R. Therefore, 1/R = 1/100 + 1/25 + 1/10 = 0.01 + 0.04 + 0.1 = 0.15. Now, to estimate the maximum error in the calculated value of R, we consider the errors in each resistance. Since each resistance has a possible error of 0.5%, we can calculate the maximum error in R by multiplying the sum of the errors in each resistance by the calculated value of R. Therefore, maximum error in R = 0.005 * 0.15 = 0.00075 Ω.

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The maximum error in the total resistance R of three parallel resistors with a potential error of 0.5% in each resistor is approximately 0.667 ohms.

The question asks to calculate the maximum error in the calculated value of total resistance R when three resistors R1, R2, and R3 are connected in parallel, where R1 = 100 Ω, R2 = 25 Ω, and R3 = 10 Ω, each with a possible error of 0.5%. The resistors in parallel have a total resistance denoted by:

1/R = 1/R1 + 1/R2 + 1/R3

To find the maximum error in the calculated value of R, we will first calculate R and then use derivative rules to estimate the maximum error considering the errors in R1, R2, and R3.

After calculating 1/R using the given resistances:

1/R = 1/100 + 1/25 + 1/10

1/R = 0.01 + 0.04 + 0.1 = 0.15

Therefore, R = 1 / 0.15 = 6.667 Ω

We calculate the maximum possible errors in resistances as:

Using the formula for the propagation of errors for functions of several independent variables, we estimate the maximum error in R (eR) as:

eR ≈ | -R² * eR1/R1² | + | -R² * eR2/R2² | + | -R² * eR3/R3² |

Plugging in the values:

eR ≈ | -6.667² * 0.5/100² | + | -6.667² * 0.125/25² | + | -6.667² * 0.05/10² |

eR ≈ | -0.04446 | + | -0.17784 | + | -0.4446 | = 0.667 Ω (Approximated to three decimal places)

The estimated maximum error in the calculated value of R is therefore approximately 0.667 Ω.

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Answer:

Depth of well 3.06m

Explanation:

We know that for a pipe closed at one end the frequencies are in ratios if 1:2:3:5:7.... to the fundamental frequency

In our case the given frequencies are in the ratio of

a)[tex]\frac{140}{196}=\frac{5}{7}[/tex]

b) [tex]\frac{196}{252}=\frac{7}{9}[/tex]

Thus the fundamental frequency can be calculated as [tex]140Hz=5n[/tex]

[tex]\therefore n=\frac{140}{5}=28Hz[/tex]

Now we know that

[tex]\lambda_{1}=4l\\\\\frac{v}{f}=4l\\\\l=\frac{v}{4f}[/tex]

Applying values we get

[tex]L=3.06m[/tex]

By identifying 56 Hz as the fundamental frequency and using the formula for the resonance of open-ended tubes, the depth of the well is calculated to be approximately 3.06 meters using the speed of sound as 343 m/s.

The student has observed standing waves at frequencies of 140 Hz, 196 Hz, and 252 Hz in a well. These frequencies represent the natural resonant frequencies or harmonics of the well. To find the depth of the well, we need to consider these frequencies as harmonics of a sound wave corresponding to the lengths of the air column in the well where each harmonic creates a standing wave.

Given the multiple frequencies, there is a constant difference of 56 Hz (196 Hz - 140 Hz, 252 Hz - 196 Hz) between consecutive frequencies. This indicates that 56 Hz is the fundamental frequency of the harmonics. We can represent these frequencies as [tex]F1 = 56 Hz \(n=1\), F2 = 112 Hz \(n=2\)[/tex]ere the observed frequencies correspond to F3, F4, and F5.

To determine the length of the well (L), we use the formula for the resonance of open-ended tubes, [tex]\(L = \frac{v}{2f}\)[/tex]d of sound (343 m/s) and 'f' is the frequency (56 Hz in this case). Therefore, [tex]L = \frac{343 m/s}{2 \times 56 Hz} = 3.0625 m[/tex]ly 3.06 meters.

Answer:

Rise in temperature is given as

[tex]\Delta T = 8.4 \times 10^{-3} ^0C[/tex]

Explanation:

Initial kinetic energy of the block is given as

[tex]KE = \frac{1}{2}mv^2[/tex]

here we will have

m = 3.3 kg

v = 3.2 m/s

now we will have

[tex]KE = \frac{1}{2}mv^2[/tex]

now we will have

[tex]KE = \frac{1}{2}(3.3)(3.2)^2[/tex]

[tex]KE = 17 J[/tex]

now we know that 74% of initial kinetic energy is absorbed as internal energy of the block

so the rise in temperature of the block is given as

[tex]KE = ms\Delta T[/tex]

[tex]0.74 \times 17 J = (3.3)(452)\Delta T[/tex]

[tex]12.5 = 1491.6 \Delta T[/tex]

[tex]\Delta T = 8.4 \times 10^{-3} ^0C[/tex]

Answer:

The temperature increases 0.0084ºC

Explanation:

Please look at the solution in the attached Word file.

Answer:

a) 250 N/m

b) 22.4 rad/s , 3.6 Hz , 0.28 sec

c) 0.3125 J

Explanation:

a)

F = force applied on the spring = 7.50 N

x = stretch of the spring from relaxed length when force "F" is applied = 3 cm = 0.03 m

k = spring constant of the spring

Since the force applied causes the spring to stretch

F = k x

7.50 = k (0.03)

k = 250 N/m

b)

m = mass of the particle attached to the spring = 0.500 kg

Angular frequency of motion is given as

[tex]w = \sqrt{\frac{k}{m}}[/tex]

[tex]w = \sqrt{\frac{250}{0.5}}[/tex]

[tex]w [/tex] = 22.4 rad/s

[tex]f[/tex] = frequency

Angular frequency is also given as

[tex]w [/tex] = 2 π [tex]f[/tex]

22.4 = 2 (3.14) f

[tex]f[/tex]  = 3.6 Hz

[tex]T[/tex] = Time period

Time period is given as

[tex]T = \frac{1}{f}[/tex]

[tex]T = \frac{1}{3.6}[/tex]

[tex]T[/tex] = 0.28 sec

c)

A = amplitude of motion = 5 cm = 0.05 m

Total energy of the spring-block system is given as

U = (0.5) k A²

U = (0.5) (250) (0.05)²

U = 0.3125 J

(a) The force constant of the spring is 250 N/m.

(b) The angular frequency of the mass oscillation is 22.36 rad/s, frequency is 3.56 Hz and the period is 0.28 s.

(c)  the total energy of the system is 0.31 J.

The force constant of the spring can be determined by applying Hooke's law as follows;

F = kx

k = F/x

k = (7.5)/0.03)

k = 250 N/m

The angular frequency of the mass oscillation is calculated as follows;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{250}{0.5} }\\\\\omega = 22.36 \ rad/s[/tex]

The angular frequency is calculated as follows;

ω = 2πf

f = ω/2π

f = (22.36)/2π

f = 3.56 Hz

The period of the oscillation is calculated as follows;

T = 1/f

T = 1/3.56

T = 0.28 s

The total energy of the system is calculated as follows;

U = ¹/₂kA²

U = ¹/₂ x 250 x (0.05)²

U = 0.31 J

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Answer: 135 grams

Explanation:

[tex]Q_{absorbed}=Q_{released}[/tex]

As we know that,  

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex]    

where,

[tex]m_1[/tex] = mass of ice = 100 g

[tex]m_2[/tex] = mass of aluminium cup =? g

[tex]T_{final}[/tex] = final temperature  =[tex]20^0C[/tex]

[tex]T_1[/tex] = temperature of ice = [tex]-10^oC[/tex]

[tex]T_2[/tex] = temperature of aluminium cup= [tex]70^oC[/tex]

[tex]c_1[/tex] = specific heat of ice= [tex]2.03J/g^0C[/tex]

[tex]c_2[/tex] = specific heat of aluminium cup = [tex] 0.902 J/g^0C[/tex]

Now put all the given values in equation (1), we get

[tex][100\times 2.03\times (20-(-10))]=-[m_2\times 0.902\times (20-70)][/tex]

[tex]m_2=135g[/tex]

Therefore, the mass of the aluminium cup was 135 g.

By calculating the heat transfer between a 100 g ice cube and an aluminum cup, the mass of the cup is found to be approximately 978.3 g.

To solve this, we will use principles of thermal equilibrium and specific heat capacities. Let's go through the steps:

Step 1: Calculate the heat required to warm the ice to 0 degrees Celsius.

The specific heat capacity of ice is 2.1 J/g°C. The formula for heat is:

Q = m * c * ΔT

m = 100.0 g (mass of ice)

c = 2.1 J/g°C (specific heat capacity of ice)

ΔT = (0°C - (-10°C)) = 10°C

Q₁ = 100 g * 2.1 J/g°C * 10°C = 2100 J

Step 2: Calculate the heat required to melt the ice at 0 degrees Celsius.

The enthalpy of fusion of ice ,as we know ,is 334 J/g.

Q₂ = m * L

m = 100.0 g

L = 334 J/g

Q₂ = 100 g * 334 J/g = 33400 J

Step 3: Calculate the heat required to warm the melted ice from 0°C to 20°C.

The specific heat capacity of water is 4.18 J/g°C.

Q₃ = m * c * ΔT

m = 100.0 g

c = 4.18 J/g°C

ΔT = (20°C - 0°C) = 20°C

Q₃ = 100 g * 4.18 J/g°C * 20°C = 8360 J

Step 4: Calculate the total heat gained by the ice.

[tex]Q_{total[/tex] = Q₁ + Q₂ + Q₃ = 2100 J + 33400 J + 8360 J = 43860 J

Step 5: Calculate the heat lost by the aluminum cup.

The specific heat capacity of aluminum is 0.897 J/g°C. Since we need the heat lost, we use:

[tex]Q_{lost[/tex] = m * c * ΔT

[tex]Q_{lost[/tex] = 43860 J

c = 0.897 J/g°C

ΔT = (70°C - 20°C) = 50°C

Rearranging for m:

m = [tex]Q_{lost[/tex] / (c * ΔT)

m = 43860 J / (0.897 J/g°C * 50°C) = 978.3 g

Therefore, the mass of the aluminum cup is approximately 978.3 g.

Answer:

[tex]F=55146000 N[/tex]

Explanation:

Given:

Pressure on the building = 170 N/m²

Pressure function P(y) = 170 + 2y

Consider a small height of the building as dy

the small force (dF) exerted on the height dy will be given as

dF = P(y)×(Area of the building face)

or

dF = (170 + 2y)(6565)dy

or

dF = 170×6565×dy + 2y×6565×dy

dF = 1116050dy + 13130ydy

integerating the function for the whole height of 40 m

[tex]\int\limits^{40}_0 {F} = \int _0^{40}\:\left(\:1116050dy\:+\:13130ydy\right)[/tex]

[tex]F = \int _0^{40}1116050dy+\int _0^{40}13130ydy[/tex]

[tex]F=\left[1116050y\right]^{40}_0+13130\left[\frac{y^2}{2}\right]^{40}_0[/tex]

[tex]F=44642000+10504000[/tex]

[tex]F=55146000 N[/tex] ; Total force exerted on the face of the building

To calculate the total force on a building due to wind pressure, integrate the pressure function over the building's height and multiply by the building's width. This calculation is vital for determining the building's required resistance to wind forces for safe and stable design.

The question asks to calculate the total force exerted on a large building by a wind blowing against one of its faces. The wind pressure increases with height as described by the function P(y)=170+2y, where y is the height in meters above the ground. To find the total force, we need to integrate this pressure function over the height of the building and multiply by the width of the building's face.

First, we can write the integral of the pressure over the height of the building:

Integrate [P(y) dy] from y=0 to y=40

= Integrate [(170+2y) dy] from y=0 to y=40

Performing this integration will give us the total pressure exerted over the entire height of the building. After integrating, we multiply the result by the width of the building (6565 m) to obtain the total force.

Using the integration results, the total force, F, can be expressed as:

F = (Integral result) × Width of the building

This force calculation is critical for architects and engineers as it provides an estimate of the resistance that the building must be designed to withstand due to wind forces, ensuring structural stability and safety.

Answer:

magnitude : 0.6 m/s²

Direction : Left

Explanation:

m = mass of the cart = 90 kg

Taking force in right direction as positive and force in left direction as negative

F₁ = Force applied by man 1 = 140 N

F₂ = Force applied by other man = - 195 N

a = acceleration of the cart

Force equation for the motion of the cart is given as

F₁ + F₂ = ma

140 + (- 195) = 90 a

a = - 0.6 m/s²

magnitude of acceleration is 0.6 m/s²

The negative sign indicates the direction of acceleration towards left

Final answer:

The magnitude of the cart's acceleration is approximately 0.611 m/s², and the direction is to the left.

Explanation:

When calculating the acceleration of the cart, we need to consider the net force acting on it and its mass. The net force is found by subtracting the smaller force from the larger force, taking into account their directions. With forces of 140 N to the right and 195 N to the left, the net force is the difference, which is 195 N - 140 N = 55 N, directed to the left since the larger force is in that direction. Using Newton's second law, acceleration (a) is the net force (Fnet) divided by the mass (m).

To find the magnitude of the acceleration, calculate a = Fnet / m. So, a = 55 N / 90 kg which equals approximately 0.611 m/s2. Since the larger force was to the left, the direction of the acceleration is also to the left.

Answer:

The change in internal energy of the system is -17746.78 J

Explanation:

Given that,

Pressure [tex]P=4.63\times10^{4}\ Pa[/tex]

Remove heat [tex]\Delta U= -1.95\times10^{4}\ J[/tex]

Radius = 0.272 m

Distance d = 0.163 m

We need to calculate the internal energy

Using thermodynamics first equation

[tex]dU=Q-W[/tex]...(I)

Where, dU = internal energy

Q = heat

W = work done

Put the value of W in equation (I)

[tex]dU=Q-PdV[/tex]

Where, W = PdV

Put the value in the equation

[tex]dU=-1.95\times10^{4}-(4.63\times10^{4}\times3.14\times(0.272)^2\times(-0.163))[/tex]

[tex]dU=-17746.78\ J[/tex]

Hence, The change in internal energy of the system is -17746.78 J

Answer:

Force of friction, F = 45.76 N

Explanation:

t is given that,

Mass of the skater, m = 64 kg

Initial velocity of the skater, u = 2.81 m/s

Finally it comes to rest, v = 0

Time, t = 3.93 s

We need to find the force of friction. According to seconds law of motion as :

F = m × a

[tex]F=m\times \dfrac{v-u}{t}[/tex]

[tex]F=64\ kg\times \dfrac{0-2.81\ m/s}{3.93\ s}[/tex]

F = −45.76 N

So, the frictional force exerting on the skater is 45.76 N. Hence, this is the required solution.

Answer:

D. 5.18 x 10⁻¹²

Explanation:

[tex]\frac{dE}{dt}[/tex] = rate at which sun radiates energy = 3.92 x 10²⁶ W

M = mass of sun = 1.99 x 10³⁰ kg

[tex]\frac{dm}{dt}[/tex] = rate at which sun's mass is lost

c = speed of light

Energy is given as

E = m c²

Taking derivative both side relative to "t"

[tex]\frac{dE}{dt}=c^{2}\frac{dm}{dt}[/tex]

[tex]3.92\times 10^{26}=(3\times 10^{8})^{2}\frac{dm}{dt}[/tex]

[tex]\frac{dm}{dt}[/tex] = 4.4 x 10⁹ kg/s

t = time interval = 75 yrs = 75 x 365 days = 75 x 365 x 24 hours = 75 x 365 x 24 x 3600 sec = 2.4 x 10⁹ sec

[tex]m[/tex] = mass lost

mass lost is given as

[tex]m = t\frac{dm}{dt}[/tex]

[tex]m = (2.4\times 10^{9})(4.4\times 10^{9})[/tex]

m = 10.56 x 10¹⁸ kg

fraction is given as

fraction = [tex]\frac{m}{M}[/tex]

fraction = [tex]\frac{10.56\times 10^{18}}{1.99\times 10^{30}}[/tex]

fraction = 5.18 x 10⁻¹²

Answer:

0.694 m

Explanation:

Case 1 : When only mass of 2.82 kg is hanged from spring

m = mass hanged from the spring = 2.82 kg

x = stretch caused in the spring = 0.331 m

k = spring constant

Using equilibrium of force in vertical direction

Spring force = weight of the mass

k x = m g

k (0.331) = (2.82) (9.8)

k = 83.5 N/m

Case 2 : When both masses are hanged from spring

m = mass hanged from the spring = 3.09 + 2.82 = 5.91 kg

x = stretch caused in the spring = ?

k = spring constant = 83.5 N/m

Using equilibrium of force in vertical direction

Spring force = weight of the mass

k x = m g

(83.5) x = (5.91) (9.8)

x = 0.694 m

180N

Using Newton's law of motion;

∑F = m x a       --------------------(i)

Where;

∑F = Resultant force

m = mass of the object (sled in this case)

a = acceleration of the sled

Calculate the resultant force;

Since the direction of motion is horizontal, the horizontal forces acting on the sled are the;

i. Applied force ([tex]F_{A}[/tex]) in one direction and;

ii. Frictional force ([tex]F_{R}[/tex]) in the other direction to oppose motion

Therefore, the resultant force ∑F is the vector sum of the two forces. i.e;

∑F = [tex]F_{A}[/tex] - [tex]F_{R}[/tex]  -----------------------(i)

Frictional force [tex]F_{R}[/tex] is the product of the coefficient of kinetic friction (μ) and weight(W) of the sled. i.e

[tex]F_{R}[/tex] = μ x W

Where;

W = mass(m) x gravity(g)

W = m x g

=> [tex]F_{R}[/tex] = μmg

Substitute [tex]F_{R}[/tex] into equation (ii)

∑F = [tex]F_{A}[/tex] - μmg

Substitute ∑F into equation (i)

[tex]F_{A}[/tex] - μmg = ma  -------------------(iii)

Since the motion is at constant speed, it means acceleration is zero (0)

Substitute a = 0 into equation (iii) to give;

[tex]F_{A}[/tex] - μmg = 0

=> [tex]F_{A}[/tex] = μmg

Substitute the values of μ = 0.3, m = 60kg and g = 10m/s² into the above equation to give;

=> [tex]F_{A}[/tex]  = 0.3 x 60 x 10

=> [tex]F_{A}[/tex] = 180N

This means that the applied force should be 180N

The amount of force the football player must apply to the sled is 176.4 Newton.

Given the following data:

We know that acceleration due to gravity (g) on Earth is equal to 9.8 [tex]m/s^2[/tex]

To find how much force the football player must apply to the sled:

Mathematically, the force of kinetic friction is given by the formula;

[tex]Fk = umg[/tex]

Where;

Substituting the given parameters into the formula, we have;

[tex]Fk = 0.30\times60\times9.8[/tex]

Force, Fk = 176.4 Newton.

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Answer:

75 m/s is faster

Explanation:

MKS units stands for meter kilogram seconds

75 miles per hour = 75 mph

1 mile = 1609.34 meters

1 hour = 60×60 = 3600 seconds

1 mph = 1609.34/3600 = 0.44704 m/s

75 mph = 75×0.44707 = 33.52792 m/s

Comparing 75 mph = 33.52792 m/s with 75 m/s it can be seen that 75 m/s is faster. Even without calculating the values you can know the answer. 75 mph means that in 1 hour the object will move 75 miles. 75 m/s means that in one second the object will cover 75 meters multiply by 3600 and you will get 270000 m/h that is 270 km/h divide it by 1.6 and you can approximately get the value in mph that will be around 168 mph which is faster than 75 mph.

Answer:

0.3375

Explanation:

w = angular speed of the DVD drive = 100.0 rpm = [tex]100.0 \frac{rev}{min}\frac{2\pi rad}{1 rev}\frac{1 min}{60 sec}[/tex] = [tex]10.5\frac{rad}{sec}[/tex]

m = mass of the dime = 2 g = 0.002 kg

r = radius = 3 cm = 0.03 m

μ = Coefficient of friction

The frictional force provides the necessary centripetal force to move in circle. hence

frictional force = centripetal force

μ mg = m r w²

μ g =  r w²

μ (9.8) =  (0.03) (10.5)²

μ = 0.3375

Answer:

The magnitude of the electric force on an electron in a uniform electric is [tex]2.4\times10^{-16}\ N[/tex] to the west.

Explanation:

Given that,

Electric field strength = 1500 N/C

We need to calculate the electric force

Using formula of electric field

[tex]F = Eq[/tex]

E = electric field strength

q = charge of electron

Electron has negative charge.

Put the value into the formula

[tex]F=1500\times(-1.6\times10^{-19})[/tex]

[tex]F=-2.4\times10^{-16}\ N[/tex]

Negative sign shows the opposite direction of the field

Hence, The magnitude of the electric force on an electron in a uniform electric is [tex]2.4\times10^{-16}\ N[/tex] to the west.

The magnitude of the electric force on an electron in a uniform electric field of strength 1500 N/C that points due east is 2.4x10⁻¹⁶ C.

We know that electric force is given by the formula,

[tex]F = E \times q[/tex]

It is given that the electric field, E = 1500 N/C,

We also know that an electron is negatively charged and has a charge of 1.60217662 × 10⁻¹⁹ C.

[tex]F = E \times q\\\\F = 1500 \times 1.6 \times 10^{-19}\\\\F = 2.4 \times 10^{-16}\rm\ N[/tex]

Hence, the magnitude of the electric force on an electron in a uniform electric field of strength 1500 N/C that points due east is 2.4x10⁻¹⁶ C.

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Answer:

Work is done by the gas  = 5.92 x 10⁵ J = 592 kJ

Explanation:

Work done at fixed pressure, W = PΔV

Pressure, P = 3.7 x 10⁵ Pa

Change in volume, ΔV = 1.6 m³

Substituting the values of pressure and change in volume we will get

Work done at fixed pressure, W = PΔV =  3.7 x 10⁵ x 1.6 = 5.92 x 10⁵ J

Work is done by the gas  = 5.92 x 10⁵ J = 592 kJ

Answer:

(a) 4 rad/s^2

(b) 18 rad

Explanation:

w0 = 0, w = 12 rad/s, t = 3 s

(a) Let α be the angular acceleration.

w = w0 + α t

12 = 0 + 3 α

α = 4 rad/s^2

(b) Let θ be the angle rotated

θ = w0 t + 1/2 α t^2

θ = 0 + 0.5 x 4 x 9

θ = 18 rad

Answer:

4.49 dollars

Explanation:

i = 9 A, V = 240 V, t = 16 h

Energy = V x i x t = 240 x 9 x 16 = 34560 W h = 34.56 kWh

The cost of 1 kWh is 13 cents.

Cost of 34.56 kWh = 13 x 34.56 = 449.28 cents = 449.28 / 100 = 4.49 dollars

The cost of running a water pump drawing 9 A at 240 V for 16 hours, with an energy cost of 13 cents per kWh, is $4.49.

To calculate the cost of running a water pump for 16 hours at an electrical energy cost of 13 cents per kWh, you need to follow these steps:

In dollars, the cost of running the pump for 16 hours is $4.49 (rounded to the nearest cent).

Answer:

a)W=20 KJ

b) ΔQ= 220 KJ

Explanation:

Given:

V₁=0.1 m^3,   P₁=200 kPa and heat is added to the system such that system expands under constant pressure.

Therefore V₂= 2V₁= 0.2 m^3

a) Work transfer W= P(V₂-V₁)= [tex]200\times(0.2-0.1)\times10^{5} = 2\times10^4 joules[/tex]

W=20 KJ

b) internal energy change ΔU= 200 KJ

from first law we know that ΔQ(net heat transfer)= ΔU + W

ΔQ= [tex]200\times10^3 +2\times10^4[/tex]

ΔQ=[tex]22\times10^4 J[/tex]

ΔQ= 220 KJ