I NEED HELP ASAPPPP PLEASE

Answer:

Explanation:

Maximum value of force will be possible when both the sphere will have same charge . In that case charge on each sphere = Q / 2 =.5Q

F( max ) = k .5Q x .5Q / R²

=.25kQ² /R²

For the second case

F = k q ( Q-q)/  R²

F = .25kQ² /3R²

.25kQ² /3R² =  k q ( Q-q)/  R²

.25 Q² = 3qQ - 3q²

3q² - 3qQ + .25 Q² = 0

q =

Answer:

9 Nm

Explanation:

The formula for torque is;

τ = lever arm * force applied

τ = 30/100 * 30

τ = 9 Nm

The torque on a bolt will be "9 Nm".

Given values are:

Force applied,

Lever arm,

As we know the formula,

→ [tex]\tau = Lever \ arm\times force \ applied[/tex]

By substituting the values, we get

→    [tex]= \frac{30}{100}\times 30[/tex]

→    [tex]= 9 \ Nm[/tex]

Thus the above response is correct.    

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Answer:

Both balloons experiences same buoyant force.

Explanation:

Buoyant force is defined as the upward force that is exerted on an object which is wholly or partly immersed in a fluid. We can also define it as the upward force exerted by any fluid upon a body immersed in it. We may also refer to this buoyant force as the upthrust.

This buoyant force is the push of air on the balloon and it is independent of the contents of the balloons. Hence, both balloons experiences same buoyant force.

Answer:

1.31×10¯⁶ N

Explanation:

From the question given above, the following data were obtained:

Mass of John (M₁) = 81 Kg

Mass of Mike (M₂) = 93 Kg

Distance apart (r) = 0.620 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force (F) =?

The gravitational force between the two students, John and Mike, can be obtained as follow:

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 81 × 93 / 0.62²

F = 6.67×10¯¹¹ × 7533 / 0.3844

F = 1.31×10¯⁶ N

Therefore, the gravitational force between the two students, John and Mike, is 1.31×10¯⁶ N

Answer:

The sun pulls with more force since the sun is more massive than the earth.

Explanation:

That your answer because the Earth is getting pulled by the gravitational pull from the Sun

Answer:

80N/cm^2

Explanation:

Given data

A car of weight 6,400N

Area= 80cm^2

The weight is distributed evenly on the 4 wheels

hence 6400/4

=1600 N

We know that

Pressure = force/Area

Pressure= 1600/80

Pressure= 20 N/cm^2

For the four wheels, the pressure is

=20*4

=80N/cm^2

Answer:

D. 45 N​

Explanation:

The weight of the block is 20 N, when the block is fully immerged in water, it weighs 15 N. Hence the loss of weight = 20 N - 15 N = 5 N.

The loss of weight is as a result of the buoyant force. The buoyant force is the upward force exerted by a fluid when an object is fully or partially immersed in a fluid.

The buoyant force of 5 N acts in the upward direction, the weight of the beaker that would be read by the scale when the beaker is immersed in water = 40 N + 5 N = 45 N

Answer:

[tex]0.15\: \mathrm{Hz}[/tex]

Explanation:

The frequency is of an object is given by [tex]f=\frac{1}{T}[/tex], where [tex]T[/tex] is the orbital period of the object.

Since the racecar makes 24 revolutions around a circular track in 162 seconds, it will take the racecar [tex]\frac{162}{24}=6.75\:\mathrm{s}[/tex] per revolution.

Therefore, the frequency of the racecar is [tex]\frac{1}{6.75}=\fbox{$0.15\:\mathrm{Hz}$}[/tex] (two significant figures).

The radius of the track is irrelevant in this problem.

Answer:

1960 joule

Explanation:

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

[tex]195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3[/tex]

(c) From any of the equations, solve for Length of the string (L);

[tex]195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m[/tex]

(a) the fundamental frequency is calculated as;

[tex]f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o = 65 \ Hz[/tex]

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

Answer:

Time period of horizontal Oscillation  = T = 2[tex]\pi[/tex][tex]\sqrt{\frac{m}{k} }[/tex]

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Explanation:

Solution:

As we know that:

F = Kx

Here,

K = Spring constant

x = displacement.

First, they are displacing it with 6 cm from its rest position for which Time period of the oscillation is T = 2 seconds.

But next, they want to know the effect on the time period of the oscillation if the displacement x is doubled from 6cm to 12 cm.

First of all, let us see the equation of the time period of the oscillation.

We need to check, if time period does depend on the displacement or not.

As we know,

Time period of horizontal Oscillation  = T = 2[tex]\pi[/tex][tex]\sqrt{\frac{m}{k} }[/tex]

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Since, K is the constant for a particular spring, we need to change the mass of the cart to change the time period.

Hence the Time period will remain same.

The time period will remain the same in both conditions. The time period of horizontal Oscillation will be [tex]2 \pi \sqrt{\frac{m}{k} }[/tex].

The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.

From Hooke's law;

F = Kx

Where,

K is the spring constant

x is the  displacement

First, they move it 6 cm from its rest position, with a T = 2 second oscillation period.

However, they want to know what influence doubling the displacement x from 6 cm to 12 cm has on the oscillation's time period.

Let's start by looking at the oscillation's time period equation. We need to see if the time period is affected by the shift.

The time period of the horizontal oscillation is given by;

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

As the equation shows, the period of oscillation is determined by the mass and the spring constant. On the displacement, no.

We must modify the mass of the cart to change the time period since K is the constant for each spring.

Hence the time period will not change.

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Answer:

Net force = 15 N

Explanation:

Given that,

Applied force on an object = 20 N

Frictional force = 5 N

We need to find the net force acting on the object.

Friction is an opposing force. It acts in the opposite direction.

Net force = Applied force - Frictional force

= 20 N - 5 N

= 15 N

Hence, the net force acting on the object is 15 N.

Answer:

hmmm thats too hard for me.

Explanation:

Answer:

The solution to this question can be defined as follows:

Explanation:

In point (a):

[tex]v_i= 100 \ mV\\\\ i_I = 100 \mu \ A\\\\ v_O = 10 \ V\\\\ R_L = 100 \ \Omega \\\\i_L = \frac{V_0}{R_L} = \frac{10}{100} = 100 \ MA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{100 \times 10^{-3}} =100 \\\\A_v(db) = 20 \lag (100) =40 \ db \\\\ A_i= \frac{i_L}{i_i} = \frac{100 \times 10^{-3}}{100 \times 10^{-6}} =1000 \\\\A_i(db) = 20 \lag (100) =60 \ db \\\\[/tex]

[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_L}{v_i i_i} = \frac{ 10(100 \times 10^{-3})}{100 \times 10^{-6} \times 100 \times 10^{-3}} =100000\\\\ A_p(db) =10 \log (100000) =50 \ db \\\\[/tex]

In point (b):

[tex]v_i = 10 \mu V\\\\ i_i = 100 \ nA \\\\ v_O =1 \ V \\\\ R_L= 10 \ k \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{1}{10 \ K} = 100 \ \muA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{10 \times 10^{-6}} =100000 \\\\A_v(db) = 20 \lag (100000) =100 \ db \\\\ A_i= \frac{i_0}{i_i} = \frac{100 \times 10^{-6}}{100 \times 10^{-9}} =1000 \\\\A_i(db) = 20 \lag (1000) =60 \ db \\\\[/tex]

[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 1 \times 100 \times 10^{-6})}{10 \times 10^{-6} \times 100 \times 10^{-9}} =100000000\\\\A_p(db) =10 \log (100000000) =80 \ db \\\\[/tex]

In point (C):

[tex]v_i =1\ V\\\\ i_I = 1 \ mA\\\\ v_O =5\ V \\\\ R_L = 10 \ \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{5}{10 } = 0.5 \ A \\\\A_v = \frac{V_0}{V_i} = \frac{5}{1} =5 \\\\A_v(db) = 20 \log 5 =13.97 \ db = 14 \db \\\\ A_i= \frac{i_0}{i_i} = \frac{0.5}{1\times 10^{-3}} =500 \\\\A_i(db) = 20 \log (500) =53.97 \ db = 54 \db \\\\[/tex]

[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 5 \times 0.5 }{1 \times 1 \times 10^{-3}} =2500\\\\A_p(db) =10 \log (2500) = 33.97 \ db = 34 \db\\\\[/tex]

Answer:

Uh, I could be wrong but doesn’t it mean that the wave and particle are reacting together to make light? I think it’s something like that... I hope this helps!

Answer:

7

Explanation:

We are given:

    w(x) = 3x   -   7

     w(x)  = 14

The problem here entails us to solve for x;

To solve for x; equate the two expressions:

          3x  - 7  = 14

           3x  = 14 + 7

           3x  = 21  

             x = 7

So the value of x  = 7

Answer:

Let Vx = 35 m/s     speed of swimmer across river

     Vy = 22 m/s     speed of river dowstream

V = (Vx^2 + Vy^2)^1/2 = 41.3 m/s      net speed of swimmer

tan theta = Vy / Vx = .629    theta = 32.2 deg

(Theta would be zero if speed of river was zero)\

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

       [tex]\tau = I * \alpha (1)[/tex]

       [tex]\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)[/tex]

        [tex]\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)[/tex]

       [tex]0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)[/tex]

       [tex]\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)[/tex]

        [tex]v = \omega * r (8)[/tex]

       [tex]v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)[/tex]

b)    

       [tex]a_{t} = \alpha * r (9)[/tex]

       [tex]a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)[/tex]

       [tex]a_{c} = \omega^{2} * r (11)[/tex]

       [tex]a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)[/tex]

       [tex]a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)[/tex]

Answer:

carbon.

Explanation:

the reaction would then create c02 as a product

Answer:

A

Explanation:

Answer:

D. Conduction

Explanation:

Convection heat transfer occurs in fluids, while conduction heat transfer occurs in solids.

Answer:

no

Explanation:

it is faster at the equator

Answer:

The weight of the girl is 250 N

Explanation:

Static Equilibrium

Static equilibrium occurs when an object is at rest, i.e., neither rotating nor translating.

In the static rotational equilibrium, the total torque is zero with respect to any rotational axis.

The torque applied by a force F perpendicular to a displacement X with respect to a reference rotating point is:

T = F*X

The seesaw will be in rotational equilibrium if the torque applied by the boy of mass m1=40 Kg at x1=2 m from the pivot is equal to the torque applied by the girl of unknown mass m2 at x2=3.2 m from the pivot.

The force applied by both children is their weight:

[tex]F_1 = W_1 = m_1g[/tex]

[tex]F_2 = W_2 = m_2g[/tex]

It must be satisfied:

[tex]m_1gx_1=m_2gx_2[/tex]

Simplifying:

[tex]m_1x_1=m_2x_2[/tex]

Solving for m2:

[tex]\displaystyle m_2=\frac{m_1x_1}{x_2}[/tex]

[tex]\displaystyle m_2=\frac{40*2}{3.2}[/tex]

[tex]m_2=25\ kg[/tex]

Her weight is:

[tex]\mathbf{W_2=25*10 = 250\ N}[/tex]

The weight of the girl is 250 N

Answer:

[tex]3.3\cdot 10^3\:\mathrm{N}[/tex]

Explanation:

Impulse on an object is given by [tex]\mathrm{[impulse]}=F\Delta t[/tex].

However, it's also given as change in momentum (impulse-momentum theorem).

Therefore, we can set the change in momentum equal to the former formula for impulse:

[tex]\Delta p=F\Delta t[/tex].

Momentum is given by [tex]p=mv[/tex]. Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:

[tex]p=2500\cdot 8.0=20,000\:\mathrm{kg\cdot m/s}[/tex].

Now we plug in our values and solve:

[tex]\Delta p=F\Delta t,\\F=\frac{\Delta p}{\Delta t},\\F=\frac{20,000}{6}=\fbox{$3.3\cdot 10^3\:\mathrm{N}$}[/tex](two significant figures).

Answer:

The charge of the negative one is 13.27 microcoulombs and the positive one has a charge of 58.27 microcoulombs.

Explanation:

Electric potential energy between two point charges is derived from concept of Work, Work-Energy Theorem and Coulomb's Law and described by the following formula:

[tex]U_{e} = \frac{k\cdot q_{A}\cdot q_{B}}{r}[/tex] (1)

Where:

[tex]U_{e}[/tex] - Electric potential energy, measured in joules.

[tex]q_{A}[/tex], [tex]q_{B}[/tex] - Electric charges, measured in coulombs.

[tex]r[/tex] - Distance between charges, measured in meters.

[tex]k[/tex] - Coulomb's constant, measured in kilogram-cubic meters per square second-square coulomb.

If we know that [tex]U_{e} = -24\,J[/tex], [tex]q_{A} = 45\times 10^{-6}\,C+ q_{B}[/tex], [tex]k = 9\times 10^{9}\,\frac{kg\cdot m^{3}}{s^{2}\cdot C^{2}}[/tex] and [tex]r = 0.29\,m[/tex], then the electric charge is:

[tex]-24\,J = -\frac{\left(9\times 10^{9}\,\frac{kg\cdot m^{3}}{s^{2}\cdot C^{2}} \right)\cdot (45\times 10^{-6}\,C+q_{B})\cdot q_{B}}{0.29\,m}[/tex]

[tex]-6.96 = -405000\cdot q_{B}-9\times 10^{9}\cdot q_{B}^{2}[/tex]

[tex]9\times 10^{9}\cdot q_{B}^{2}+405000\cdot q_{B} -6.96 = 0[/tex] (2)

Roots of the polynomial are found by Quadratic Formula:

[tex]q_{B,1} = 1.327\times 10^{-5}\,C[/tex], [tex]q_{B,2} \approx -5.827\times 10^{-5}\,C[/tex]

Only the first roots offer a solution that is physically reasonable. The charge of the negative one is 13.27 microcoulombs and the positive one has a charge of 58.27 microcoulombs.

Answer:

The angular velocity of a person standing on the equator is approximately [tex]7.272\times 10^{-5}[/tex] radians per second.

Explanation:

The Earth rotates at constant speed. From Rotational Physics, the angular velocity ([tex]\omega[/tex]), measured in radians per second, is defined by the following formula:

[tex]\omega = \frac{2\pi}{T}[/tex] (1)

Where [tex]T[/tex] is the period of rotation of the Earth, measured in seconds.

If we know that [tex]T = 86400\,s[/tex], then the angular velocity of a person standing on the equator is:

[tex]\omega = \frac{2\pi}{86400\,s}[/tex]

[tex]\omega \approx 7.272\times 10^{-5}\,\frac{rad}{s}[/tex]

The angular velocity of a person standing on the equator is approximately [tex]7.272\times 10^{-5}[/tex] radians per second.

The object Earth, Sun, and Blood are constantly in motion. The correct option is A, D, and E.

if a body changes its position with respect to its surroundings in a given interval of time, Then the body is said to be in motion

.

Motion is generally classified as follows.:

i) Rectilinear motion.

ii) Circular motion.

iii) Rotational motion.

iv) Periodic motion.

The Earth is continuously in motion because it continuously revolves around The Sun in an elliptical orbit, due to which a year is 365 days. and also The earth rotates about its own axis once a day.

The Sun also revolves around the galactic center of our Milkyway galaxy. and it also rotates about its own axis continuously. so that the sun is also continuously in motion.

The Human blood is continuously in motion Because our blood is continuously circulating whole over the body with the help of our heart. The heart continuously pumps our blood and circulates it inside the human body.

Hence the Earth, Sun, and blood are continuously in motion.

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Answer:

C

Explanation:

Kinetic energy is the energy of motion. It has the most potential energy at the top but the most kinetic at the bottom after it's accelerated fully down the slope.

Answer: Surface water is replaced by the water cycle.

Explanation: The water cycle is a cycle that describes the movement of water.

Answer:

it is water cycle when it's one of the process occurs and process is precipation this replaces surface water

Answer:

A is the best character that defines an electron.