the force of attraction between any two objects that have mass is

Answer:

both

Explanation:

baby for fun, toddler for vengence

Answer:

0.25m

Explanation:

Given parameters:

Spring constant , K  = 120N/m

Work done  = 3.75J

Unknown:

magnitude of extension = ?

Solution:

To solve this problem;

           Work done  = [tex]\frac{1}{2}[/tex]kx²  

K is the spring constant

x is the extension

               3.75  =  [tex]\frac{1}{2}[/tex] x 120x²

               3.75  = 60x²

                x²  = 0.06

                x = √0.06  = 0.25m

Answer:

it would be 49.03325 Newton.

Answer:

lo

Explanation:

Answer:

An experiment is a procedure carried out to support, refute, or validate a hypothesis. Experiments provide insight into cause-and-effect by demonstrating what outcome occurs when a particular factor is manipulated.

Explanation:

Complete question:

Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor? If the potential difference due to electric field between the two ends of the resistor is 10 V.

Answer:

The electric field inside this metal resistor is 3125 V/m

Explanation:

Given;

length of the wire, L = 3.2 mm = 3.2 x 10⁻³ m

diameter of the wire, d = 0.4 mm = 0.4 x 10⁻³ m

the potential difference due to electric field between the two ends of the resistor, V = 10 V

The electric field inside this metal resistor is given by;

ΔV = EL

where;

ΔV is change in electric potential

E = ΔV / L

E = 10 / (3.2 x 10⁻³ )

E = 3125 V/m

Therefore, the electric field inside this metal resistor is 3125 V/m

Answer:

delivery truck

Explanation:

because i got it right

Answer:

2872.8 N

Explanation:

We have the following information

m =n72kg

Δy = 18m

t = 0.95s.

From here we use the equation

Δy=1/2at2 in order to solve for the acceleration.

So a

=( 2x 18m)/(0.95s²)

= 36/0.9025

= 39.9m/s2.

From there we use the equation

F = ma

F=(72kg) x (39.9)

= 2872.8N.

2872.8N is the average net force exerted on him in the barrel of the cannon.

Thank you!

Answer:

D. [tex]\vec F_{B} = -20\,\hat{i}-15\,\hat{j}\,\,\,[N][/tex]

Explanation:

The statement is not correctly written, the correct form is now described:

A particle with charge [tex]q = -1\,C[/tex] is moving in the positive z-direction at 5 meters per second. The magnetic field at its position is [tex]\vec B = 3\,\hat{i}-4\,\hat{j}\,\,\,[T][/tex]. What is the magnetic force on the particle?

From classic theory on Magnetism, we remember that the magnetic force  exerted on a particle ([tex]\vec F_{B}[/tex]), measured in newtons, is determined by the following vectorial formula:

[tex]\vec F_{B} = q\cdot \vec v \,\times \,\vec B[/tex] (1)

Where:

[tex]q[/tex] - Electric charge, measured in coulombs.

[tex]\vec v[/tex] - Velocity of the particle, measured in meters per second.

[tex]\vec B[/tex] - Magnetic field, measured in teslas.

If we know that [tex]q = -1\,C[/tex], [tex]\vec v = 5\,\hat{k}\,\,\,\left[\frac{m}{s} \right][/tex] and [tex]\vec B = 3\,\hat{i}-4\,\hat{j}\,\,\,[T][/tex], then the magnetic force on the particle is:

[tex]\vec F_{B} = \left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\0\,\frac{C\cdot m}{s}&0\,\frac{C\cdot m}{s} &(-1\,C)\cdot (5\,\frac{m}{s} ) \\3\,T&-4\,T&0\,T\end{array}\right|[/tex]

[tex]\vec F_{B} = -(-4\,T)\cdot (-1\,C)\cdot \left(5\,\frac{m}{s} \right)\,\hat{i}+(-1\,C)\cdot\left(5\,\frac{m}{s} \right)\cdot (3\,T)\,\hat{j}[/tex]

[tex]\vec F_{B} = -20\,\hat{i}-15\,\hat{j}\,\,\,[N][/tex]

Which corresponds to option D.

Answer:

[tex]q_1=5.64\times 10^{-7}\ \text{C}[/tex] and [tex]q_2=2.32\times 10^{-6}\ \text{C}[/tex]

Explanation:

[tex]F_1=0.072\ \text{N}[/tex]

[tex]F_2=0.115\ \text{N}[/tex]

r = Distance between shells = 40.4 cm

[tex]q_1[/tex] and [tex]q_2[/tex] are the charges

[tex]k[/tex] = Coulomb constant = [tex]8.99\times10^{9}\ \text{Nm}^2/\text{C}^2[/tex]

Force is given by

[tex]F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow q_1q_2=\dfrac{F_1r^2}{k}\\\Rightarrow q_1q_2=\dfrac{0.072\times 0.404^2}{8.99\times 10^{9}}\\\Rightarrow q_1q_2=1.307\times 10^{-12}\\\Rightarrow q_1=\dfrac{1.307\times 10^{-12}}{q_2}[/tex]

[tex]F_2=\dfrac{kq^2}{r^2}\\\Rightarrow q=\sqrt{\dfrac{F_2r^2}{k}}\\\Rightarrow q=\sqrt{\dfrac{0.115\times 0.404^2}{8.99\times 10^{9}}}\\\Rightarrow q=1.44\times 10^{-6}\ \text{C}[/tex]

[tex]q=\dfrac{q_1+q_2}{2}\\\Rightarrow q_1+q_2=2q\\\Rightarrow q_1+q_2=2\times1.44\times 10^{-6}\\\Rightarrow q_1+q_2=2.88\times 10^{-6}[/tex]

Substituting the above value of [tex]q_1[/tex] we get

[tex]\dfrac{1.307\times 10^{-12}}{q_2}+q_2=2.88\times 10^{-6}\\\Rightarrow q_2^2-2.88\times 10^{-6}q_2+1.307\times 10^{-12}=0\\\Rightarrow \frac{-\left(-0.00000288\right)\pm \sqrt{\left(-0.00000288\right)^2-4\times \:1\times \:1.307\times 10^{-12}}}{2\times \:1}\\\Rightarrow q_2=2.32\times 10^{-6}, 5.64\times 10^{-7}[/tex]

[tex]q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{2.32\times 10^{-6}}\\\Rightarrow q_1=5.63\times 10^{-7}[/tex]

[tex]q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{5.64\times 10^{-7}}\\\Rightarrow q_1=2.32\times 10^{-6}[/tex]

Since we know [tex]q_1<q_2[/tex]

[tex]q_1=5.64\times 10^{-7}\ \text{C}[/tex] and [tex]q_2=2.32\times 10^{-6}\ \text{C}[/tex].

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the formula to be used here is

ω = 2π/T

Where ω is the angular frequency (in rad/s)

T is the period - the time taken for Block A to complete one oscillation and return to it's original position.

To solve for this period T, the formula below should be used

T = 2π√m/k

where m is the mass of the object (Block A) and k is the spring constant (281 J/m²)

Answer:

x = 1127 [m]

Explanation:

In order to solve this problem, we must use the equations of kinematics. With the first equation, we must find the acceleration and with the second equation we must find the distance.

[tex]v_{f} =v_{o} -a*t[/tex]

where:

Vf = final velocity = 9 [m/s]

Vo = initial velocity = 55 [m/s]

a = acceleration o desacceleration [m/s²]

t = time = 49 [s]

Now replacing:

9 = 55 - a*49

a*49 = 55 + 9

a = 1.306 [m/s²]

Note: The negative sign in the above equation means that the speed decreases.

Now using the second equation.

[tex]v_{f}^{2} =v_{o}^{2} -2*a*x[/tex]

(9)² = (55)² - 2*(1.306)*x

2944 = 2.612*x

x = 1127 [m]

Answer:

Because mechanical energy (the sum of potential and kinetic energy) is conserved, as the kinetic energy increases, the potential energy decreases. The maximum kinetic energy is achieved when the pendulum passes through the lowest point, and the maximum potential energy is achieved at the highest point.

Answer:

it makes me wish I was a cartoon

Answer:

goofy and stupid and act like a kid

Explanation:

Answer:

a

[tex]\lambda_{long} = 288.5 \ nm[/tex]

b

The velocity is  [tex]v = 3.7 *0^{5} \ m/s[/tex]

Explanation:

From the question we are told that

   The work function of Zinc is  [tex]W = 4.3 eV[/tex]

Generally the work function can be mathematically represented as

     [tex]E_o = \frac{hc}{\lambda_{long}}[/tex]

=>   [tex]\lambda_{long} = \frac{hc}{E_o}[/tex]

Here  h is the Planck constant with the value  [tex]h = 4.1357 * 10^{-15} eV s[/tex]

  and c is the speed of light with value  [tex]c = 3.0 *10^{8} \ m/s[/tex]

So

     [tex]\lambda_{long} = \frac{4.1357 * 10^{-15} * 3.0 *10^{8}}{4.3}[/tex]

=>  [tex]\lambda_{long} = 2.885 *10^{-7} \ m[/tex]

=>  [tex]\lambda_{long} = 288.5 \ nm[/tex]

Generally the kinetic energy of the emitted electron is mathematically represented as

      [tex]K = E -E_o[/tex]

Here  E is the energy of the photon that strikes the surface

So

    [tex]E- E_o = \frac{1}{2} m * v^2[/tex]

Here m is the mass of electron with value  [tex]m = 9.11*10^{-31 } \ kg[/tex]

Generally  [tex]1 ev = 1.60 *10^{-19} \ J[/tex]

=>   [tex]v = \sqrt{ \frac{2 (E - E_o ) }{ m } }[/tex]

=>    [tex]v = \sqrt{ \frac{2 (4.7 - 4.3 )* 1.60 *10^{-19} }{ 9.11 *10^{-31} } }[/tex]

=>    [tex]v = 3.7 *0^{5} \ m/s[/tex]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s  

Explanation:

1) We can find the speed of the object by conservation of energy:

[tex] E_{i} = E_{f} [/tex]

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

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]                              

[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s [/tex]

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

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

[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]      

[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s [/tex]      

3) When the object is at the equilibrium position, the speed of the object is:

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

[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]      

[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s [/tex]

I hope it helps you!                                                                                        

(1) the speed of the object when compression of the spring is 8 cm is 0.449 m/s

(2) the speed of the object when compression of the spring is 5 cm is 0.648 m/s

(3) the speed of the object when the spring is at equilibrium is 0.748 m/s

Given that the mass of the object, m = 5 kg

spring constant, k = 280 N/m

compression of the spring , x = 10 cm = 0.1m

(i) the spring compression is at d = 8cm

according to the conservation of energy:

[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]

where v is the speed at the given compression of the spring.

[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.08)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.2016[/tex]

v = 0.449 m/s

(ii) the spring compression is at d = 5cm

according to the conservation of energy:

[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]

where v is the speed at the given compression of the spring.

[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.05)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.42[/tex]

v = 0.648 m/s

(iii) the spring is at equilibrium so compression is at d = 0cm

according to the conservation of energy:

[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]

where v is the speed at the given compression of the spring.

[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.56[/tex]

v = 0.748 m/s

Learn more about conservation of energy:

https://brainly.com/question/14245799?referrer=searchResults

Answer:

A. The current in the wire increases.

Explanation:

Increasing the speed at which a magnet moves in and out of a wire coil increases the current in the wire.

This phenomenon shows the inter-relationship between electricity and magnetic fields.

Answer: Natural science can be divided into two main branches

Explanation:

life science and physical science. Life science is alternatively known as biology, and physical science is subdivided into branches: physics, chemistry, astronomy and Earth science.

Answer:

349 Hz

Explanation:

We are told that the guitar string produces 3 beats/s when sounded with a 352 Hz tuning fork and 8 beat/s when sounded with a 357 Hz tuning fork.

This means that for the 352 Hz tuning fork, the vibrational frequency is;

f = 352 ± 3

f = (352 + 3) or (352 - 3)

f = 355 Hz or 349Hz

For the 357 Hz tuning fork, the vibrational frequency is;

f = 357 ± 8

f = (357 + 8) or (357 - 8)

f = 365 Hz or 349 Hz

In both cases, 349 Hz is common;

Thus, the vibrational frequency of the string = 349 Hz

Answer:

1.7333333m/s²

Explanation:

Tension of the line = the weight + force from pulling up the fish

30N = mg + ma

30 = (6)(9.8) + (6)a

10.4 = 6a

∴ a = 1.7333333m/s²

You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 30 N, the maximum acceleration is  1.7333333 m/s².

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration.

Tension of the line = the weight + force from pulling up the fish

30 N = mg + ma

30 = (6)(9.8) + (6)a

10.4 = 6 a

a = 1.7333333 m/s²

You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 30 N, the maximum acceleration is  1.7333333 m/s².

To learn more about acceleration refer to the link:

brainly.com/question/12550364

#SPJ2

Answer:

The number is  [tex]N = 300[/tex]

Explanation:

From the question we are told that

   The  net charge is  [tex]Q = -4.8 *10^{-17 } \ C[/tex]

Generally the charge on a electron is [tex]e = - 1.60 *10^{-19 } \ C[/tex]

Generally the number of excess electrons is mathematically represented as

      [tex]N = \frac{Q}{e}[/tex]

=>  [tex]N = \frac{-4.8 *10^{-17}}{-1.60 *10^{-19}}[/tex]

=>  [tex]N = 300[/tex]

Answer:

Explanation:

Step one:

given data

mass = 7kg

acceleration =2m/s^2

time= 9seconds

acceleration = velocity/time

velocity= acceleration *time

velocity=2*9

velocity= 18m/s

distance moved= velocity* time

distance= 18*9

distance=162m

we also know that the force on impulse is given as

Ft=mv

F=mv/t

F=7*18/9

F=126/9

F=14N

work done = Force* distance

work done=14*162

work=2268Joules

work= 2.27kJ

Answer:

C. a liter of salt water.

Explanation:

Defination of Solution =>

a liquid mixture in which the minor component (the solute) is uniformly distributed within the major component (the solvent).

Answer:

The angular speed is 0.13 rev/s

Explanation:

From the formula

[tex]\tau = I\alpha[/tex]

Where [tex]\tau[/tex] is the torque

[tex]I[/tex] is the moment of inertia

[tex]\alpha[/tex] is the angular acceleration

But, the angular acceleration is given by

[tex]\alpha = \frac{\omega}{t}[/tex]

Where [tex]\omega[/tex] is the angular speed

and [tex]t[/tex] is time

Then, we can write that

[tex]\tau = \frac{I\omega}{t}[/tex]

Hence,

[tex]\omega = \frac{\tau t}{I}[/tex]

Now, to determine the angular speed, we first determine the Torque [tex]\tau[/tex] and the moment of inertia [tex]I[/tex].

Here, The torque is given by,

[tex]\tau = rF[/tex]

Where r is the radius

and F is the force

From the question

r = 3.00 m

F = 195 N

∴ [tex]\tau = 3.00 \times 195[/tex]

[tex]\tau = 585[/tex] Nm

For the moment of inertia,

The moment of inertia of the solid disk is given by

[tex]I = \frac{1}{2}MR^{2}[/tex]

Where M is the mass and

R is the radius

∴[tex]I = \frac{1}{2} \times 325 \times (3.00)^{2}[/tex]

[tex]I = 1462.5[/tex] kgm²

From the question, time t = 2.05 s.

Putting the values into the equation,

[tex]\omega = \frac{\tau t}{I}[/tex]

[tex]\omega = \frac{585 \times 2.05}{1462.5}[/tex]

[tex]\omega = 0.82[/tex] rad/s

Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π

0.82 rad/s = 0.82/2π rev/s

= 0.13 rev/s

Hence, the angular speed is 0.13 rev/s,

Answer:

This question is incomplete.

Explanation:

This question is incomplete. However, it should be noted that the voltage, V, across resistors in parallel is the same (although there currents are not the same). Thus, if a voltage has been provided, it remains the same but if not provided, you can solve for it using the formulas below

V = IR

where V is the voltage. I is the current and R is the resistance

R in parallel can be calculated as R = 1/R₁ + 1/R₂ + 1/R₃ + ......

Answer:A

Explanation:

Answer:

A. -2.83 x 10-7N

Explanation:

Answer:

1. 6.15m/s

2. 820N

Explanation:

The total upward force

= 410x2

= 820

g = 9.81

a = v²/r

= 2xT - msg = m x v²/r

= 820-37*9.81 = 37v²/3.06

= 820-362.97 = 37v²/3.06

= 457.03 = 12.09v²

To get v²

V² = 457.03/12.09

V² = 37.8

V = √37.8

V = 6.15m/s

B. We already have the answer to this question

The force exerted is simply gotten by this calculation

2x410

= 820N