Yea, gonna need some help. Thanks

[tex]F = 5.93×10^{13}\:\text{N}[/tex]

Explanation:

Given:

[tex]m_1= 2×10^{16}\:\text{kg}[/tex]

[tex]m_2= 4×10^{22}\:\text{kg}[/tex]

[tex]r = 30000\:\text{km} = 3×10^7\:\text{m}[/tex]

Using Newton's universal law of gravitation, we can write

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

[tex]\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}[/tex]

[tex]\:\:\:\:= 5.93×10^{13}\:\text{N}[/tex]

Answer:

4

Explanation:

Answer:

d = 6.32 m

Explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2[/tex]

Let d is the distance moved in 2.25 s. Using second equation of motion,

[tex]d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m[/tex]

So, it will move 6.32 m from rest in 2.25 seconds.

Answer:

350

Explanation:

Since it travels 350 meters per second, the jet will travel 350 meters in one second.

Answer:

pendulum, quartz

Explanation:

Answer:

effect on motation.effect on direction

Explanation:

Lasers produce a narrow beam of light in which all of the light waves have very similar wavelengths. The laser's light waves travel together with their peaks all lined up, or in phase. This is why laser beams are very narrow, very bright, and can be focused into a very tiny spot.

Answer:

[tex]u=13.3cm[/tex]

Explanation:

From the question we are told that:

Focal Length [tex]F=10.0cm[/tex]

Distance [tex]d=40cm[/tex]

Generally the equation for Focal length  is mathematically given by

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]

[tex]\frac{1}{10}=\frac{1}{u}+\frac{1}{40}[/tex]

[tex]\frac{1}{u}=\frac{3}{40}[/tex]

[tex]u=13.3cm[/tex]

Focal length is the distance from the center of the lens to principle foci.  The distance of the from the lens to the light bulb is 13.3 cm.

The distance can be determined by the formula,

[tex]\bold {\dfrac 1{f} = \dfrac 1{u} + \dfrac 1{v} }[/tex]

Where,

f - focal length = 10 cm

u - distance of object = ?

v = distance of image = 40 cm

Put the values in the equation,

[tex]\bold {\dfrac 1{10} = \dfrac 1{u} + \dfrac 1{40} }\\\\\bold {\dfrac 1{u} = \dfrac 3{40}}\\\\\bold {\dfrac 1{u} = 13.3 cm}[/tex]

Therefore, the distance of the from the lens to the light bulb is 13.3 cm.

To know more about the focal length,

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Answer:

15.88m/s

Explanation:

At the top of the roller coaster you will have three forces acting on the roller-coaster. See the image below. Fc is the centripetal force (for an object in circular motion), Fg is the gravitational force, and Fn is the normal force. To achieve the minimum speed we assume the roller-coaster is barely touching the vertical loop and so the normal force is zero. This leaves two acting forces.

[tex]F_g = F_c\\mg = \frac{m\times v^2}{r}\\v = \sqrt{gr} = \sqrt{9.81 \times 25.7} = 15.88 m/s[/tex]

Solution :

Let the positive [tex]x-axis[/tex] is along the East and the positive [tex]y[/tex] direction is along the north.

Given :

Mass of the Tesla car, [tex]m_1[/tex] = [tex]750 \ kg[/tex]

Mass of the Ford car, [tex]m_2 = 1250 \ kg[/tex]

Now let the initial velocity of Tesla car in the south direction be = [tex]-v_1j[/tex]

The initial momentum of Tesla car, [tex]p_1 = -750 \ v_1[/tex]

Let the initial velocity of Ford car in the east direction be = [tex]v_2 \ i[/tex]

So the initial momentum of the Ford car is [tex]p_2=1250\ v_2 \ i[/tex]

Therefore, the initial velocity of both the cars is [tex]p_i = p_1+p_2[/tex]

                                                                  [tex]=1250 \ v_2 \ i - 750\ v_1 \ j[/tex]

Now the final velocity of both the cars is [tex]v = 18 \ m/s[/tex]

So the vector form is :

[tex]v = 18\cos 32\ i-18 \sin 32 \ j[/tex]

  [tex]= 15.26 \ i - 9.54 \ j[/tex]

Therefore the momentum after the accident is

[tex]p_f=(m_1+m_2) \times v[/tex]

    [tex]=(750+1250) \times (15.26 \ i - 9.54 \ j)[/tex]

    [tex]= 30520\ i -19080\ j[/tex]

According to the law of conservation of momentum, we know

[tex]p_i = p_f[/tex]

[tex]1250 \ v_2 \ i - 750\ v_1 \ j[/tex]  [tex]= 30520\ i -19080\ j[/tex]

[tex]1250 \ v_2 = 30520[/tex]

[tex]v_2=24.4 \ m/s[/tex]

From, [tex]750\ v_1 = 19080[/tex]

We get, [tex]v_1=25.4 \ m/s[/tex]

Therefore the speed of Tesla car before collision = 25.4 m/s

The speed of ford car before collision = 24.4 m/s

Answer:

Magnetic Forces on Moving Charges. The magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field with direction given by the right hand rule.

The direction of the charge where it does not experience magnetic force is left and right. The correct option is C and D.

A magnetic field is a region in space where a magnetic force can be observed. It is created by moving electric charges, such as electrons, and is characterized by the direction and strength of the force it exerts on other magnetic materials or moving charges.

Magnetic field lines are used to visualize the direction and strength of the magnetic field. They represent the path that a small magnetic north pole would follow if placed in the magnetic field. The direction of the magnetic field is given by the direction in which the north pole of a compass needle would point if placed in the field.

Magnetic flux is the measure of the strength of the magnetic field passing through a surface. It is given by the product of the magnetic field strength and the area of the surface, as well as the cosine of the angle between the magnetic field and the surface normal. The unit of magnetic flux is Weber (Wb).

Magnetic flux is important in many applications, such as electric motors and generators, where the interaction between the magnetic field and moving charges produces electrical energy. It is also used in magnetic imaging techniques, such as MRI, to visualize the internal structures of the human body.

Here in the question,

Options A (up), B (down), E (into the screen), and F (out of the screen) are all perpendicular to the direction of the magnetic field and so will experience a magnetic force.

Therefore, options C and D are correct i. e left and right.as they are parallel and anti-parallel to the direction of the magnetic field, respectively.

To learn about Ohm's law click:

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Explanation:

The density of earth [tex]\rho_E[/tex] is given by

[tex]\rho_E = \dfrac{M_E}{\left(\frac{4\pi}{3}R_E^3\right)}[/tex]

and in terms of this density, we can write the acceleration due to gravity on earth as

[tex]g_E =G\dfrac{M_E}{R_E^2} = \dfrac{4\pi G}{3}\rho_ER_E[/tex]

Similarly, the acceleration due to gravity [tex]g_P[/tex] on this new planet is given by

[tex]g_P = G\dfrac{M_P}{R_P^2} = G\dfrac{\frac{4\pi}{3}R_p^3\rho_P}{R_P^2}[/tex]

[tex]\:\:\:\:\:= \dfrac{4\pi G}{3}\rho_PR_P[/tex]

We know that this planet has the same density as earth and has a radius 2 times as large. We can then rewrite [tex]g_P[/tex] as

[tex]g_P = \dfrac{4\pi G}{3}\rho_E(2R_E)[/tex]

[tex]\:\:\:\:\:= 2\left(\dfrac{4\pi G}{3}\rho_ER_E\right) = 2g_E[/tex]

[tex]\:\:\:\:\:= 2(9.8\:\text{m/s}^2) = 19.6\:\text{m/s}^2[/tex]

Answer:

Resistance of the armature coil = 6.61 ohms

Back emf developed at normal speed = 98.62 V (Approx.)

Current drawn by the motor at one-third normal speed = 12.73 A

Explanation:

Given:

Potential difference V = 117 V

Current = 17.7 A

Motor drawn current = 2.78 A

Find:

Resistance of the armature coil

Back emf developed at normal speed

Current drawn by the motor at one-third normal speed

Computation:

A] Resistance of the armature coil R = V/ I

Resistance of the armature coil = 117 / 17.7

Resistance of the armature coil = 6.61 ohms

B] Back emf developed at normal speed  = V- IR

Back emf developed at normal speed = 117 V - (2.78 A)(6.61 ohms)

Back emf developed at normal speed = 117 V - 18.37

Back emf developed at normal speed = 98.62 V (Approx.)

C] Current drawn by the motor at one-third normal speed = 17.7 A - (98.62/3)/(6.61 ohms)

Current drawn by the motor at one-third normal speed = 17.7 - 4.97

Current drawn by the motor at one-third normal speed = 12.73 A

The question is incomplete. The complete question is :

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 .

Find the energy U1 of the dielectric-filled capacitor. I got U1=2.99*10^-10 J which I know is correct. Now I need these:

1. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.

Solution :

Given :

[tex]A = 10 \ cm^2[/tex]

   [tex]$=0.0010 \ m^2$[/tex]

d = 10 mm

  = 0.010 m

Then, Capacitance,

[tex]$C=\frac{k \epsilon_0 A}{d}$[/tex]

[tex]$C=\frac{8.85 \times 10^{12} \times 3 \times 0.0010}{0.010}$[/tex]

[tex]$C=2.655 \times 10^{12} \ F$[/tex]

[tex]$U_1 = \frac{1}{2}CV^2$[/tex]

[tex]$U_1 = \frac{1}{2} \times 2.655 \times 10^{-12} \times (15V)^2$[/tex]

[tex]$U_1=2.987 \times 10^{-10}\ J$[/tex]

Now,

[tex]$C_k=\frac{1}{2} \frac{k \epsilon_0}{d} \times \frac{A}{2}$[/tex]

And

[tex]$C_{air}=\frac{1}{2} \frac{\epsilon_0}{d} \times \frac{A}{2}$[/tex]

In parallel combination,

[tex]$C_{eq}= C_k + C_{air}$[/tex]

[tex]$C_{eq} = \frac{1}{2} \frac{\epsilon_0 A}{d}(1+k)$[/tex]

[tex]$C_{eq} = \frac{1}{2} \times \frac{8.85 \times 10^{-12} \times 0.0010}{0.01} \times (1+3)$[/tex]

[tex]$C_{eq} = 1.77 \times 10^{-12}\ F$[/tex]

Then energy,

[tex]$U_2 =\frac{1}{2} C_{eq} V^2$[/tex]

[tex]$U_2=\frac{1}{2} \times 1.77 \times 10^{-12} \times (15V)^2$[/tex]

[tex]$U_2=1.99 \times 10^{-10} \ J$[/tex]

b). Now the charge on the [tex]\text{capacitor}[/tex] is :

[tex]$Q=C_{eq} V$[/tex]

[tex]$Q = 1.77 \times 10^{-12} \times 15 V$[/tex]

[tex]$Q = 26.55 \times 10^{-12} \ C$[/tex]

Now when the capacitor gets disconnected from battery and the [tex]\text{dielectric}[/tex] is slowly [tex]\text{removed the rest}[/tex] of the way out of the [tex]\text{capacitor}[/tex] is :

[tex]$C_3=\frac{A \epsilon_0}{d}$[/tex]

[tex]$C_3 = \frac{0.0010 \times 8.85 \times 10^{-12}}{0.01}$[/tex]

[tex]$C_3=0.885 \times 10^{-12} \ F$[/tex]

[tex]$C_3 = 0.885 \times 10^{-12} \ F$[/tex]

Without the dielectric,

[tex]$U_3=\frac{1}{2} \frac{Q^2}{C}$[/tex]

[tex]$U_3=\frac{1}{2} \times \frac{(25.55 \times 10^{-12})^2}{0.885 \times 10^{-12}}$[/tex]

[tex]$U_3=3.98 \times 10^{-10} \ J$[/tex]

Answer:

a) Directamente proporcional

Explanation:

El peso se puede definir como la fuerza que actúa sobre un cuerpo o un objeto como resultado de la gravedad.

Matemáticamente, el peso de un objeto viene dado por la fórmula;

[tex] Peso = mg [/tex]

Donde;

m es la masa del objeto.

g es la aceleración debida a la gravedad.

De la expresión matemática, podemos deducir que el valor del peso de un objeto es directamente proporcional a la masa del objeto.

Por lo tanto, un aumento en la masa de un objeto provocaría un aumento en el peso del objeto y viceversa.

Complete question:

A bullet 2 cm long is fired at 420m/s and passes straight through a 10.0 cm thick board, exiting at 280m/s? What is the average acceleration of the bullet through the board?

Answer:

The average acceleration of the bullet through the board is -4.083 x 10⁵ m/s²

Explanation:

Given;

initial velocity of the bullet, u = 420 m/s

final velocity of the bullet, v = 280 m/s

length of the bullet, d₁ = 2 cm

thickness of the board, d₂ = 10 cm

total distance penetrated by the bullet through the board;

d = d₁ + d₂ = 2 cm + 10 cm = 12 cm = 0.12 m

The average acceleration of the bullet through the board is calculated as;

[tex]v^2 = u^2 + 2ad\\\\2ad = v^2 - u^2\\\\a = \frac{v^2 - u^2}{2d} \\\\a = \frac{(280^2) - (420^2)}{2(0.12)} = -4.083 \times 10^{5} \ m/s^2[/tex]

Therefore, the average acceleration of the bullet through the board is -4.083 x 10⁵ m/s²

Answer:

true

Explanation:

The kinetic energy of a gas is directly proportional to the temperature of the gas.because temperature is the average kinetic energy of a substance

I hope this helps

Answer:

The correct answer is - 8.99N/C

Explanation:

[tex]dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k[/tex]

Answer:

the electric field and the electric potential increase 5 times

Explanation:

The electric field created by a point charge is

         E = k q / r²

in this case the charge changes from q₁ = 1 10⁺⁰ C to q₂ = 5 10⁻⁹ C

with the electric field is proportional to the charge

         E₅ = 5 E₁

the electrical power for a point charge is

         V = k q / r

as the electric power is proportional to the charge

         V₅ = 5 V₁

consequently both the electric field and the electric potential increase 5 times

An electric field is produced by a charged object:

[tex]\to E = \frac{k q}{r^2}\\\\[/tex]

In this situation, the charge shifts:

[tex]\to q_1 = 1 10^{\circ}\ C \\\\ \to q_2 = 5 \times 10^{-9}\ C[/tex]

so with electric field remaining proportional to the charge

[tex]\to E_5 = 5 E_1[/tex]

The electrical power consumed for a point charge:

 [tex]\to V = \frac{k q}{ r}[/tex]

since the electric power is related to the charge

[tex]\to V_5 = 5 V_1[/tex]

As a result, both the electric field as well as the electric potential increase by 5 times.

Learn more:

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Answer:

0A.Output force divided by input force.

Answer:

T = 1/f = 1/50(s)

ω = 2πf = 100π (rad/s)

(vote 5 sao nhó :3 )

Answer:

1.37 mm²

Explanation:

From the image attached below:

Let's take a look at the two rays r and r' hitting the same mirror from two different positions.

Let x be the distance between these rays.

[tex]d_o =[/tex] distance between object as well as the mirror

[tex]d_{eye}[/tex] = distance between mirror as well as the eye

Thus, the formula for determining the distance between these rays can be expressed as:

[tex]x = 2d_o tan \theta[/tex]

where; the distance between the eye of the observer and the image is:

[tex]s = d_o + d_{eye}[/tex]

Then, the tangent of the angle θ is:

[tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex]

replacing [tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex] into [tex]x = 2d_o tan \theta[/tex], we have:

[tex]x = 2d_o \Big( \dfrac{R}{d_o+d_{eye}}\Big)[/tex]

[tex]x = 2(10) \Big( \dfrac{0.25}{10+28}\Big)[/tex]

[tex]x = 20\Big( \dfrac{0.25}{38}\Big) cm[/tex]

x = (0.13157 × 10) mm

x = 1.32 mm

Finally, the area A = π r²

[tex]A = \pi(\frac{x}{2})^2[/tex]

[tex]A = \pi(\frac{1.32}{2})^2[/tex]

A = 1.37 mm²

Answer:

a)  fem = - 2.1514 10⁻⁴ V,  b) I = - 64.0 10⁻³ A, c)    P = 1.38  10⁻⁶ W

Explanation:

This exercise is about Faraday's law

         fem = [tex]- \frac{ d \Phi_B}{dt}[/tex]

where the magnetic flux is

        Ф = B x A

the bold are vectors

        A = π r²

we assume that the angle between the magnetic field and the normal to the area is zero

         fem = - B π 2r dr/dt = - 2π B r v

linear and angular velocity are related

        v = w r

        w = 2π f

        v = 2π f r

we substitute

        fem = - 2π B r (2π f r)

        fem = -4π² B f r²

For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T

we reduce the magnitudes to the SI system

       f = 2 rev / s (2π rad / 1 rev) = 4π Hz

we calculate

       fem = - 4π² 428 10⁻⁶ 4π 0.10²

       fem = - 16π³ 428 10⁻⁶ 0.010

       fem = - 2.1514 10⁻⁴ V

for the current let's use Ohm's law

        V = I R

        I = V / R

         I = -2.1514 10⁻⁴ / 0.00336

         I = - 64.0 10⁻³ A

Electric power is

        P = V I

        P = 2.1514 10⁻⁴ 64.0 10⁻³

        P = 1.38  10⁻⁶ W

Answer:

gshshs

Explanation:

hshsksksksbsbbshd

Answer:

About 37.70 radians.

Explanation:

1 revolution = 2[tex]\pi[/tex] radians

∴ 6 revolutions = (6)(2[tex]\pi[/tex] radians)

6 revolutions = 37.6991 or ≈ 37.70 radians

Answer

Delta Q = change in thermal energy = c M * change in temperature

change in temperature = Q / (c * M)

change in temperature = -12 J  / (390 J / Kg*deg * .012 kg

change in temp = -12 / (390 * .012) =  - 2.56 deg C

Answer:

Explanation:

Equilibrium is a state in which the algebraic sum of all forces acting on an object is zero. Thus the object has no force acting on it. The types are: stable, unstable and neutral equilibrium. While a torque is a turning force which are equal but acts in an opposite direction. When applied to on object, it constitute a turning effect. Example is the force applied on a tap, handle wheel of a car etc.

In the given question, the condition stated shows that the stick would experience a torque, thus not in equilibrium. Since the forces at its ends are in opposite directions, then it continues to rotate about its axis.

Answer:

The correct solution is:

(a) 1.66

(b) 1.05

Explanation:

Given:

Bending stress,

[tex]\sigma_b = 25 \ kpsi[/tex]

Torsional stress,

[tex]\tau= 15 \ kpsi[/tex]

Yield stress of steel bar,

[tex]\delta_y = 60 \ kpsi[/tex]

As we know,

⇒ [tex]\sigma_{max}^' \ = \sqrt{\sigma_b^2 + 3 \gamma^2}[/tex]

        [tex]= \sqrt{(25)^2+3(15)^2}[/tex]

        [tex]=36.055 \ kpsi[/tex]

(a)

The factor of safety against static failure will be:

⇒ [tex]\eta_y = \frac{\delta_y}{\sigma_{max}^'}[/tex]

By putting the values, we get

        [tex]=\frac{60}{36.055}[/tex]

        [tex]=1.66[/tex]

(b)

According to the Goodman line failure,

[tex]\sigma_a = \sigma_b = 25 \ kpsi[/tex]

[tex]S_e = 40 \ kpsi[/tex]

[tex]\sigma_m = \sqrt{3} \tau[/tex]

     [tex]=\sqrt{3}\times 15[/tex]

     [tex]=26 \ kpsi[/tex]

[tex]Sut = 80 \ kpsi[/tex]

⇒ [tex]\frac{\sigma_a}{S_e} +\frac{\sigma_m}{Sut} =\frac{1}{\eta_y}[/tex]

      [tex]\frac{25}{40}+\frac{26}{80}=\frac{1}{\eta_y}[/tex]

              [tex]\eta_y = 1.05[/tex]

Answer:

Explanation:

The best way to do this is to remember the rule about the halfway mark in a parabolic path. At a trajectory's half way point in its travels, it will be at its max height. To get the total time in the air, we take that time at half way and double it. Here's what we know that we are told:

initial velocity is 20 m/s

Here's what we know that we are NOT told:

a = -9.8 m/s/s and

final velocity is 0 at an object's max height in parabolic motion.

We will use the equation:

[tex]v=v_0+at[/tex] where v is final velocity and v0 is initial velocity. Filling in:

0 = 20 + (-9.8)t and

-20 = -9.8t so

t = 2 seconds. The stone reaches its max height 2 seconds after it is thrown; that means that after another 2 seconds it will be on the ground. Total air time is 4 seconds.