If the moon started it's orbit around the Earth from a spot in line with a certain star, it will return to that same spot in about _______.​

Answer:

x ’= 368.61 m,  y ’= 258.11 m

Explanation:

To solve this problem we must find the projections of the point on the new vectors of the rotated system  θ = 35º

            x’= R cos 35

            y’= R sin 35

The modulus vector can be found using the Pythagorean theorem

            R² = x² + y²

            R = 450 m

we calculate

            x ’= 450 cos 35

            x ’= 368.61 m

            y ’= 450 sin 35

            y ’= 258.11 m

Answer:

E = 3495.96 J

Explanation:

From the law of conservation of energy:

Total Mechanical Energy at h1 = Total Mechanical Energy at h2

Kinetic energy at h1 + potential energy at h1 = Kinetic energy at h2 + potential energy at h2 + Mechanical Energy Lost due to Friction

[tex]K.E_{h1}+P.E_{h1} = K.E_{h2}+P.E_{h2} + E\\\\\frac{1}{2}mv_1^2\ J + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2 + E\\\\\frac{1}{2}(36\ kg)(0\ m/s)_1^2\ J + (36\ kg)(9.81\ m/s^2)(24\ m)_1 = \frac{1}{2}(36\ kg)(7.8\ m/s)_2^2 + (36\ kg)(9.81\ m/s^2)(11\ m)_2 + E\\\\0\ J + 8475.84\ J = 1095.12\ J + 3884.76\ J + E\\E = 8475.84\ J - 1095.12\ J - 3884.76\ J\\[/tex]

E = 3495.96 J

Answer:

Look at explanation

Explanation:

a)Only force acting on the object is gravity, so a=-g (consider up to be positive)

use: v^2=v0^2+2a(y-y0)

plug in givens, at max height v=0

0=400-19.6(H)

Solve for H

H= 20.41m

b) Use: y=y0+v0t+1/2at^2

Plug in givens

0=0+20t-4.9t^2

solve for t

t=4.08 seconds

c) v=v0+at

v=20-39.984= -19.984m/s

Answer:

Converging (convex) lens.

Explanation:

A lens can be defined as a transparent optical instrument that refracts rays of light to produce a real image.

Basically, there are two (2) main types of lens and these includes;

I. Diverging (concave) lens.

II. Converging (convex) lens.

A converging (convex) lens refers to a type of lens that typically causes parallel rays of light with respect to its principal axis to come to a focus (converge) and form a real image. Thus, this type of lens is usually thin at the lower and upper edges and thick across the middle.

Basically, the image of the object formed by a converging (convex) lens. lens is real, enlarged and inverted.

Answer:

It is all a thermodynamic system that is highly related to each other.

Explanation:

Because they are in the physics of thermodynamics it is not wrong to say they follow the same thermodynamic rules and has highly the same properties of energy.

Answer:

Wheel and axle

Explanation:

Which simple machine is shown in the diagram?

a wheel and axle

From the given diagram, the machine shown is actually a wheel and axle

Description of wheel and axle

The wheel and axle is a machine consisting of a wheel attached to a smaller axle so that these two parts rotate together in which a force is transferred from one to the other.

Answer:

Wheel and axle

Explanation:

Answer:

A) q = 1.67 × 10^(-7) C

B) q = 1.67 × 10^(-10) C

Explanation:

We are given;

Potential; V = 30 KV = 30000 V

Radius of sphere; r = diameter/2 = 0.1/2 = 0.05 m

A) To find the charge of the sphere, we will use the formula;

V = kq/r

Where;

q is the charge

k is electric force constant = 9 × 10^(9) N.m²/C²

Thus;

q = Vr/k

q = (30000 × 0.05)/(9 × 10^(9))

q = 1.67 × 10^(-7) C

B) Now, potential energy here is a formula; U = qV

However, for the drop of paint to move, the potential energy will be equal to the kinetic energy. Thus;

qV = ½mv²

q = mv²/2V

Where;

v is speed = 10 m/s

V = 30000 V

m = mass = 0.100 mg = 0.1 × 10^(-6) Thus;

q = (0.1 × 10^(-6) × 10²)/(2 × 30000)

q = 1.67 × 10^(-10) C

Hi there!

[tex]\large\boxed{\approx 35.29 C}[/tex]

Use the following formula:

E = F / C, where:

E = electric field (N/C)

F = force (N)

C = Charge (C)

Thus:

6.8 × 10⁻⁵ = 2.4 × 10⁻³ / C

Isolate for C:

C = 2.4 × 10⁻³  / 6.8 × 10⁻⁵

Solve:

≈ 35.29 C

Explanation:

[tex]\underline{\boxed{\large{\bf{Option \; A!! }}}} [/tex]

Here,

We have to find the equivalent resistance of the circuit.

Here, [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] are connected in series, so their combined resistance will be given by,

[tex]\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] is connected in parallel combination with [tex]\rm { R_3} [/tex], so their combined resistance will be given by,

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ [/tex]

Reciprocating both sides,

[tex]\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex], [tex]\rm { R_2} [/tex] and [tex]\rm { R_3} [/tex] is connected in series combination with [tex]\rm { R_4} [/tex]. So, equivalent resistance will be given by,

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ [/tex]

Henceforth, Option A is correct.

[tex]\underline{\boxed{\large{\bf{Option \; B!! }}}} [/tex]

Here, we have to find the amount of flow of current in the circuit. By using ohm's law,

[tex] \longrightarrow [/tex] V = IR

[tex] \longrightarrow [/tex] 3 = I × 3.33

[tex] \longrightarrow [/tex] 3 ÷ 3.33 = I

[tex] \longrightarrow [/tex] 0.90 Ampere = I

Henceforth, Option B is correct.

[tex] \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ [/tex]

Answer:

f = 347.08 N

Explanation:

The frictional force exerted by the floor on the refrigerator is given as follows:

[tex]f = \mu R = \mu W[/tex]

where,

f = frictional force = ?

μ = coefficient of static friction = 0.58

W = Weight of refrigerator = mg

m = mass of refrigerator = 61 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]f = \mu mg\\f = (0.58)(61\ kg)(9.81\ m/s^2)\\[/tex]

f = 347.08 N

Answer:

82.25 moles of He

Explanation:

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

Volume (V) = 10 L

Mass of He = 0.329 Kg

Temperature (T) = 28.0 °C

Molar mass of He = 4 g/mol

Mole of He =?

Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:

1 Kg = 1000 g

Therefore,

0.329 Kg = 0.329 Kg × 1000 g / 1 Kg

0.329 Kg = 329 g

Thus, 0.329 Kg is equivalent to 329 g.

Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:

Mass of He = 329 g

Molar mass of He = 4 g/mol

Mole of He =?

Mole = mass / molar mass

Mole of He = 329 / 4

Mole of He = 82.25 moles

Therefore, there are 82.25 moles of He in the tank.

Answer:

Normal force exerted on the rover would be greater at a point on the surface of the planet where the weight of the rover is experienced to be greater.

Explanation:

Since weight is a vector quantity, it can vary with position. Weight is the amount of force the planet exerts on the rover centered towards the planet.

Such a force is the result of gravitational pull and is quantified as:

[tex]F=G\times \frac{M.m}{R^2}[/tex]

and [tex]M=\rho\times \frac{4\pi.r^3}{3}[/tex]

where:

R = distance between the center of mass of the two bodies (here planet & rover)

G = universal gravitational constant

M = mass of the planet

m = mass of the rover

This gravitational pull varies from place to place on the planet because the planet may not be perfectly spherical so the distance R varies from place to place and also the density of the planet may not be uniform hence there is variation in weight.

Weight is basically a force that a mass on the surface of the planet experiences.

According to Newton's third law the there is an equal and opposite reaction force on the body (here rover) which is the normal force.

Answer: The total rate of heat transfer from the container to its surroundings ignoring radiation is 332.67 W.

Explanation:

Given: Inner diameter = 0.9 m

q = 872 [tex]W/m^{3}[/tex]

Now, radii is calculated as follows.

[tex]r = \frac{diameter}{2}\\= \frac{0.9}{2}\\= 0.45 m[/tex]

Hence, the rate of heat transfer is as follows.

[tex]Q = q \times V[/tex]

where,

V = volume of sphere = [tex]\frac{4}{3} \pi r^{3}[/tex]

Substitute the values into above formula as follows.

[tex]Q = q \times \frac{4}{3} \pi r^{3}\\= 872 W/m^{3} \times \frac{4}{3} \times 3.14 \times (0.45 m)^{3}\\= 332.67 W[/tex]

Thus, we can conclude that the total rate of heat transfer from the container to its surroundings ignoring radiation is 332.67 W.

Answer:

a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more

b) If it is spilled in water the weight of the two beakers is the same

Explanation:

The beaker weight is

 beaker A

          W_total = W_ empty + W_water

Beaker B

            W_total = W_ empty + W_water + W_roca

a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more

b) If it is spilled in water, the weight of the two beakers is the same because the amount of liquid spilled and equal to the weight of the stone, therefore the two beakers weigh the same

Answer:

one thousand-millionth of a second.

A nanosecond is an SI unit of time equal to one billionth of a second, that is, ​¹⁄₁ ₀₀₀ ₀₀₀ ₀₀₀ of a second, or 10⁻⁹ seconds. The term combines the prefix nano- with the basic unit for one-sixtieth of a minute. A nanosecond is equal to 1000 picoseconds or ​¹⁄₁₀₀₀ microsecond.  

Answer:

b)

Explanation:

GCMs (general circulation models) are useful instruments for gaining a quantitative knowledge of climate processes. Physical processes in the atmosphere, cryosphere, and land surface are represented by them. They are used for modeling the global climate system's reaction to rising greenhouse gas concentrations available at the moment by utilizing spectral models based on the energy emitted by the biosphere and clouds.

Answer:

[tex]a_s=4.8\times 10^{-2}~m^2[/tex]

Explanation:

Given:

cross sectional area of the bone, [tex]a=4.8 \times 10^{-4} ~m^2[/tex]

factor of up-scaling the dimensions, [tex]s=10[/tex]

Since we need to find the upscaled area having two degrees of the dimension therefore the scaling factor gets squared for the area being it in 2-dimensions.

The scaled up area is:

[tex]a_s=a\times s^2[/tex]

[tex]a_s=[4.8 \times 10^{-4}]\times 10^2[/tex]

[tex]a_s=4.8\times 10^{-2}~m^2[/tex]

The area is defined as the space covered by an object in 2 d dimension. For a rectangle, it is a product of length and breadth. The new cross-section area will be 4.8×10⁻² m².

The area is defined as the space covered by an object in 2 d dimension. For a rectangle, it is a product of length and breadth. Its unit is m².

Given data in the problem

a is the crossectional area of conical bone = 4.8×10⁻⁴m².

s is the factor of up-scaling the dimensions =10

For two degrees of dimension, the upscaled area will be square of the given area.

The scaled-up area will be

[tex]\rm a_s=a\times s^2\\\\ a_s= 4.8\times10^{-4}\times {10}^2\\\\\ \rm a_s=4.8\times10^{-2}\;m^2[/tex]

Hence the new cross-section area will be 4.8×10⁻² m².

To learn more about the area refer to the link;

https://brainly.com/question/1631786

Explanation:

someone to check if the answer is correct

Answer:

hope it helps

explanation:

Explanation:

The area of a circle of radius r is given by

[tex]A = \pi r^2[/tex]

Taking the derivative of A with respect to time t, we get

[tex]\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}[/tex]

We also know that

[tex]\dfrac{dr}{dt} = 2\:\text{m/s}\:\text{at}\:r = 39\:\text{m}[/tex]

[tex]\dfrac{dA}{dt} = 2\pi (39\:\text{m})(2\:\text{m/s})= 490\:\text{m}^2\text{/s}[/tex]

Answer:

(i)8m/s²(ii)40m/s

Explanation:

according to the formula

½at²=s.

then substituting the data

½a•5²=100

a=8m/s²

v=at=8•5=40m/s

Answer:

(I)

[tex]{ \bf{s = ut + \frac{1}{2} a {t}^{2} }} \\ 100 = (0 \times 5) + \frac{1}{2} \times a \times {5}^{2} \\ 200 = 25a \\ { \tt{acceleration = 8 \: m {s}^{ -2} }}[/tex]

(ii)

[tex]{ \bf{v = u + at}} \\ v = 0 + (8 \times 5) \\ { \tt{final \: velocity = 40 \: m {s}^{ - 1} }}[/tex]

By the work-energy theorem, the total work done on the mass by the spring is equal to the change in the mass's kinetic energy:

W = ∆K

and the work done by a spring with constant k as it gets compressed a distance x is -1/2 kx ²; the work it does is negative because the restoring force of the spring points opposite the direction in which it's getting compressed.

So we have

-1/2 k (0.15 m)² = 0 - 1/2 (2.0 kg) (3.0 m/s)²

Solve for k to get k = 800 N/m.

Answer:

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Explanation:

If you accelerate at 2 m/s^2 for 10 seconds, at the end of the 10 seconds you are moving at a rate of 20 m/s.

V(f) = V(i) + a*t

Final velocity = initial velocity + acceleration x time

Your average velocity will be half of your final, because you accelerated at a constant rate. So your average velocity is 10 m/s.

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Answer:

100 m

Explanation:

Given,

Initial velocity ( u ) = 0 m/s

Acceleration ( a ) = 2 m/s^2

Time ( t ) = 10 sec s

To find : Displacement ( s ) = ?

By 2nd equation of motion,

s = ut + at^2 / 2

= ( 0 ) ( 10 ) + ( 2 ) ( 10 )^2 / 2

= 0 + ( 2 ) ( 100 ) / 2

= 200 / 2

s = 100 m

Answer:

a. C = 0.5 m/s²

b. F = 35 Newton

Explanation:

Given the following data;

Radius, r = 200 m

Velocity, v = 10 m/s

Mass, m = 70 kg

a. To find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

C = v²/r

Where:

C is the centripetal acceleration

v is the velocity

r is the radius

Substituting into the formula, we have;

C = 10²/200

C = 100/200

C = 0.5 m/s²

b. To find the force;

F = mv²/r

F = (70*10²)/200

F = (70 * 100)/200

F = 7000/200

F = 35 Newton

Answer:

-45,597.07

Explanation:

if not in scientific calculator and yung answer nung sa scientific sa comment na lang dinadownload ko ka eh

Answer:

a)  T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] ,  b) T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]

c)  x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex],  d)  m₂ = m₁  ( [tex]\frac{ L}{2d} -1[/tex])

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

      ∑ τ = 0

      T_A d - W₂ x -W₁ L/2 = 0

      T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]

b) the tension in string B

we write the expression of the translational equilibrium

       ∑ F = 0

       T_A - W₂ - W₁ - T_B = 0

       T_B = T_A -W₂ - W₁

       T_ B =   [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]  - g m₂ - g m₁

       T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

         T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

          g m₂ (d -x) -  g m₁  (0.5 L -d) + 0 = 0

          m₂ (d-x) = m₁ (0.5 L- d)

          m₂ x = m₂ d - m₁ (0.5 L- d)

          x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex]

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

           0 = d - \frac{m_1}{m_2} \  \frac{L}{2d}

         [tex]\frac{m_1}{m_2} \ (0.5L -d) = d[/tex]

          [tex]\frac{m_1}{m_2} = \frac{ d}{0.5L-d}[/tex]

          m₂ = m₁  [tex]\frac{0.5 L -d}{d}[/tex]

          m₂ = m₁  ( [tex]\frac{ L}{2d} -1[/tex])

Answer:

300 meters

Explanation:

a= 2m/s^2

t= 5 min

Convert into seconds, 5*60= 300seconds

v0= 0

x0=0

use x-x0= v0t + 1/2at^2

plug in values

x= 1/2*2*(300)

Solve

x= 300 meters

Answer:

a)    v = 517.99 m / s,  b) θ = 296.3º

Explanation:

This is an exercise in kinematics, we are going to solve each axis independently

X axis

the acceleration is aₓ = 5.10 1 / S², they are on for t = 670 s and reaches a speed of vₓ=  3670 m / s, let's use the relation

           vₓ = v₀ₓ + aₓ t

           v₀ₓ = vₓ - aₓ t

           v₀ₓ = 3670 - 5.10 670

           v₀ₓ = 253 m / s

Y axis  

the acceleration is ay = 7.30 m / s², with a velocity of 4378 m / s after

t = 670 s

          v_y = v_{oy} + a_y t

          v_{oy} = v_y - a_y t

          v_oy} = 4378 - 7.30 670

          v_{oy}  = -513 m / s

to find the velocity modulus we use the Pythagorean theorem

          v = [tex]\sqrt{v_o_x^2 + v_o_y^2}[/tex]

          v = [tex]\sqrt{253^2 +513^2}[/tex]

          v = 517.99 m / s

to find the direction we use trigonometry

         tan θ ’= [tex]\frac{v_o_y}{v_o_x}[/tex]

         θ'= tan⁻¹  [tex]\frac{voy}{voy}[/tex]  

         θ'= tan⁻¹ (-513/253)

         tea '= -63.7

the negative sign indicates that it is below the ax axis, in the fourth quadrant

to give this angle from the positive side of the axis ax

          θ = 360 -   θ  

          θ = 360 - 63.7

          θ = 296.3º

Answer:

[tex]M_2=0.79kg[/tex]

Explanation:

From the question we are told that:

Period [tex]T=0.517s[/tex]

Trial 1

Spring constant [tex]\mu=117N/m[/tex]

Period [tex]T_1=0.37[/tex]

Mass [tex]m=0.400kg[/tex]

Trial 2

Period [tex]T_2=0.52[/tex]

Generally the equation for Spring Constant  is mathematically given by

\mu=\frac{4 \pi^2 M}{T^2}

Since

[tex]\mu _1=\mu_2[/tex]

Therefore

[tex]\frac{4 \pi^2 M_1}{T_1^2}=\frac{4 \pi^2 M_2}{T_2^2}[/tex]

[tex]M_2=M_1*(\frac{T_2}{T_1})^2[/tex]

[tex]M_2=0.400*(\frac{0.52}{0.37}})^2[/tex]

[tex]M_2=0.79kg[/tex]