What is the biggest planet in the solar system

Answer:

  K_b = 78 J

Explanation:

For this exercise we can use the conservation of energy relations

starting point. Lowest of the trajectory

        Em₀ = K = ½ mv²

final point. When it is at tea = 50º

        Em_f = K + U

        Em_f = ½ m v_b² + m g h

where h is the height from the lowest point

        h = L - L cos 50

        Em_f = ½ m v_b² + mg L (1 - cos50)

energy be conserve

        Em₀ = Em_f

         ½ mv² = ½ m v_b² + mg L (1 - cos50)

         K_b = ½ m v_b² + mg L (1 - cos50)

let's calculate

          K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)

          K_b = 36 +42.0

          K_b = 78 J

Answer:

Greater.

Explanation:

This pressure difference will be greater if the scuba diver were in seawater and went to the same depth because the seawater have salts which increases the density of water as compared to freshwater. Salt in water increases the density which automatically increases the pressure on the diver so that's why we can say that the pressure will be increases for the scuba diver in seawater as compared to freshwater.

Answer:

7.5 kg

Explanation:

We are given that

[tex]m_1=10 kg[/tex]

Length of plank, l=3 m

Distance of fulcrum from one end of the plank=1 m

[tex]m_2=20 kg[/tex]

We have to find the mass must be on the other end if the plank remains balanced.

Let m be the mass must be on the other end if the plank remains balanced.

In balance condition

[tex]20\times 1=10\times (1.5-1)+m\times (1.5+0.5)[/tex]

[tex]20=10(0.5)+2m[/tex]

[tex]20=5+2m[/tex]

[tex]2m=20-5=15[/tex]

[tex]\implies m=\frac{15}{2}[/tex]

[tex]m=7.5 kg[/tex]

Hence, mass 7.5 kg   must be on the other end if the plank remains balanced.

Answer:

The mass at the other end is 7.5 kg.

Explanation:

Let the mass is m.

Take the moments about the fulcrum.

20 x 1 = 10 x 0.5 + m x 2

20 = 5 + 2 m

2 m = 15

m = 7.5 kg

Answer: The focal length of the lens is 2.60 cm

Explanation:

The equation for lens formula follows:

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

where,

f = focal length = ? cm

v = image distance = 2.9 cm

u = Object distance = -25 cm

Putting values in above equation, we get:

[tex]\frac{1}{f}=\frac{1}{2.9}-\frac{1}{(-25)}\\\\\frac{1}{f}=\frac{1}{2.9}+\frac{1}{(25)}\\\\\frac{1}{f}=\frac{25+2.9}{2.9\times 25}\\\\f=\frac{72.5}{27.9}=2.60cm[/tex]

Hence, the focal length of the lens is 2.60 cm

The acceleration will increase by 61.3%.

Explanation:

The centripetal acceleration [tex]a_c[/tex] is given by

[tex]a_c = \dfrac{v^2}{r}[/tex]

If the velocity of the object increases by 27.0%, then its new velocity v' becomes

[tex]v' = 1.270v[/tex]

The new centripetal acceleration [tex]a'_c[/tex] becomes

[tex]a'_c = \dfrac{(1.270v)^2}{r} = 1.613 \left(\dfrac{v^2}{r} \right)[/tex]

[tex]\:\:\:\:\:\:\:\:\:= 1.613a_c[/tex]

Answer:

 v = -v₀ / 2

Explanation:

For this exercise let's use kinematics relations.

Let's use the initial conditions to find the acceleration of the electron

            v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

           a = v₀² / 2y

now they tell us that the initial velocity is half

          v’² = v₀’² - 2 a y’

          v₀ ’= v₀ / 2

at the point where turn v = 0              

          0 = v₀² /4  - 2 a y '

          v₀² /4 = 2 (v₀² / 2y)  y’

          y = 4 y'

          y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

         v² = v₀² -2a y’

         v² = 0 - 2 (v₀² / 2y) y / 4

         v² = -v₀² / 4

         v = -v₀ / 2

We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.

Answer:

0.013 m/min

Explanation:

Applying,

dV/dt = (dh/dt)(dV/dh)............. Equation 1

Where

V = πr²h................ Equation 2

Where V = volume of the tank, r = radius, h = height.

dV/dh = πr²............ Equation 3

Substitute equation 3 into equation 1

dV/dt = πr²(dh/dt)

From the question,

Given: dV/dt = 2 m³/min, r = 7 m, π = 3.14

Substitute these values into equation 3

2 = (3.14)(7²)(dh/dt)

dh/dt = 2/(3.14×7²)

dh/dt = 0.013 m/min

Answer:

[tex]6.63\times 10^8\ N/C[/tex]

Explanation:

Given that,

The magnitude of magnetic field, B = 2.21

We need to find the magnitude of the electric field. Let it is E. So,

[tex]\dfrac{E}{B}=c\\\\E=Bc[/tex]

Put all the values,

[tex]E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C[/tex]

So, the magnitude of the electric field is equal to [tex]6.63\times 10^8\ N/C[/tex].

Answer:

increase

Explanation:

hope it will help you:)

Answer:

t = 2.7 s

Explanation:

This is a kinematics problem.

How the elevator reduces its speed to zero by a distance equal to the length of the cylinder

        y = 2π r = 2 π d / 2

        y = π d

        y = π 1.1

        y = 3,456 m

now we can look for the acceleration of the system

        v² = v₀² - 2 a y

        0 = v₀² - 2 a y

        a = v₀² / 2y

        a = 2.6² / 2 3.456

        a = 0.978 m / s²

now let's calculate the time

        v = v₀ - a t

        0 = v₀ - at

         t = v₀ / a

         t = 2.6 /0.978

         t = 2.658 s

ask for the result with two significant figures

         t = 2.7 s

Answer:

Pressure is more in the open container than the closed one.

Explanation:

The pressure due to the fluid at a depth is given by

Pressure = depth x density of fluid x gravity

So, when the container is open, the atmospheric pressure is also add  up but when  the container is closed only the pressure due to the fluid is there.

So, when the container is open, the pressure is atmospheric pressure + pressure due to the fluid.

hen the container is closed only the pressure due to the fluid is there.

Explanation:

Here,we've been given that,

We've to find the acceleration of the object. By using the third equation of motion,

→ v² - u² = 2as

→ (420)² - (0)² = 2 × a × 145

→ 176400 - 0 = 290a

→ 176400 = 290a

→ 176400 ÷ 290 = a

→ 608.275862 m/s² = a

If you know initial speed and final speed, you can find the average speed.  Then, knowing distance, you can find the time.

KimYurii posted the first answer to this question.  

That answer is well organized, well presented, elegant and correct, and it deserves to be awarded "Brainliest" and several merit badges.

My problem is that I can never remember all the different formulas.  I guess I had to work with so many uvum in all the Physics, Geometry, and Calculus classes that I took, I filled up all the memory slots with formulas, and over the years they all eventually merged into a big glob of goo.  Now, the only formulas I can remember are the ones I had to use as an Electrical Engineer.

When I see this kind of question, I can only remember one or two simple formulas, and I reason it out like this:

Starting speed . . . zero

Ending speed . . . 420 m/s

Formula:  Average speed . . . (1/2)·(0 + 420) = 210 m/s

Distance covered . . . 145 m

Formula: Time taken = (distance) / (average speed) = (145/210) second

(Now you have the time.)

Formula: Distance = (1/2)·(acceleration)·(time²)

145 m = (1/2)·(acceleration)·(145/210 sec)²

Acceleration = 290 m / (145/210 s)²

Acceleration = 608.28 m/s²

Answer:

Smaller and upside down

Explanation:

To answer the question, we must recognise that the characteristics of the image formed by a convex lens depends on the position of the object from the lens.

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

1. The image is smaller than the object.

2. The image is inverted i.e upside down.

3. The image is closer to the lens

4. The image between 2f and f

Now, considering the options given in question above, the correct answer to the question is:

The image is smaller and upside down.

Answer:

(a) 1 : 2

(b) same

Explanation:

Let the mass of puck A is m and the mass of puck B is 2 m.

initial speed for both the pucks is same as u and the distance is same for both is s.

let the tension is T for same.

The kinetic energy is given by

[tex]K = 0.5 mv^2[/tex]

(a) As the speed is same, so the kinetic energy depends on the mass.

So, kinetic energy of A : Kinetic energy of B = m : 2m  = 1 : 2

(b) A the distance s same so the final velocities are also same.

(a)  The kinetic energy of puck B is 2 times the kinetic energy of puck A.

(b)  The final speed of both the puck A and B are same.

Let the mass of puck A is m and the mass of puck B is 2 m.

Initial speed for both the pucks is same as u and the distance is same for both is s.

Let the tension is T for same.

Then, the kinetic energy is given as,

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

(a)

As the speed is same, so the kinetic energy depends on the mass.

Then,

[tex]\dfrac{KE_{A}}{KE_{B}} = \dfrac{1/2 \times mv^{2}}{1/2 \times (2m)v^{2}}\\\\\\\dfrac{KE_{A}}{KE_{B}} =\dfrac{1}{2}[/tex]

So, kinetic energy of A : Kinetic energy of B = 1 : 2.

Thus, we can conclude that the kinetic energy of puck B is 2 times the kinetic energy of puck A.

(b)

The final speed for the puck is given as,

v = s/t

here, s is the distance covered.

Since, both pucks are pulled the same distance across frictionless ice. Then, the final speed of each puck is also same.

Thus, we can conclude that the final speed of both the puck A and B are same.

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Answer:

a = 6.67 m/s²

Explanation:

F = 10.0 N

m = 1.50 kg

a = ?

F = ma

10.0 = (1.50)a

6.67 = a

Answer:

non biodegradable

Explanation:

It is non biodegradable because plastic cannot dispose off easily ..

Answer:

I = 0.0857 A

Explanation:

Given that,

Power consumed by the cellphone, P = 300 mW

The voltage of the battery, V = 3.5 V

Let I is the current flowing through the cell-phone. We know that,

P = VI

Where

I is the current

So,

[tex]I=\dfrac{P}{V}\\\\I=\dfrac{300\times 10^{-3}}{3.5}\\\\I=0.0857\ A[/tex]

So, the current flowing the cell-phone is 0.0857 A.

Answer:

There's no risk of animals or bad weather interfering with your campsite, either. You don't even really need a tent. A sleeping pad, sleeping bag and a mindful eye to pick up everything you brought in is all you really need to enjoy overnight caving. Do your research

Explanation:

Answer:

C.  Rotational motion

Explanation:

The kinematics of rotational motion describes the relationships between the angle of rotation, angular velocity, angular acceleration, and time. It only describes motion—it does not include any forces or masses that may affect rotation (these are part of dynamics). Recall the kinematics equation for linear motion: v = v+at (constant a).

Rotational motion is mutually exclusive with each of the others. Hence, option (C) is correct.

"The motion of an object around a circular route, in a fixed orbit, is referred to as rotational motion."

Rotational motion dynamics are identical to linear or translational dynamics in every way. The motion equations for linear motion share many similarities with the equations for the mechanics of rotating objects. Rotational motion only takes stiff bodies into account. A massed object that maintains a rigid shape is referred to as a rigid body.

Curvilinear motion is the movement of an object along a curved route. Example: A stone hurled at an angle into the air.

The motion of a moving particle that follows a predetermined or known curve is referred to as curvilinear motion. Two coordinate systems—one for planar motion and the other for cylindrical motion—are used to examine this type of motion.

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Answer:

E_total = - 50 N / A

Explanation:

The electric field is a vector magnitude whereby

          E_total = Eₐ + E_b

where the bold letters indicate vectors, in this case the charges of the two objects A and B are the same and they are on the same line

         E_total = - E_a - E_b

the electric field for a point charge is

        E_a = [tex]k \ \frac{q_a}{r_a^2 }[/tex]

        qₐ= Eₐ rₐ² / k

indicates that Eₐ = 40.0 N / C

        qₐ = 40.0 0.250²/9 10⁹

        qₐ = 2.777 10⁻¹⁰ C

indicates that the charge of the two points is the same

        qₐ = q_b

      E_total = - k qₐ / rₐ² - k qₐ / (2 rₐ)²

      E_total = [tex]-k \ \frac{q_a}{r_a^2} \ ( 1 + \frac{1}{4} )[/tex]

we calculate

       E_total = - 40.0 (5/4)

       E_total = - 50 N / A

Answer:

The work done by the spring is 2.025 J

Explanation:

Given;

mass on the spring, m = 0.5 kg

spring constant, k = 5 N/m

extension of the spring, x = 0.9 m

The work done by the spring is calculated as;

[tex]W = \frac{1}{2} kx^2\\\\W = \frac{1}{2} \times 5 \times (0.9)^2\\\\W = 2.025 \ J[/tex]

Therefore, the work done by the spring is 2.025 J

Answer:

a)   W_total = mg (2h + d)   , b)     E_total = - mg (h + d)

Explanation:

a) We must solve this problem in two parts, the first for the accelerated movement and the second for the movement with constant speed

Let's look for work for the part that is in free fall

        y = y₀ + v₀ t - ½ g t²

when he jumps out of a plane his vertical speed is zero

        y =y₀ - ½ g t²

        dy = 0 - ½ g 2t dt

the work in this first part is

        W₁ = ∫ F dy

        W₁ = mg ∫ g t dt

        W₁ = m g² t² / 2

the time it takes to travel the distance y₀-y = h is

         y₀-y = ½ g t²

         t =[tex]\sqrt{2h/g}[/tex]

we substitute

          W₁ = m g² 2h / g

          W₁ = m g 2h

now we look for the work for the part with constant speed

since the velocity is constant let's use the uniform motion ratio

          W₂ = F d

           W₂ = mg d

the total work is

           W_total = W₁ + W₂

           W_total = 2mgh + m gd

           W_total = mg (2h + d)

b) The change in gravitational potential energy

           U = mg Δy

in the part with accelerated movement

           U₁ = mg h

in the part with uniform movement

            U₂ = mg d

the total potential energy is

           E_total = U₁ + U₂

           E_total = - mg (h + d)

Answer:

Gradual shifting of a time series to relatively higher or lower values over a long period of time is called a Trend.

Answer:

Hydroelectricity

Explanation:

Because of flooding of water, we can assume that the electricity was generated by Water which is known as Hydroelectricity.

We can presume that the energy was produced by water because of the flooding of the water, which is a process known as hydroelectricity.

Hydroelectric power, often known as hydropower, is the name given to electricity generated by turbines that turn the potential energy of falling or swiftly running water into mechanical energy. As of 2019, hydropower accounted for more than 18% of the world's total power generation capacity, giving it the most frequently used renewable power source in the early 21st century.

When water is used to produce energy, it is first gathered or stored at a higher altitude and then transported through extensive pipelines or tunnels (called pen stocks) to a lower level; the difference between these two altitudes is referred to as the head. The falling water triggers the rotation of turbines at the bottom of its descent through the pipes.

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Answer:

[tex]R1/R2=\frac{2}{3}[/tex]

Explanation:

From the question we are told that:

1st's Length [tex]l=L[/tex]

1st's radius [tex]r=r[/tex]

2nd's Length [tex]l=6L[/tex]

2nd's radius [tex]r=2r[/tex]

Generally the equation for Resistance R is mathematically given by

 [tex]R=\frac{\rho L}{\pi r^2}[/tex]

Therefore

 [tex]R_1=\frac{\rho L}{\pi r^2}[/tex]

And

 [tex]R_2=\frac{\rho 6L}{\pi (2r)^2}[/tex]

Therefore

 [tex]R1/R2=\frac{\frac{\rho L}{\pi r^2}}{\frac{\rho 6L}{\pi (2r)^2}}[/tex]

 [tex]R1/R2=\frac{2}{3}[/tex]

Answer:

I am pretty sure it is C

Explanation:

It can be found all over the universe

Answer:

C

Explanation:

20 cm = 0.2m

since uncertainty is 0.1 cm (0.001 m), should be recorded to same number of decimal place as uncertainty

therefore it's 0.200m

Answer:

The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.

Explanation:

The angular velocity of the object ([tex]\omega[/tex]), in radians per second, is given by the following expression:

[tex]\omega(t) = 5\cdot t^{2}[/tex] (1)

Where [tex]t[/tex] is the time, measured in seconds.

The change in the angular displacement ([tex]\Delta \theta[/tex]), in radians, is found by means of the following definite integral:

[tex]\Delta \theta = \int\limits^{4}_{2} {5\cdot t^{2}} \, dt[/tex] (2)

Then we proceed to integrate on the function in time:

[tex]\Delta \theta = \frac{5}{3}\cdot (4^{2}-2^{2})[/tex]

[tex]\Delta \theta = 20\,rad[/tex]

The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.

Answer:

The work done on the block by the spring as it accelerates the block is 4kx².

Explanation:

Let initial distance is x.

It was compressed three times farther and then the block is released, new distance is 3x.

The work done in compressing the spring is given by :

[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)[/tex]

[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\W=\dfrac{1}{2}k((3x)^2-x^2)\\\\W=\dfrac{1}{2}k((9x^2-x^2)\\\\W=\dfrac{1}{2}k\times 8x^2\\\\W=4kx^2[/tex]

So, the work done on the block by the spring as it accelerates the block is 4kx².