How much energy is transferred in 10 seconds with a current of 13 amperes and a potential difference of 230 volts?​

Answer:

pp i hate khan academy pls help me

If you are not advised to evacuate or are unable to do so safely, stay indoors, away from windows, skylights and doors. Continue to monitor weather reports and do not go outside until the storm has passed. And, to protect your children from seeing too many sights and images of the hurricane, limit media exposure.

Answer:

Impulse = 50,000 J

Explanation:

Given: force = 100,000 kg ms^2 or N; time = 0.5 s

Unknown: impulse = ?

Equation: impulse = (force)(time)

Substitute: impulse = (100,000 kg ms^2 or N)(0.5 s)

Solution: impusle = 50,000 kg ms^2 or J

Answer:

1.84 m/s

Explanation:

Applying,

The kinetic energy of the pinball = Elastic energy of the spring

mv²/2 = ke²/2

mv² = ke².................. Equation 1

Where m = mass of the pin ball, v = velocity of the pin ball, k = force constant of the spring, e = extension of the spring.

make v the subject of the equation

v = √(ke²/m)................ Equation 2

From the question,

Given: e = 0.15 m, k = 30 N/m, m = 0.20 kg

Substitute these values into equation 2

v = √[(30×0.15²)/0.2]

v = 1.84 m/s

Answer:

im guessing red giant-

Answer:    

Explanation:

Answer:

Cardboard, Paper towels, Food waste, and wooden based items

Answer:

How fast do things biodegrade?

Vegetables 5 days –1 month

Aluminium cans 80–100 years

Glass bottles 1 million years

Styrofoam cup 500 years to forever

Plastic bags 500 years to forever

Answer:

A. Increase the distance between the objects.

D. Decrease the mass of one of the objects.

Explanation:

The smaller the mass of an object the less gravity it has and the farther away two objects are the less gravitational force

First, calculate the components of its initial velocity v0:

v0x = v0cos10 = 24.6 m/s

v0y = v0sin10 = 4.34 m/s

The ball reaches its peak when vy = 0. Let's calculate the time it takes for vy to become zero:

vy = v0y - gt ---> t = v0y/g = 0.44 s

The horizontal distance it travels in this time is

x = v0xt = (24.6 m/s)(0.44 s)

= 10.8 m

Note that the net is 15 m away. After traveling a horizontal distance of 10.8 m, the height of the ball is

y = -(1/2)gt^2 + v0yt + 2.1

= -(4.9 m/s^2)(0.44 s)^2 + (4.34 m/s)(0.44 s) + 2.1 m

= -0.95 m + 1.9 m + 2.1 m

= 3.05 m

Note that this is the height of the ball at its peak. While the ball is well above the net at its peak, it is well short of its required horizontal distance to clear it. Instead, let's find the time it takes for the tennis ball to travel a horizontal distance of 15 m first:

x = v0xt ----> t = x/v0x = (15 m)/(24.6 m/s) = 0.61 s

Then calculate the height y when t = 0.61 s. If y > 0.9 m (height of the net), then the ball will clear the net.

y = -4.9t^2 + v0yt + 2.1

= -4.9(0.61 s)^2 + (4.34 m/s)(0.61 s) + 2.1 m

= -1.82 m + 2.65 m + 2.1 m

= 2.93 m

Yes, the ball will clear the net by 2.03 m.

Answer:

Sedimentary rocks are porous.

Explanation:

So option A) Sedimentary is your answer. Have a great summer

Answer:

Thus, the time for the first lamp is 44 minutes.

Explanation:

Power of first lamp, P' = 1000 W

Power of second lamp, P'' = 4400 W

time for second lamp, t'' = 10 minutes

Let the time for first lamp is t'.

As the energy is same, so,

P' x t' = P'' x t''

1000 x t' = 4400 x 10

t' = 44 minutes

Answer: A: 70/2=35

B: 35/2=17

C: 17.5/2=8.75

D: 8.75 of C-14 will be left

E: 5,730 years

F: 5,631

Explanation:

that’s all I got, hope I helped kinda

Answer:

The velocity with which the sand throw is 24.2 m/s.

Explanation:

acceleration due to gravity, a =  3.9 m/s2

height, h = 75 m

final velocity, v = 0

Let the initial  velocity at the time of throw is u.

Use third equation of motion

[tex]v^{2} = u^{2} - 2 g h \\\\0 = u^{2} -2\times 3.9\times 75\\\\u=24.2m/s[/tex]

The velocity with which the sand throw is 24.2 m/s.

Answer:

Explanation:

The formula to find Potential Energy is PE = mgh, where m is the mass of the object, g is gravity, and h is the height from which the object can potentially fall. Because this is linear, then PE will increase as either the mass or the height increase (gravity is constant at 9.8 m/s/s). If the mass of the ball being dropped doesn't change, then the only thing that determines this ball's max PE is the height from which it is dropped; max PE ALWAYS occurs at the highest point from which an object can potentially fall. So your answer is "At the top".

Answer:

Option D

Explanation:

The options for the question are

a) How do galaxies rotate?

b) What is the weather on Neptune?

c) What is the core of Saturn made of?

d) What other solar systems have planets?

Solution

The Hubble space telescope was designed and integrated into the extraterrestrial system in order to capture information about the surrounding universe. If the  Keck Observatory has a better observation capacity than the Hubble space telescope then the scientist would be interested to know the surrounding planets in the solar system.

Hence, option D is correct

Answer:

[tex]W=17085KJ[/tex]

Explanation:

From the question we are told that:

Height [tex]H=16m[/tex]

Radius [tex]R=3[/tex]

Height of water [tex]H_w=9m[/tex]

Gravity [tex]g=9.8m/s[/tex]

Density of water [tex]\rho=1000kg/m^3[/tex]

Generally the equation for Volume of water is mathematically given by

 [tex]dv=\pi*r^2dy[/tex]

 [tex]dv=\frac{\piR^2}{H^2}(H-y)^2dy[/tex]

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 [tex]dw=(pdv)g (H-y)[/tex]

Substituting dv

 [tex]dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)[/tex]

 [tex]dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy[/tex]

Therefore

 [tex]W=\int dw[/tex]

 [tex]W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy[/tex]

 [tex]W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)[/tex]

 [tex]W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0[/tex]

 [tex]W=3420.84*0.25[2401-65536][/tex]

 [tex]W=17084965.5J[/tex]

 [tex]W=17085KJ[/tex]

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Answer:

3.2075*10^16

Explanation:

Q=P/V just search up a converter and youll get 30V and so you do 15/30 which is a half and a single coulomb is 6.415*10^16 so you half it. I belive this is correct if you dont belive me wait for someone else smarter to answer and compare.

Explanation:

.Flame retar,dant plastic additi,ves are compounds added to plastics and other materials to inhibit, suppress or de,lay combustion. These compounds are useful in impending burning in the ign,ition phase of fire. They do not prevent char,ring or melting nor do they increase the heat resistance of a material

Answer:

Explanation:

Magnitude of frictional force = μ mg

μ is either static or kinetic friction.

To start the crate moving , static friction is calculated .

a ) To start crate moving , force required = μ mg where μ is coefficient of static friction .

force required =.517 x 56.6  x 9.8 = 286.76  N .

b ) to  slide the crate across the dock at a constant speed , force required

= μ mg where μ is coefficient of kinetic  friction , where μ is kinetic friction

= .26 x 56.6  x 9.8 = 144.21 N .

Answer:

391.5 J

Explanation:

The amount of work done can be calculated using the formula:

Looking at the formula, we can see that the mass of the object does not affect the work done on it.

Substitute the force applied and the displacement of the object into the equation.

The amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.

Mass doesn't matter on amount of work done .We can calculate amount of work done through Force and Displacement

[tex]\boxed{\sf W=Fd}[/tex]

[tex]\\ \sf\longmapsto Work\:Done=87(4.5)[/tex]

[tex]\\ \sf\longmapsto Work\:done=391.5J[/tex]

Answer:

 a = 10 ρ_air g [tex]\frac{\Delta P}{m P_o}[/tex]

Explanation:

Let's solve this problem in parts, with the initial data the camera rises slowly, so we can assume that at constant speed, we apply the equilibrium condition

         B - W = 0

The thrust is given by Archimedes' law

         B = rho_aire g V_body

The volume of the body can be found from the ideal gas ratio

         P V = n R T

let's use the subscript "o" for the initial concisions

         V₀ = (nR) T₀/P₀               1

we substitute

          10 ρ_air g (nR) T₀ /P₀ = W

when the storm approaches the pressure decreases, P

If we use the ideal gas equation

           V = (nR) T₀ / P

we combine this equation with equation 1

            V = V₀P₀ / P

if we write the pressure

             P = P₀ -ΔP

we substitute

            V = [tex]\frac{V_oP_o}{P_o - \Delta P} =V_o \ ( 1 - \frac{\Delta P}{P_o} )^{-1}[/tex]

we expand serie  and eliminate higher order terms

            V = V₀ ([tex]1+ \frac{\Delta P}{Po}[/tex])

with this expression we can write the thrust

             B = B₀ + ΔB

Newton's second law for the new conditions is

             B - W = m a

             (B₀ + ΔB) - W = ma

              ΔB = m a

              a = ΔB / m

              a = 10 ρ_air g [tex]\frac{\Delta P}{m P_o}[/tex]

This is the initial acceleration of the camera

Answer:

Explanation:

We will use the KE equation you wrote here and fill in what we are given:

[tex]36=\frac{1}{2}m(12)^2[/tex] and isolating the m:

[tex]m=\frac{2(36)}{12^2}[/tex] which gives us

m = .50 kg

Answer:

Explanation:

The frictionless surface implies that the speed of the spring is at a max. When the speed of the spring is at its max, the potential energy in the spring is 0. Use the equation for the Total Energy in a Spring/Mass System:

KE + PE = [tex]\frac{1}{2}kA^2[/tex] where KE is the Kinetic Energy available to the spring, PE is the potential energy available to the spring, and the sum of those is equal to one-half times the spring constant, k, times the amplitude of the spring's movement away from the equilibrium position. Sometimes this amplitude is the same as the displacement of the spring. This can be tricky. But since we are only given one value for the distance, we are going to use it as an amplitude. Keeping in mind that the PE is 0 when KE is at its max, then the equation becomes

KE + 0 = [tex]\frac{1}{2}kA^2[/tex] or to put it simpler terms:

KE = [tex]\frac{1}{2}kA^2[/tex] We need to find the value for KE before we can fully solve the problem we are being tasked with.

Filling in using the info given:

[tex]KE=\frac{1}{2}(2.0)(.06)^2[/tex] Notice I added another place of significance to the 2 because 1 simply isn't enough and the physics teacher in me can't handle that. Simplifying a bit:

[tex]KE=(.06)^2[/tex] because the k = 2 cancels out the 2 in the denominator of the 1/2. So

KE = 3.6 × [tex]10^{-3[/tex]

Now plug that in for KE and solve for v:

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

[tex]3.6*10^{-3}=\frac{1}{2}(5.0)v^2[/tex] and

[tex]v=\sqrt{\frac{2(3.6*10^{-3})}{5.0} }[/tex] gives us a velocity of

v= [tex]3.8*10^{-2[/tex]

Answer:

Explanation:

During an earthquake, there are both transverse and longitudinal wave. The answer is c.

Answer:

Explanation:

ans is c. Both transverse and longitudinal

Answer:

It’s like medium uk

Explanation:

The size of electron charge is estimated to be 1.6 × 10-¹⁹

Answer:

The given potential energy of the spring is expressed as follows;

ΔUsp = -(WB + WW)

Where;

WB = Th work done by the spring on the block to which it is attached

WW = The work done by the spring on the wall

We recall that work done, W = Force applied × Distance moved in the direction of the force

The work done by the spring on the block, WB = The spring force × The distance the block moves

The work done by the spring on the wall, WW = The spring force × The distance the wall moves

However, given that the wall does not move, we have;

The distance the wall moves = 0

∴ The work done by the spring on the wall, WW = The spring force × 0 = 0 J

Therefore, WW = 0 J, and the spring does not do work on the wall, and WW can be ignored in the next subsequent) steps

Explanation:

Answer:

No.

Explanation:

A feather is less dense and thus less force exerted while a rock is very dense thus exerting more force .