A.All three bulbs will go out
B.Bulbs 2 and 3 will go out, but bulb 1 will remain lit
C. All three bulbs will go out
D. Bulb 3 will go out, but bulbs 1 and 2 remain lit

Answer:

Answer:

-2.5 m/s²

Explanation:

The acceleration of a body is the change in it's velocity with time.

The change in velocity with time can be obtained as the slope of a velocity time graph ;

Acceleration = (change in velocity / change in time)

Taking the slope :

Change in Velocity = △y = y2 - y1

Change in time = △x = x2 - x1

(10, 15) ; (0, 40)

△y / △x = y2 - y1 / x2 - x1 = (40 - 15) / (0 - 10)

△y / △x = 25 / - 10 = - 2.5 m/s²

Answer:

weight

Explanation:

Weight is the force of gravity with which a body is attracted towards the earth.

Answer:

A = 70cm³

B= 805 grams

Explanation:

(a) V = L×B×H

= 7cm × 5cm × 2cm

= 35cm × 2cm

= 70cm³

(b) Mass= volume × Density

= 70cm³ × 11.5g/cm³

=805 grams

I hope this will help you ...

Answer:

The outline of the energy transfer are;

a) Kinetic energy → Clockwork spring → Potential energy

b) Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run

c) Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy

Please find attached the drawings of the energy transfer created with MS Visio

Explanation:

The energy transfer diagrams are diagrams that can be used to indicate the part of a system where energy is stored and the form and location to which the energy is transferred

a) The energy transfer diagram for the winding up a clockwork car is given as follows;

Mechanical kinetic energy is used to wind up (turn) the clockwork car such that the kinetic energy is transformed into potential energy and stored in the wound up clockwork as follows;

Kinetic energy → Clockwork spring → Potential energy

b) Letting a wound up clockwork car run results in the conversion of mechanical potential energy into kinetic (energy due tom motion) energy as follows;

Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run

c) The energy stored in the battery of a battery powered car is chemical potential energy. When the battery powered car runs, the chemical potential energy produces an electromotive force which is converted into kinetic energy as electric current flows from the batteries

Therefore, we have;

Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy

Answer:

so you gotta bring the 5 and add to 7

Answer: [tex]36.22\ V[/tex]

Explanation:

Given

Energy in a capacitor [tex]E=7.77\times 10^{-7}\ J[/tex]

Charge [tex]Q=4.29\times 10^{-8}\ C[/tex]

Energy of a capacitor is given by

[tex]E=\dfrac{1}{2}CV^2=\dfrac{1}{2}QV\quad [\text{Q=CV}][/tex]

Insert the values

[tex]E=\dfrac{1}{2}QV\\\\\Rightarrow 7.77\times 10^{-7}=\dfrac{1}{2}\times 4.29\times 10^{-8}\times V\\\\\Rightarrow V=3.622\times 10\\\\\Rightarrow V=36.22\ V[/tex]

Thus, the voltage around the capacitor is [tex]36.22\ V[/tex]

Answer:

The final velocity of the skier and boat is 0.33 m/s to the east.

Explanation:

We can find the final velocity of the skier by conservation of linear momentum:

[tex] m_{s}v_{s_{i}} + m_{b}v_{b_{i}} = m_{s}v_{s_{f}} + m_{b}v_{b_{f}} [/tex]

Where:  

[tex]m_{s}[/tex]: is the mass of the water skier = 62.0 kg

[tex]m_{b}[/tex]: is the mass of the boat = 775 kg

[tex]v_{s_{i}}[/tex]: is the initial velocity of the skier = 4.50 m/s (as she leaves the dock)

[tex]v_{b_{i}}[/tex]: is the initial velocity of the boat = 0 (it is at rest)

[tex]v_{s_{f}}[/tex]: is the final velocity of the skier =?

[tex]v_{b_{f}}[/tex]: is the final velocity of the boat =?

Since the final velocity of the skier is the same that the velocity of the boat ([tex]v_{f}[/tex]) we have:

[tex] m_{s}v_{s_{i}} + 0 = v_{f}(m_{s} + m_{b}) [/tex]

[tex]v_{f} = \frac{m_{s}v_{s_{i}}}{m_{s} + m_{b}} = \frac{62.0 kg*4.50 m/s}{62.0 kg + 775 kg} = 0.33 m/s                  

Therefore, the final velocity of the skier and boat is 0.33 m/s to the east.

I hope it helps you!                                                          

Answer:

152,155 J

Explanation:

115,333 + 36,822 = 152,155J

D. The new electric potential energy is 1/2as strong as the original

electric potential energy.

Given W (work) = 300J

F (force) = 75N

s (displacement) = ?

We know that, W = Fs

So,

300 J = 75N × s

= 300J/75N = s

= 4m = s

So, I traveled 4m.

Answer:

4

Explanation:

4-3

Answer:

19m/s

Explanation:

U= 4m/s

a = 0.5m/s²

t= 30s

V= U+at

= 4+15

=19m/s

Answer:

Option C

Explanation:

From the question we are told that:

Mass [tex]m=0.015kg[/tex]

Radius [tex]r=0.5m[/tex]

Time [tex]t=0.70[/tex]

Generally the equation for Tension is mathematically given by

 [tex]T=\frac{4mr}{t^2} \pi[/tex]

 [tex]T=\frac{4*0.015*0.5}{(0.70)^2} (3.142)[/tex]

 [tex]T=0.60N[/tex]

Therefore

[tex]T=0.60 N[/tex], toward the center of the circle

Option C

Answer:

0.60 N, toward the center of the circle

Explanation:

Got it right on the test.

Answer:

wats the following items

Answer:

0.5 m/s

Explanation:

In  solving this, we would be applying the principle of conservation of momentum. This states that momentum is conserved, and thus, Initial momentum has to be equal to the final momentum. This also means that it can be related by the formula

M * u = m * v, where

M = mass of the astronaut

m = mass of the wrench

u = speed of the astronaut

v = final speed of the wrench

70 * u m/s = 1.6 * 22 m/s

70u = 35.2 m/s

u = 35.2/70

u = 0.5 m/s

Answer:

Power

step by step explanation:

Answer:

1 is false 2 is true 3 is false

Answer:

A cyclotron is a type of particle accelerator.

Explanation:

Cyclotron frequency is the frequency of a charged particle moving perpendicular to the direction of a uniform magnetic field B, since that motion is always circular, the cyclotron frequency is given by equality of centripetal force and magnetic Lorentz force.

Answer:

0.16 h

Explanation:

Speed: 25km per hour

Speed=distance/time

25=4/t

t=4/25

t=0.16 hour

t=9.6 minuites

Brainliest please~~

Answer: Milky Way

Explanation:

Answer:

Look at explanation

Explanation:

a) Weight is another term for how much gravity is on an object and can be calculated by using the local gravity*mass so in order to find weight divide 1200 by g

m= 1200/9.8= 122.45kg

b) We know Fnet=ma and Fnet=Fapp+Fresistivity so Fapp+Fresistivity=ma

Plug in values

500+Fresistivity=122.45*2

solve for Fresistivty= 244.9-500=-255.1N (the reason it is negative is because it is in the opposite direction

c)Power= ΔE/Δt and we also know ΔE=ΔWork so Power= ΔWork/Δt. If a person pulls harder, they have a greater force and since mass is constant, acceleration is greater and since the amount of time needed to cover A to B is reduced since x-x0=v0t+1/2at²(v0=0, when you solve for t it will be lower because acceleration increases). If t decreases than Power increases by inverse proportionality. Work =Fd if the amount of Force increases by distance travelled remains constant than work also increases so power will also increase.

Answer:

Explanation:

This is a Law of Momentum Conservbation problem, where the total energy of the system cannot increase or decrease, only change form. The total energy equation for this situation is

TE = PE + KE where TE is total energy, PE is potential energy, and KE is kinetic energy. We begin by realizing that the go-kart is motionless at the top of a hill. If the kart isn't moving, then it has no KE, but if it is up off the ground and has the potential to fall to a point lower than it is curremtly, it has potential energy. That means that the total energy available to this go-kart is found in its potential energy and will not change throughout the trip's entirety. Thus,

TE = PE + 0 and

TE = (50.0)(9.8)(10.0) so

TE = 4900 J and since that's the total energy available throughout the trip, and we are looking to find the height of the next hill where this is both potential and kinetic energy, then

4900 = PE + KE and

4900 = (50.0)(9.8)(h) + [tex]\frac{1}{2}(50.0)(1.0)^2[/tex] and

4900 = 490h + 25.0 and

4875 = 490h so

h = 9.9 m (I kinda ignored the rules for significant digits at the end, which goes against every teacher's bone in my body, but nonetheless, there's your answer!)

Answer:

no

Explanation:

the equation can't be balanced because it doesn't have the same elements on each side of the equal sign.

The given equation is not a balanced chemical equation because it does not contain the same elements on both sides of the equation.

A chemical equation can be explained as the representation of a chemical reaction in terms of symbols of the substances. A balanced equation containing the same number of atoms of each element on either side of the equation.

The law of mass conservation is followed by every balanced chemical equation. By obeying this law, the total mass on the reactant side should be equal to the total mass on the product side in a balanced equation.

On both sides of the chemical equation, the same elements as well as an equal number of elements are present as the chemical reaction does not alter the identity of the elements. Therefore, the given equation is not a balanced chemical equation.

Learn more about the balanced chemical equation, here:

brainly.com/question/28294176

#SPJ2

Answer:

16 s

Explanation:

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

Mass (m) = 40 Kg

Velocity (v) = 4 m/s

Power (P) = 20 W

Time (t) =?

Next, we shall determine the energy. This can be obtained as follow:

Mass (m) = 40 Kg

Velocity (v) = 4 m/s

Energy (E) =?

E = ½mv²

E = ½ × 40 × 4²

E = 20 × 16

E = 320 J

Finally, we shall determine the time. This can be obtained as follow:

Power (P) = 20 W

Energy (E) = 320 J

Time (t) =?

E = Pt

320 = 20 × t

Divide both side by 20

t = 320 / 20

t = 16 s

Thus, the time is 16 s

Answer:

Please find detailed explanation of second class levers below

Explanation:

Levers are one of the classes of machine that possesses three levels namely: first class, second class and third claas. A second class lever is the level of levers in which the load (L) is in between the pivot (F) and the effort (E).

Examples of second class levers include; wheelbarrow, a bottle opener etc. In the bottle opener for example, the bottle lid (load) is in between the pivot of the opener and the hand opening it (effort).