Three lights are connected to a power source by a series circuit. All three lights are turned on. However, when one of the lights is removed, the other two go out. Why does this happen? In a series circuit, all of the lights share the same power source. So when one light is removed, there is an imbalance in the power to the other lightsbIn a series circuit, all the lights can be treated as one very bright light. So when one light is removed, the other lights have to dim to compensate for itcIn a series circuit, all the lights are connected on their own separate paths. So when one light goes out, the others go dark to remove it from the path it was ondIn a series circuit, all the lights are on the same wire. So when one light is removed, the path is broken and the current cannot get to the other lightsI think its c

Every location you measure from, it's not accelerating, can be regarded of as part of a frame of reference. Motion is essentially another method for that various perspectives just motion single item occasionally exist.

The position of an object was expressed in reference to both the perceiver's point of view and another item's position. That cat, the house, as well as the perceiver themselves are all three reference points in the sentence "The cat is what's left of the house."

The vantage point from which you look at and gauge things is known as a reference frame. It is employed to describe how or where an object is moving. The identical circumstance can be viewed through two different reference frames by two persons. If this occurs, they will view an object's motion and position in a different way.

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Answer:

20*365=7300

Explanation:

1 Year=365day

20years = x days

Answer:

D: unconscious

Explanation:

Apex have a nice day :)

Answer: D. Unconscious

Explanation:

a p e x

At the highest attitude, the  velocity of  the ball is 0 m/s, so the kinetic energy is 0 as well.

Hence the answer is potential energy because it doesn't depend on velocity .

Answer:

\(\theta= (-74.42)^{\circ} C\)

Explanation:

Horizontal speed of water, \(v_{xf}=9\ m/s\)

Height, h = -53 (below pool)

We can find firstly the final vertical speed of the water using third equation of kinematics. So

\(v^2_{yf}=u^2_{yi}+2(-g)h\\\\v^2_{yf}=2\times -9.8\times -53\\\\v_{yf}=32.23\ m/s\)

Let \(\theta\) is the angle where the falling water moving as it enters the  pool. So,

\(\tan\theta=\dfrac{v_{yf}}{v_{xf}}\\\\=\dfrac{-32.3}{9}\\\\=-74.42^{\circ} C\)

Hence, the angle is (-74.42)°C.

The velocity of the combined lump after the collision is 39.5 cm/s, which is the average velocity of the two lumps before the collision.

To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of a system before a collision is equal to the total momentum of the system after the collision, provided there are no external horizontal forces acting on the system.

The momentum of an object is equal to its mass multiplied by its velocity. Therefore, we can calculate the total momentum of the system before the collision as:

Total momentum before collision = (0.50 g) × (-20.0 cm/s) + (70.0 g) × 40.0 cm/s

= -10.0 g·cm/s + 2800.0 g·cm/s

= 2790.0 g·cm/s

Since the two lumps stick together after the collision, their masses combine to form a single lump. Let's call the velocity of the combined lump after the collision "v". We can then calculate the total momentum of the system after the collision as:

Total momentum after collision = (0.50 g + 70.0 g) × v

= 70.50 g × v

According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Therefore, we can equate these two expressions and solve for "v":

Total momentum before collision = Total momentum after collision

2790.0 g·cm/s = 70.50 g × v

v = 2790.0 g·cm/s ÷ 70.50 g

v = 39.5 cm/s

This result can be explained by the fact that, in the absence of external horizontal forces, the momentum of the system is conserved, and the total mass of the system remains constant.

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Answer:

Reduction of order delivery times.

More efficient order fulfilment.

Lower number of incomplete orders.

Improved service in emergency situations.

Reduction of errors in the composition of orders and deliveries.

Better management of internal quality levels.

Greater flexibility (wider range of services on offer to the client)

Explanation:

Answer:

B. False

Explanation:

Health-related components of fitness are not just important for people who do not play competitive sports.

Answer:

The frequency of light can be calculated using the formula:

`c = λv`

Where `c` is the speed of light in a vacuum, `λ` is the wavelength of light, and `v` is the frequency of light.

The speed of light in a vacuum is `3.00 × 10^8 m/s`.

To convert the wavelength from nanometers to meters, we need to divide by `1 × 10^9`.

Thus, the frequency of light with a wavelength of 655 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(655 × 10^-9 m)`

`v = 4.58 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`.

Similarly, the frequency of light with a wavelength of 515 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(515 × 10^-9 m)`

`v = 5.83 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz`.

Finally, the frequency of light with a wavelength of 475 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(475 × 10^-9 m)`

`v = 6.32 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.

So, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz` and the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.

Answer:

115.12

Explanation:

-2.8=v-122.4/2.6

v=115.12

The distance of your hometown from school is 318.64 km.

Given data

Distance S = 190 km

Speed v = 97 km / h

Now, the time taken to travel S distance is calculated as,

t = S / v

t = 180 / 95

t = 1.958 hrs

Now, the total time is given as

Total time T = 4.0 h

So, remaining time t ' = T - t = 4 hrs - 1.958 hrs = 2.042 hrs

Now, the velocity after travel 190 km is v ' = 63 km/ h

Distance travel in this velocity S ' = v ' t '

= 128.6 km

Now, the distance of your hometown from school , S " = S + S '

= 190 km + 128.6 km = 318.64 km

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When a 2.5 kg sledgehammer hit a cement block with a force of 6.0 Newtons. The force the sledgehammer exerted on the cement block is greater in magnitude and opposite in direction to the force the cement block exerted on the sledgehammer.

This refers to forces that act on an object in opposite directions. The net force is gotten by solving for the difference between the two forces.

When the opposing forces are equal or balanced, the net force is zero.  The sledgehammer hits with a force and the cement block is receiving the impart as a stationary object.

Obviously, the force the sledgehammer exerted on the cement block is greater in magnitude and opposite in direction to the force the cement block exerted on the sledgehammer.

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The concept of an inverse squared relationship explains the behavior of intensity of light and distance of the light source. As the distance of the light source increases, the intensity of light decreases.

The law of inverse square law says that intensity equals the inverse of the square of the distance from the source.

Mathematically, the formula for the inverse square law is given as;

I ∝ 1/r²

where;

The intensity of light can be defined as the measure of the amount of power emitted by a given surface or reflected by the surface.

Thus, the inverse square law is all about the inverse relation between the  intensity of light or power output of a given surface to the distance of the light source.

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The velocity of the punk at time t = 1.5 s is equal to 2.47 m/s.

The object will have two velocities during the projectile motion, one is horizontal velocity and the other is vertical velocity. The horizontal velocity of the object is constant throughout the projectile motion but the vertical velocity is changed in the projectile motion.

Given that the velocity vector is making an angle θ with a horizontal table, θ = 35°

The vertical component of the velocity of the puck can be determined as:\(u_x = u Sin 35^o = 3.60 \times 0.573 =2.06 m/s\)

The horizontal component of the velocity of the puck can be determined as:

\(u_y = u cos 35^o = 3.60 \times 0.82=2.95 m/s\)

Given the acceleration aₓ = -0.360 m/s² and a y = -0.980 m/s²

The final vertical components of the velocity of the punk are:

\(v_x = u_x + a_xt\)

\(v_x = 2.95 - (0.360) (1.5)\\v_x = 2.41 m/s\)

and, the horizontal velocity of the punk is:

\(v_y = u_y +a_yt\\v_y = 2.06 - (0.980) (1.5)\\v_y = 0.59 m/s\)

The velocity of the punk can be determined as:

\(v = \sqrt{v_x^2 + v_y ^2}\\ v =\sqrt{(5.80)^2 + (0.35)^2}\\ v = 2.47 m/s\)

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Answer:

4.41N or 4.5N (check explanation)

Explanation:

450g = 0.45kg

F = ma

Using 10m/s² = 10(0.45) = 4.5N

Using 9.8m/s² = 9.8(0.45) = 4.41N

Answer:

dont die bro

Explanation:

Answer:

Two cars of equal weight and braking ability are travelling along the same road but combined with other factors it could mean the difference between life.

The potential energy of the roller coaster is 176,400 J (joules).

The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.

In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).

Using the formula, we can calculate the potential energy:

PE = mgh

= (900 kg)(9.8 \(m/s^2\))(20 m)

= 176,400 J

Therefore, the potential energy of the roller coaster is 176,400 J (joules).

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A. The average speed you can use to pull the safe is 1.02 m/s

B. The time needed to pull the safe up is 17.65 s

First, we shall obtain the force. This is shown below:

F = mg

F = 60 × 9.8

F = 588 N

Finally, we shall obtain the average speed. Details below:

Power = force × average speed

600 = 588 × average speed

Divide both sides by 588

Average speed = 600 / 588

Average speed = 1.02 m/s

The time needed to pull the safe up can be obtained as follow:

Time = Total distance / average speed

Time = 18 / 1.02

Time = 17.65 s

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Complete question:

You know you can provide 600 W

of power to move large objects. You need to move a 60-kg

safe up to a storage loft, 18 m

above the floor.

Part A

With what average speed can you pull the safe straight up?

Part B

What is the time needed to pull the safe up?

Answer:

there is no net force acting on the object

Explanation:

If an object is moving with a constant velocity, then by definition it has zero acceleration. So there is no net force acting on the object. The total work done on the object is thus 0 (that's not to say that there isn't work done by individual forces on the object, but the sum is 0 ).

thank ya :>

For a photographer that wishes to determine the color of light that he can use in a dark room that will not expose the films he is processing, having used a Blue Incandescent bulb, he should proceed to use a Red Incandescent bulb for the next trial.

The photographer in question is performing an experiment. For these kinds of experiments it is important to identify the variables present, which can be of three kinds:

For this experiment, the dependent variable is the exposure of the light onto the films, given that this is what we wish to measure. The independent variable will be the color of the light being used which is what will affect the dependent variable.

The remaining variable must be the control variable. Unlike the previous variables, we can have more than one of these. The control variable is there to make sure that only the dependent variable is affecting the outcome. We do this by keeping the control variable the same through each trial, which is why the photographer should not change the type of bulb in the second experiment, changing only the color of the light.

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Answer:

F = 64800 N

Explanation:

Given that,

Initial speed of a car, u = 70 km/h = 19.44 m/s

Finally it comes to a stop, v = 0

Time,t = 450 ms

The mass of the car is 1500 kg

We need to find the force exerted on the car. The force exerted on an object is given by :

F = ma

So,

\(F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{1500\times (19.44-0)}{450\times 10^{-3}}\\\\F=64800\ N\)

So, the required force is equal to 64800 N.

Answer:

-54,200 m/s^2

Explanation:

a=(vf-vi)/t

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

The near point of a human eye is about a distance of 25 cm.

The closest distance that an object may be viewed clearly without straining is known as the near point of the eye.

This distance (the shortest at which a distinct image may be seen) is 25 cm for a typical human eye.

The closest point within the accommodation range of the eye at which an object may be positioned while still forming a focused picture on the retina is also referred to as the near point.

In order to focus on an item at the average near point distance, a person with hyperopia must have a near point that is further away than the typical near point for someone of their age.

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To set an object in motion on a surface, it usually requires overcoming several factors. The primary requirement is to overcome the force of static friction between the object and the surface.

Static friction is the force that resists the motion of an object when it is at rest. The magnitude of static friction depends on the nature of the surfaces in contact and the normal force acting perpendicular to the surface. In order to overcome static friction and initiate motion, an external force must be applied to the object.

The minimum amount of force required to overcome static friction is known as the threshold force. This force must exceed the maximum static frictional force. Once the threshold force is applied, the object transitions from rest to motion, and the static friction changes to kinetic friction, which is typically lower than static friction.

The exact amount of force required to initiate motion varies depending on factors such as the weight of the object, the roughness of the surfaces, and any other external forces acting on the object. Smooth surfaces and lighter objects generally require less force to set them in motion compared to rough surfaces and heavier objects.

To overcome the force of static friction and set an object in motion, one can apply a force in the direction of desired motion, either by pushing, pulling, or applying a torque. Once the object is in motion, it continues to move with a constant velocity unless acted upon by external forces such as friction or gravity.

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Answer:

The atomic nucleus is the small, dense region consisting of protons and [[]]s at the center of an atom

Explanation:

Answer: The capacitor is fully charged when the voltage of the power supply is equal to that at the capacitor terminals. This is called capacitor charging; and the charging phase is over when current stops flowing through the electrical circuit.

Answer:

That is Melon Collision and the second blank is Lone Collision.

Explanation:

Answer:

Approximately \(8.17\; \rm m \cdot s^{-1}\) assuming that the effect of gravity on the box can be ignored.

Explanation:

If the force \(F\) is constant, then the work would be found with \(W = F \cdot \Delta x\). However, this equation won't work for this question since the

\(\displaystyle W = \int\limits_{x_0}^{x_1} F\, d x\),

For this particular question, \(x_0 = 0\; \rm m\) and \(x_1 = 14.0\; \rm m\). Apply this equation:

\(\begin{aligned}W &= \int\limits_{x_0}^{x_1} F\, d x \\ &= \int\limits_{0\; \rm m}^{14.0\; \rm m} \left[{18.0\; \rm N} - {\left(0.530\; {\rm N \cdot m^{-1}}\right)}\cdot x \right]\, d x \\ &= \left[{(18.0\; \rm N)}\cdot x - \frac{1}{2}\;{\left(0.530\; {\rm N \cdot m^{-1}}\right)}\cdot x^2\right]_{x = 0\; \rm m}^{x = 14.0\; \rm m} \approx 200.06\; \rm N \cdot m\end{aligned}\).

(Side note: keep in mind that \(1\; \rm J = 1\; \rm N\cdot m\).)

Since friction is ignored, all these work should have been converted to the mechanical energy of this object.

Assume that the effect of gravity on this box can also be ignored. That way, there won't be a change in the gravitational potential energy of this object. Hence, all these extra mechanical energy would be in the form of the kinetic energy of this box.

That is:

\(\begin{aligned}& \text{Kinetic energy of this object} \\ =& \text{Initial Kinetic Energy} + \text{Change in Kinetic Energy} \\ =& \text{Initial Kinetic Energy} + \text{Change in Mechanical Energy} \\ =& \text{Initial Kinetic Energy} + \text{External Work} \\=& 0\; \rm N \cdot m + 200.06\; \rm N \cdot m \\ =& 200.06\; \rm N \cdot m \end{aligned}\).

Keep in mind that the kinetic energy of an object of mass \(m\) and speed \(v\) is:

\(\displaystyle \frac{1}{2}\, m \cdot v^{2}\).

Therefore:

\(\begin{aligned}v &= \sqrt{\frac{2\, (\text{Kinetic energy})}{m}} \\ &= \sqrt{\frac{2\times 200.06\; \rm N \cdot m}{6.00\; \rm kg}} \approx 8.17\; \rm m \cdot s^{-1}\end{aligned}\).