How does the current in a resistor change if the voltage across the resistor is
increased by a factor of 2?

Answer:

d.

50 m

Explanation:

v² = u² + 2as

s = (v² - u²) / 2a

s =(25² - 0²) / (2(6.25))

s = 50 m

Explanation:

The sun is what makes the water cycle work. The sun provides what almost everything on Earth needs to go—energy, or heat. Heat causes liquid and frozen water to evaporate into water vapor gas, which rises high in the sky to form clouds... clouds that move over the globe and drop rain and snow.

Answer:

"The sun is what makes the water cycle work. ... Heat causes liquid and frozen water to evaporate into water vapor gas, which rises high in the sky to form clouds... clouds that move over the globe and drop rain and snow. This process is a large part of the water cycle."

Explanation:

Average Velocity

v = s/t

v = 300 km/(1 hour + 40/60 hour)

v = 300 km/(1.66 hour)

v = 180 km/hour

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a) Las balas de cañon disparadas desde el puerto deben tener una ángulo de 27.39° para que puedan impactar al barco, con una velocidad inicial de 82 m/s.

b) El tiempo de vuelo de las balas de cañon para alcanzar al barco que está a 560 m de distancia es de 7.69 s.

c) Sabiendo que el ángulo calculado en el inciso a) es para una distancia de 560 m, el barco debe estar a una distancia mayor para que las balas no lo alcancen.

a)

Podemos usar las ecuaciones de tiro parabólico para encontrar el ángulo que permita derribar al barco invasor.

[tex]x=\frac{v_{i}^{2}sin(2\alpha)}{g}[/tex] (1)

Donde:

Lo que debemos hacer es depejar α de la ecuación 1

[tex]sin(2\alpha)=\frac{xg}{v_{i}^{2}}[/tex]

[tex]sin(2\alpha)=\frac{560*9.81}{82^{2}}[/tex]

[tex]sin(2\alpha)=0.82[/tex]

[tex]\alpha=\frac{sin^{-1}(0.82)}{2}[/tex]

[tex]\alpha=27.39^{\circ}[/tex]

Por lo tanto, el ángulo para que el cañón impacte en el barco es de 27.39 °.

b)

Sabemos que la componente de la velocidad en el eje x es constante, así que podemo usar la siguiente ecuación.

[tex]v_{x}=\frac{x}{t}[/tex]

La componente x de la velocidad es V(x) = V(i)cos(α) y sabiendo la distancia total de 560 m, el tiempo será:

[tex]v_{i}cos(\alpha)=\frac{x}{t}[/tex]

[tex]t=\frac{x}{v_{i}cos(\alpha)}[/tex]

[tex]t=\frac{560}{82cos(27.39)}[/tex]

[tex]t=7.69\: s[/tex]

El tiempo total de vuelo de las balas de cañon es de 7.69 s.

c)

Sabemos que el ángulo calculado en el inciso a) es de 27.39 °, y ese valor fue calcualdo para una distancia de 560 m, por lo tanto el barco debe estar a una distancia mayor que esa para que las balas no lo alcancen.

Puedes encontrar más información sobre tiro parabólico aquí:

https://brainly.lat/tarea/3605927

Espero te haya sido de ayuda!

Answer:

None of the above

Explanation:

None of the above

P.s this language is: Viatnamese :)

Answer:

This is just to help the other readers help you. Or so that you understand:

Đây là để giúp bạn đọc giúp bạn!

What they tried to say:

Choose the wrong sentence:

A. The velocity-time graph of uniform linear motion is a line parallel to the horizontal axis Ot

B. In uniform linear motion, the time graphs of the coordinates and of the velocity are both straight lines

C. The time-coordinate graph of uniform rectilinear motion is always a straight line

D. The coordinate-time graph of uniform linear motion is an oblique line

Is what they asked you.

Thankyou and your welcome

Vì tôi đã giúp đỡ, tôi có thể vui lòng có cám ơn nào không?

Have a great day!

~Happy helper~

Answer:

deception

Explanation:

Answer:

D-it's never acceptable to ignore rules and etiquette

Explanation:

We should know that it's never acceptable to ignore rules and etiquette.

The rules of etiquette in sports are the rules of the sports that have been document by the various sports federations.

We should know that it's never acceptable to ignore rules and etiquette. These will make the the sports a lawless act and full of chaos.

Learn more about etiquette in sports:https://brainly.com/question/979191

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Answer:2.02s

Explanation:

At the y component, the tract of the ball is free falling body. The initial velocity at y component is 0 m/s.

Height = 1/2 * g * t^2

20 = 1/2 * 9.8 * t^2

t = (20 * 2 / 9.8)^1/2, t = 2.02s

Answer:

5 912 500 000 000 = 5.91x10^12 miles

Explanation:

The first thing we need to do is convert the light year distance to kilometres, since we have the kilometre to mile conversion. There are 1000m in a kilometre, so we need to divide the light year value that we have by 1000, giving us 9.46x10^12 km for 1 light year.

Now that we have our value in kilometres, we simply need to divide it by 1.6 to get it in miles. Doing so gives us 5 912 500 000 000 miles, or 5.91x10^12 miles.

Hope that made sense, and let me know if you have any further questions!

Answer:

Explanation:

jshshjabsjbrnejdjdhjsndjdhfisjshsidsnsndgjsbdjdvsjsvdudhwjgdusbsuwgshidsbhsgsuydisksvdudh

Answer:

You can't use energy conservation when mass is removed. This has required energy that you are not taking into account.

Answer:

because some time it dose not need to be i equation becasue it is not need

Explanation:

Answer:

2.5 × 10-⁴¹ Ns

Explanation:

Impulse

I = F × t

I = 5000 N × 5 × 10-⁴⁵ s

I = 25 × 10-⁴² Ns

I = 2.5 × 10-⁴¹ Ns

#LearnWithEXO

Explanation:

What are you searching for?

Potential energy?

Answer:

3018.4

Explanation:

So this velocity just before reaching the ground, the final equals square root of two times G times H. So gs 9.8 m per second squared H is 2.8 m. So it gives the velocity of 7.41 m per second squared. That's the velocity right before touching the ground. So to find the impulse required to come to rest here, we just do mass times the velocity minus zero. So zero is the final momentum, final velocity zero. So that makes the final momentum zero. So we multiply the mass with the velocity before touching the ground and after. And subject zero. That's the momentum after touching the ground. So we multiply mass 55 kg. At 7.41 velocity we got 407.45.  kilograms meters per second of impulse. we know that the force will be larger for stiff stiff flex compared to bend flex. Because the stiff flex. Bring the object to rest in a short amount of time and distance. So in part C, that's calculate for stiff flex. The stopping distance, Delta Y. Is 0.1 m. This one centimeter. So we can calculate the acceleration of the body when it comes to rest.  From the moment it touches the ground.  equals v squared over two times Delta Y. So  we use the square is 54.88. The 7.41 square divided by two times 0.01 gives us an explosion of 2744 m/s square. So you multiply this acceleration by the mass to get the force so force equals mass times acceleration gives us a large force,  159 20 newtons. Then for bent legs, repeat the same process with a different delta Y, delta Y of 200.5 m. So that's 50 centimeters and calculate acceleration the same process. Just use a different value of delta Y. And we get as much smaller acceleration 54.88 m per second square. And the force experience would just be mass times acceleration. That gives us 3018.4 newton's a lot smaller than for stiff flex.

Answer:

8. Friction

9. Net force

Explanation:

Hope this helps you. Have a nice day. :)

I HOPE IT WILL HELP YOU.

Thank you.

TEDDYLOVER

Answer:

Explanation:

I'm going to ASSUME that point A is at the bottom of a circular curve.

The track must support the weight of the car and also supply the needed centripetal force

F = mg + mv²/R

F = 585(9.81 + 18.0²/10) = 24,692.85

F = 24.7 kN

I'm going to ASSUME that point B is at the top of a circular curve.

The track must support the weight of the car and also supply the needed centripetal force.

At a minimum, the weight can be reduced to zero and all the gravity acceleration acts as centripetal acceleration.

0 = mg - mv²/R

        g = v²/R

         v = √Rg = √(15(9.81) = 12.1305399...

        v = 12.1 m/s

Answer:

graph b

Explanation:

Answer:

Explanation:

tanθ = opp/adj

adj = opp/tanθ

adj = (60 - 40)/tan55 = 14 m

Answer:

15kg

(i actually did this)

Explanation:

k12

hope this helped!

i hope you ace it!

Answer: B) 5g

I took the 4.11 Unit Test: Nutrition quiz.

24.8 m/s

Explanation:

You don't have to solve for time first. You can use the equation below to solve for the velocity before impact:

[tex]v_y^2 = v_{0y}^2 + 2a_yy[/tex]

We know that [tex]v_{0y} = 0[/tex] so the equation above becomes

[tex]v_y^2 = 2a_yy \Rightarrow v_y = \sqrt{-2a_yy}[/tex]

Plugging in the numbers, we get

[tex]v_y = \sqrt{-2(-9.8\:\text{m/s}^2)(31.3\:\text{m})}[/tex]

[tex]\:\:\:\:\:=24.8\:\text{m/s}[/tex]

METHOD 2:

If you insist on using that equation for d, we can do that too. So solving for t from the equation for d. we get

[tex]t = \sqrt{\dfrac{2d}{g}} = \sqrt{\dfrac{2(31.3\:\text{m})}{9.8\:\text{m/s}^2}}[/tex]

[tex]\:\:\:\:= 2.53\:\text{s}[/tex]

So using the equation [tex]v_y = gt,[/tex] we get

[tex]v_y = (9.8\:\text{m/s}^2)(2.53\:\text{s}) = 24.8\:\text{m/s}[/tex]

So you get the same result regardless of the method you use.

[tex]\\ \sf\longmapsto W=Fd[/tex]

[tex]\\ \sf\longmapsto d=\dfrac{W}{F}[/tex]

[tex]\\ \sf\longmapsto d=\dfrac{9}{12}=0.72m[/tex]

Answer:

Anne pushed a cart by 12 N.

● Force = 12 N

Work done is 9 J

● Work = 9 J

=> Work = Force × distance

[tex]=> distance = \frac{work}{force} [/tex]

[tex]=> distance = \frac{9}{2} [/tex]

[tex]=> distance = 0.72m[/tex]

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Explanation:

I will definitely explain you if you follow me please

50 mL of water have to poured to the glass bottle of 100 g to increase the stretching of the spring from 1.8 cm to 2.7 cm.

A spring can hang when it loads a weight. The stretching of spring is the change in length of the spring end from its initial length.  Spring balance works with the principle of Hooks law.

The stretching is made by the elasticity of the spring which make it in a wave like motion.

Given that the glass bottle with 100 g weight when loaded on the spring scale, it will stretch down to 1.8 cm. Then the weight required to make a stretch of 2.7 cm is calculated as follows:

weight required  = (100 g × 1.8 cm) /2.7 cm

                            = 150 g.

A total of 150 g is needed to stretch into 2.7 cm. Thus we have to add 50 g of water to the glass weighing 100 g . 50 g is equivalent to 50 ml, since density of water is 1 g/L.

Hence, 50 ml of water should be poured to the glass.

To find more about spring scale, refer the link below:

https://brainly.com/question/2568071

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The Correct choice is :

This question involves the concept of tension, centripetal force, centrifugal force, and weight.

The tension of the stone at the highest point is "0 N".

The centrifugal force always acts away from the circle. The tension always acts toward the support that is the center of the circle in this case. When the stone is at its lowest point, the centrifugal force, and the weight of the stone, both act downward. But the tension is pointing in the upward direction. Hence, the equilibrium equation will be:

[tex]Tension = Weight\ + \ Cenrifugal\ Force\\T = W + F_c\\T = 2\ W\ (given)\\2\ W = W+F_c\\F_c = 2\ W - W\\F_c = W[/tex]

Now, we consider the stone at the top position. Here, the centrifugal force will act in the upward direction (away from the center of the circle), while the weight and tension will be acting in the downward direction. Thus, the equilibrium equation, in this case, will be:

[tex]F_c = W+T\\T=F_c-W\\T = W-W\\[/tex]

T = 0 N

Learn more about tension here:

https://brainly.com/question/2287912?referrer=searchResults

The attached picture shows the free body diagram of the stone at different locations.