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3 years ago

What happens to a circuit's resistance (R), voltage (V), and current (1) when

Physics

2 answers:

klemol [59]3 years ago

7 0

Answer:

R and I will change, V will remain constant

Explanation:

:) I just did the quiz (and have screenshots for proof)

Naya [18.7K]3 years ago

4 0

Answer:

Explanation:

Hope this helps

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Two equally charged, 2.807 g spheres are placed with 3.711 cm between their centers. When released, each begins to accelerate at

statuscvo [17]

$\begin{gathered} m=\text{ 2.807 g} \\ d=\text{ 3.711cm} \\ a=260.125\text{ m/s}^2 \\ m=\text{ mass of both the spheres} \\ d=\text{ distance between the centers of sphere.} \\ a=\text{ acceleration of spheres.} \end{gathered}$ $\begin{gathered} force\text{ due to the sphere having charge q, outside its surface is given by } \\ \vec{F}=\frac{1}{4\pi\epsilon_o}\frac{q_1q_2}{r^2}\hat{r} \\ q_1=charge\text{ on the source object.} \\ q_2=charge\text{ of the object in which we are observing the force.} \\ F=\text{ the force on the charged particle outside the sphere} \\ r=\text{ distance of the charged particle from the center of the sphere} \\ \hat{r}\text{= direction of the force acting on the charged particle} \end{gathered}$ $\begin{gathered} from\text{ Newton's second law} \\ F=ma \\ F=\text{ force acting on the particle.} \\ m=\text{ mass of the object.} \\ a=\text{ acceleration of the object.} \end{gathered}$ $\begin{gathered} from\text{ both the equation } \\ ma=\frac{1}{4\pi\epsilon_o}\frac{q_1q_{\frac{2}{}}}{r^2}\hat{r} \\ here\text{ q}_1\text{ and q}_2\text{ are the same, according to the question.} \end{gathered}$ $\begin{gathered} converting\text{ all the values in s.i. unit} \\ m=2.807*10^{-3}kg \\ d=3.711*10^{-2}m \\ according\text{ to the question q}_1=\text{ q}_2 \\ value\text{ of }\frac{1}{4\pi\epsilon_o}=9*10^9\text{ Nm}^2\text{/C}^2 \\ now\text{ put all the values in the above equation } \\ 2.807*10^{-3}kg*260.125\text{ m/s\textasciicircum2}=9*10^9Nm^2\text{/C}^2*\frac{q^2}{3.711*10^{-2}m} \\ \end{gathered}$ $\begin{gathered} by\text{ trasformation} \\ q=\sqrt{\frac{2.807*10^{-3}kg*260.125m/s^2*3.711*^10^{-2}m}{9*10^9Nm^2\text{/C}^2}} \\ by\text{ solving this we get } \\ q=17.3514*10^{-7}C \\ q=1.73514\text{ micro coulombs.} \end{gathered}$ Hence the correct answer is q= 1.73514 micro coulombs.

5 0

1 year ago

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizo

Norma-Jean [14]

Answer:

0.528m

Explanation:

a)58.7 cm = 0.587 m

Let g = 9.8m/s2. When the frog jumps from ground to the highest point its kinetic energy is converted to potential energy:

$E_p = E_k$

$mgh = mv^2/2$

where m is the frog mass and h is the vertical distance traveled, v is the frog velocity at take-off

$v^2 = 2gh = 2*9.8*0.587 = 11.5$

$v = \sqrt{11.5} = 3.4 m/s$

b) Vertical and horizontal components of the velocity are

$v_v = vsin(\alpha) = 3.4sin(58^0) = 2.877 m/s$

$v_h = vcos(\alpha) = 3.4cos(58^0) = 1.8 m/s$

The time it takes for the vertical speed to reach 0 (highest point) under gravitational acceleration g = -9.8m/s2 is

$\Delta t = \Delta v / g = \frac{0 - 2.877}{-9.8} = 0.293s$

This is also the time it takes to travel horizontally, we can multiply this with the horizontal speed to get the horizontal distance it travels

$s_h = v_ht = 1.8*0.293 = 0.528 m$

3 0

3 years ago

The diagram below shows the reactants and products of a chemical reaction. What property of reactions does this diagram illustra

Paladinen [302]

Based on the chemical reaction, Carbon reacts with Oxygen would yield Carbon Dioxide, we can imply that the "<span>Atoms are always conserved during chemical reactions." It is one of the fundamental concepts in chemistry wherein the mass is always conserved in a reaction since it can't be created nor destroyed.</span>

5 0

3 years ago

• (-/1 Points) DETAILS OSCOLPHYS2016 9.5.WA.040.

Tanya [424]

The force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.

<h3>Force exerted outside the wheel</h3>

The force exerted on the outside of the wheel can be determined by applying the principle of conservation of angular momentum as shown below.

∑τ = 0

Let the distance traveled by the load = 1.5 m
Let the radius of the wheel or position of the force = 0.45 m

∑τ = R(mg) - r(F)

rF = R(mg)

0.45F = 1.5(21,200 x 9.8)

F = 6.925 x 10⁵ N.

Thus, the force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.

Learn more about angular momentum here: brainly.com/question/7538238

6 0

3 years ago

If it takes a planet 2.8 Ã— 108 s to orbit a star with a mass of 6.2 Ã— 1030 kg, what is the average distance between the planet

Shtirlitz [24]

The average distance between the planet and the star is:

R=9.36*10^11 m

Orbital velocity v=√{(G*M)/R},

G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,

M = mass of the star

R =distance from the planet to the star.

v=ωR, with ω as the angular velocity and R the radius

ωR=√{(G*M)/R},

ω=2π/T,

T = orbital period of the planet

To get R we write the formula by making R the subject of the equation

(2π/T)*R=√{(G*M)/R}

{(2π/T)*R}²=[√{(G*M)/R}]²,

(4π²/T²)*R²=(G*M)/R,

(4π²/T²)*R³=G*M,

R³=(G*M*T²)/4π²,

R=∛{(G*M*T²)/4π²},

Substitute values

R=9.36*10^11 m

As was already said, Earth is located roughly 150 million kilometres (93 million miles) from the Sun on average. It is 1 AU. Mars is on our fictitious football field's three-yard line. On average, the distance between the Sun and the red planet is around 142 million miles (228 million kilometres).

Learn more about average distance:

brainly.com/question/18366547

#SPJ4

The complete question is ''If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 10^30 kg, what is the average distance between the planet and the star? 1.43 × 10^9 m 9.36 × 10^11 m 5.42 × 10^13 m 9.06 × 10^17 m''.

4 0

1 year ago