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2 years ago

An electric current through a wire is 6.00 C every 2.00 s. What is the magnitude of this current?

Physics

1 answer:

musickatia [10]2 years ago

6 0

Answer:

I=6 lakh

T=2 sec

Q=?

Now,

according to formula,

I=Q/T

6=Q/2

6 multiple by 2= Q

hence,Q=12 answer..

You might be interested in

which has a greater kinetic energy, a bowling ball that has a mass of 5kg travelling at 6m/s, or a ship that has a mass of 12000

AVprozaik [17]

Answer:bowling ball has greater kinetic energy

Explanation:

Kinetic energy of bowling ball:

mass=m=5kg

Velocity=v=6m/s

Kinetic energy =ke

Ke=0.5 x m x v x v

Ke=0.5 x 5 x 6 x 6

Ke=90J

Kinetic energy of ship:

mass=m=120000kg

velocity=v=0.02m/s

Ke=0.5 x m x v x v

Ke=0.5 x 120000 x 0.02 x 0.02

Ke=24J

6 0

2 years ago

Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory and notices that it lands,

Elenna [48]

Let $u$ be the initial velocity of the soccer ball at an angle of inclination of $\theta_0$ with the positive x-axis.

Given that:

$\theta_0=45^{\circ}$

The horizontal distance covered by the projectile=20 m

Time of flight, $t_f=2$ seconds

Acceleration due to gravity, $g= 10 m/s^2$ downward.

As "north" and "up" as the positive x ‑ and y ‑directions, respectively.

So, $g= -10 m/s^2$

As the acceleration due to gravity is in the vertical direction, so the horizontal component of the initial velocity remains unchanged.

The x-component of the initial velocity, $u_x=u\cos\theta_0$ .

The horizontal distance covered by the projectile $= u_x\times t_f$

$\Rightarrow u_x\times t_f=20$

$\Rightarrow u_x\times 2=20$

$\Rightarrow u_x=10$ m/s

So, the horizontal component of the velocity is 10 m/s which is constant and the graph has been shown in the figure (i).

Now, $u\cos(45^{\circ})=10$ [as $u_x=u\cos\theta_0$ ]

$\Rightarrow u=10\sqrt{2}$ m/s.

The vertical component of the initial velocity,

$u_y= u\sin\theta_0$

$\Rightarrow u_y=10\sqrt{2}\sin(45^{\circ})$

$\Rightarrow u_y=10$ m/s

Let v be the vertical component of the velocity at any time instant t.

From the equation of motion,

$v=u+at$

where u: initial velocity, v: final velocity, a: constant acceleration, and t: time taken to change the velocity from u to v.

In this case, we have $u=u_y, a= -10 m/s^2$ .

So at any time instant, t.

$v=u_y+(-10)t$

$\Rightarrow v=10-10t$

The vertical component of the velocity, v, is the function of time and related as $v=10-10t$ .

This is a linear equation.

At 2 second, the vertical component of the velocity

v=10-10x2=-10 m/s.

The graph has been shown in figure (ii).

7 0

2 years ago

At 1 atm pressure, the heat of sublimation of gallium is 277 kj/mol and the heat of vaporization is 271 kj/mol. to the correct n

Andreyy89

Below are the choices that can be found elsewhere:

a. 268 kJ
<span>b. 271 kJ </span>
<span>c. 9 kJ </span>
<span>d. 6 kJ
</span>
So the key thing to realize here is what the information given to you actually means. Sublimation is going from a sold to a gas. Vaporization is going from a liquid to a gas. Hence you can create two equations from the information that you have:

<span>Ga (s) --> Ga (g) delta H = 277 kJ/mol </span>

<span>Ga (l) --> Ga (g) delta H = 271 kJ/mol </span>

<span>From these two equations, you can then infer how to get the melting equation be simply finding the difference between the sublimation (two steps) and vaporization (one step). </span>

<span>Ga (s) --> Ga (l) delta H = 6 kJ/mol </span>

<span>At this point, all you need to do is a bit of stoichiometry. You start with 1.50 mol and multiply by the amount of energy per mole (6 kJ/mol). </span>

<span>*ANSWER* </span>
<span>9 kJ/mol (C)</span>

7 0

2 years ago

Which statement is not a good practice when working inside a computer case?

lozanna [386]

Answer:

a. be sure to hold expansion cards by the edge connectors

Explanation:

Removal of loose jewelry is a good safety practice. Also not touching a microchip with a magnetized screwdriver is also a good practice.

But holding expansion cards by the edge connectors is not a good practice, so it is the odd one in the question. Therefore answer option a provides the correct and best answer to the question

3 0

2 years ago

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

2 years ago