What is the is the 94th element of the periodic table

A current of 2.0A flows through a light bulb. What is the amount of charge flowing through the light bulb in 40 seconds?

lidiya [134]

Answer:

Quantity of charge = 80 Coulombs

Explanation:

Given the following data;

Current = 2 A

Time = 40 seconds

To find the amount of charge flowing through the light bulb;

Mathematically, the quantity of charge passing through a conductor is given by the formula;

Quantity of charge = current * time

Substituting into the formula, we have;

Quantity of charge = 2 * 40

Quantity of charge = 80 Coulombs

5 0

2 years ago

The parking brake on a 1200kg automobile has broken, and the vehicle has reached a momentum of 7800kg.M/s. What is the velocity

AysviL [449]

Ok so the equation for momentum is:
v=p/m

So you would do:
7800/1200=6.5

So the answer is:
6.5 m/s

Hope this helps :)

6 0

3 years ago

DOES ANYONE KNOW WAHTS NOISE POLLUSION POEM?

solmaris [256]

Noise pollution, also known as environmental noise or sound pollution, is the propagation of noise with ranging impacts on the activity of human or animal life, most of them harmful to a degree.

7 0

2 years ago

Light with a frequency of 7.30 x 1014 hz lies in the violet region of the visible spectrum. What is the wavelength of this frequ

Leona [35]

From the theory we know that:

c = λ / T

f = 1 / T

Where:

c = 3. $10^{8}$ / m (the speed of light)

λ is the wavelengh (in meters)

T is the period (in seconds)

f is the frequency (in Hz)

We were told that:

f = 7.30 . $10^{14}$

And we want to find out the value of λ.

c = λ / T

c = λ . 1/T

Swaping 1/T = f

c = λ . f

λ = c / f

λ = 3 . $10^{8}$ / 7.30 . $10^{14}$

λ = 4.12 $10^{-7}$ m

Response: 4.12 $10^{-7}$ m = 412 nm

:-)

6 0

3 years ago

Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 32.0°below the horizontal. If it st

scoray [572]

The figure of the problem is missing: find in attachment.

(a) 1.64 s

The ball follows a projectile motion path. The horizontal displacement is given by

$x(t) = v_0 cos \theta t$

where

$v_0$ is the initial speed

t is the time

$\theta=32.0^{\circ}$ is the angle below the horizontal

We can rewrite this equation as

$t=\frac{x(t)}{v_0 cos \theta}$ (1)

The vertical displacement instead is given by

$y(t) = -v_0 sin \theta t - \frac{1}{2}gt^2$ (2)

where

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting (1) into (2),

$y(t) = -x(t) tan \theta - \frac{1}{2}gt^2$

We know that for t = time of flight, the horizontal displacement is

$x(t) =50.8 m$

We also know that the vertical displacement is

$y(t) = -45 m$

Substituting everything into the equation, we can find the time of flight:

$\frac{1}{2}gt^2=-y -x tan \theta\\t=\sqrt{\frac{2(-y-xtan \theta)}{g}}=\sqrt{\frac{2(-(-45)-50.8 tan 32.0^{\circ})}{9.8}}=1.64 s$

(b) 36.5 m/s

We can now find the initial speed directly by using the equation for the horizontal displacement:

$x(t) = v_0 cos \theta t$

where we have

x = 50.8 m

$\theta=32.0^{\circ}$

Substituting the time of flight,

t = 1.64 s

We find:

$v_0 = \frac{x}{t cos \theta}=\frac{50.8}{(1.64)(cos 32.0^{\circ})}=36.5 m/s$

(c) 47.1 m/s at 48.8 degrees below the horizontal

As the ball follows a projectile motion, its horizontal velocity does not change, so its value remains equal to

$v_x = v_0 cos \theta = (36.5)(cos 32.0^{\circ})=31.0 m/s$

The initial vertical velocity is instead

$u_y = -v_0 sin \theta = -(36.5)(sin 32.0^{\circ})=-19.3 m/s$

And it changes according to the equation

$v_y = u_y -gt$

So at t = 1.64 s (when the ball hits the ground),

$v_y = -19.3 - (9.8)(1.64)=-35.4 m/s$

So the impact speed is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(31.0)^2+(-35.4)^2}=47.1 m/s$

While the direction is:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-35.4}{31.0})=-48.8^{\circ}$

8 0

3 years ago