Can someone plz help me on this, it would me alot to me <br> thx

What needs to be done to this circuit so that the light bulb lights up?

Pachacha [2.7K]

The answer is the first one a

5 0

3 years ago

70 POINTS!!!! PLZ ANSWER!!!!! The Newton-Clark moraine is located on the volcano Mount Hood. It contains approximately 950 milli

Strike441 [17]

A jumble of relatively young volcanic debris, some of it located where it fell in Mount Hood’s eruptive past, some of it moved here by the colossal advance of the Newton Clark Glacier during the last ice age.
Newton Clark Moraine
As a result, the rocks making up the moraine are sharp and raw, not rounded, and the debris is largely unsorted. Giant boulders perch precariously atop loose rubble, making the moraine one of the most unstable places on the mountain.

8 0

4 years ago

Read 2 more answers

NEED ANSWER ASAP!!!

xenn [34]

Answer:

d

Explanation:

it has high pressure of speed

3 0

3 years ago

A 1.20 kg solid ball of radius 40 cm rolls down a 5.20 m long incline of 25 degrees. Ignoring any loss due to friction, how fast

sladkih [1.3K]

Answer:

1.6s

Explanation:

Given that A 1.20 kg solid ball of radius 40 cm rolls down a 5.20 m long incline of 25 degrees. Ignoring any loss due to friction,

To know how fast the ball will roll when it reaches the bottom of the incline, we need to calculate the acceleration at which it is rolling.

Since the frictional force is negligible, at the top of the incline plane, the potential energy = mgh

Where h = 5.2sin25

h = 2.2 m

P.E = 1.2 × 9.8 × 2.2

P.E = 25.84 j

At the bottom, K.E = P.E

1/2mv^2 = 25.84

Substitutes mass into the formula

1.2 × V^2 = 51.69

V^2 = 51.69/1.2

V^2 = 43.07

V = 6.56 m/s

Using the third equation of motion

V^2 = U^2 + 2as

Since the object started from rest,

U = 0

6.56^2 = 2 × a × 5.2

43.07 = 10.4a

a = 43.07/10.4

a = 4.14 m/s^2

Using the first equation of motion,

V = U + at

Where U = 0

6.56 = 4.14t

t = 6.56/4.14

t = 1.58s

Therefore, the time the ball rolls when it reaches the bottom of the incline is approximately 1.6s

6 0

3 years ago

A 2.0-mm-diameter copper ball is charged to 40 nC . What fraction of its electrons have been removed? The density of copper is 8

murzikaleks [220]

Answer:

0.02442 × 10⁻⁹

Explanation:

Given:

Diameter of copper ball = 2.00 mm = 0.002 m

Charge on ball = 40 nC = 40 × 10⁻⁹ C

Density of copper = 8900 Kg/m³

Now,

The number of electrons removed, n = $\frac{\textup{Charge on ball}}{\textup{Charge of an electron}}$

also, charge on electron = 1.6 × 10⁻¹⁹ C

Thus,

n = $\frac{40\times10^{-9}}{1.6\times10^{-19}}$

or

n = 25 × 10¹⁰ Electrons

Now,

Mass of copper ball = volume × density

Or

Mass of copper ball = $\frac{4}{3}\pi(\frac{d}{2})^3$ × 8900

or

Mass of copper ball = $\frac{4}{3}\pi(\frac{0.002}{2})^3$ × 8900

or

Mass of copper ball = 0.03726 grams

Also,

molar mass of copper = 63.546 g/mol

Therefore,

Number of mol of copper in 0.03726 grams = $\frac{ 0.03726}{63.546}$

or

Number of mol of copper in 0.03726 grams = 5.86 × 10⁻⁴ mol

and,

1 mol of a substance contains = 6.022 × 10²³ atoms

Therefore,

5.86 × 10⁻⁴ mol of copper contains = 5.86 × 10⁻⁴ × 6.022 × 10²³ atoms.

or

5.86 × 10⁻⁴ mol of copper contains = 35.88 × 10¹⁹ atoms

Now,

A neutral copper atom has 29 electrons.

Therefore,

Number of electrons in ball = 29 × 35.88 × 10¹⁹ = 1023.37 × 10¹⁹ electrons.

Hence,

The fraction of electrons removed = $\frac{25\times10^{10}}{1023.37\times10^{19}}$

or

The fraction of electrons removed = 0.02442 × 10⁻⁹

3 0

3 years ago

Can someone plz help me on this, it would me alot to me thx