All categories

3 years ago

Are u god tell me u have to tell me are u a god

Physics

1 answer:

Allushta [10]3 years ago

7 0

They tell me I'm a god, I'm lost in the facade

Six-feet off the ground at all times, I think I'm feelin' odd

No matter what I make, they never see mistakes

Makin' so much bread, I don't care that they're just being fake

They tell me they're below me, I act like I'm above

The people blend together but I would be lost without their love

Can you heal me? Have I gained too much?

When you become untouchable, you're unable to touch

Is there a real me? Pop the champagne

It hurts me just to think and I don't do pain

Stayin' still, eyes closed

Let the world just pass me by

Pain pills, nice clothes

If I fall, I think I'll fly

Touch me, Midas

Make me part of your design

None to guide us

I feel fear for the very last time.

You might be interested in

NEED HELP ASAP PLEASE!

katrin [286]

Answer:

Here is the solution hope it helps:)

5 0

2 years ago

Which kind of pigment reflects two primary light colors and absorbs one?1) primary pigment2) secondary pigment3) complementary p

weqwewe [10]

The primary colors of light are red, blue and green.

There are the pigments like yellow, magenta and cyan that are the mixture of two primary colors.

For example, magenta is a mixture of red and blue color. Thus, it reflects the red and blue color. Also, magneta absorbs the green color.

These type of colors that reflects two primary colors and absorb one color are known as secondary pigments.

Hence, 2nd option is the correct answer.

8 0

8 months ago

A hydraulic lift has two connected pistons with cross-sectional areas 15 cm2 and 670 cm2. It is filled with oil of density 510 k

BigorU [14]

Answer:

Explanation:

Given

Cross-sectional area of two areas is

$A_1=15\ cm^2$

$A_2=670\ cm^2$

It is filled with oil of density $\rho _0=510\ kg/m^3$

mass of car place on Large area $M=1100\ kg$

Suppose a mass of m kg is placed on smaller area

According to pascal law's intensity of pressure is same at every point on Liquid

$P_1=P_2$

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

$\frac{mg}{15}=\frac{Mg}{670}$

$m=1100\times \frac{15}{670}$

$m=24.62\ kg$

4 0

3 years ago

A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes

Nitella [24]

Answer:

The value of F= - 830 N

Since the force is negative, it implies direction of the force applied was due south.

Explanation:

Given data:

Mass = 1000-kg

Distance, d = 240 m

Initial velocity, v1 = 20.0 m/s

Final velocity, v2 = 0 (since the car came to rest after brake was applied)

v2²= v1² + 2ad (using one of the equation of motion)

0= 20² + (2 x a x 240)

0= 400 + 480 a

a = - 400/480

a = - 0.83 m/s²

Then, imputing the value of a into

F = ma

F = 1000 kg x ( - 0.83 m/s²)

F= - 830 N

The car was driving toward the north, and since the force is negative, it implies direction of the force applied was due south.

3 0

2 years ago

The engine starter and a headlight of a car are connected in parallel to the 12.0-V car battery. In this situation, the headligh

stepladder [879]

Answer:

The total power they will consume in series is approximately 2.257 W

Explanation:

The connection arrangement of the headlight and the engine starter = Parallel to the battery

The voltage of the battery, V = 12.0 V

The power at which the headlight operates in parallel, $P_{headlight}$ = 38 W

The power at which the kick starter operates in parallel, $P_{kick \ starter}$ = 2.40 kW

We have;

P = V²/R

Where;

R = The resistance

V = The voltage = 12 V (The voltage is the same in parallel circuit)

For the headlight, we have;

R₁ = V²/ $P_{headlight}$ = 12²/38 = 72/19

R₁ = 72/19 Ω

For the kick starter, we have;

R₂ = V²/ $P_{kick \ starter}$ = 12²/2.4 = 60

R₂ = 60 Ω

When the headlight and kick starter are rewired to be in series, we have;

Total resistance, R = R₁ + R₂

Therefore;

R = ((72/19) + 60) Ω = (1212/19) Ω

The current flowing, I = V/R

∴ I = 12 V/(1212/19) Ω = (19/101) A

We note that power, P = I²R

In the series connection, we have;

$P_{headlight}$ = I² × R₁

∴ $P_{headlight}$ = ((19/101) A)² × 72/19 Ω = 1368/10201 W ≈ 0.134 W

The power at which the headlight operates in series, $P_{headlight, S}$ ≈ 0.134 W

$P_{kick \ starter}$ = ((19/101) A)² × 60 Ω = 21660/10201 W ≈ 2.123 W

The power at which the kick starter operates in series, $P_{kick \ starter, S}$ ≈ 2.123 W

The total power they will consume, $P_{Total}$ = $P_{headlight, S}$ + $P_{kick \ starter, S}$

Therefore;

$P_{Total}$ ≈ 0.134 W + 2.123 W = 2.257 W

4 0

3 years ago