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3 years ago

PLZZ I WILL DO ANY THING

Physics

1 answer:

zhuklara [117]3 years ago

3 0

Answer: Universe.

Explanation: In addition to the solar system, the heliocentric model stated that the sun was the center of the UNIVERSE.

Every plant orbits around the sun, which makes our universe whole and bright.

Hope this helps you out! ☺

-Karleif-

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A 2200 kg car doubles its speed from 50 km/hr to 100 km/hr. By how many times does the kinetic energy from the car’s forward mot

Firlakuza [10]

Answer: 4

Explanation:

Given that:

Mass of car M = 2200 kg

Initial speed Vi = 50 km/hr

Final speed Vf = 100 km/hr

Kinetic energy is the energy possessed by a moving object. It is measured in joules, and depends on the mass (m) of the object and the speed (v) by which it moves

i.e K.E = 1/2MV^2

So, when traveling at 50 km/h

KE = 1/2x 2200kg x (50km/h)^2

KE = 0.5 x 2200 x 2500

KE1 = 2750000J

So, when traveling at 100 km/h

KE = 1/2x 2200 x (100 km/h)^2

KE = 0.5 x 2200 x 10000

KE2 = 11000000J

Thus, the number of times kinetic energy increases is obtained by dividing KE2 by KE1

i.e 11000000J / 2750000J

= 4

Thus, the kinetic energy from the car’s forward motion increase 4 times

5 0

3 years ago

Read 2 more answers

In this circuit (see picture), which resistor will draw the least power?

Basile [38]

A few different ways to do this:

Way #1:
The current in the series loop is (12 V) / (total resistance) .
(Turns out to be 2 Amperes, but the question isn't asking for that.)

In a series loop, the current is the same at every point, so it's
the same current through each resistor.

The power dissipated by a resistor is (current)² · (resistance),
and the current is the same everywhere in the circuit, so the
smallest resistance will dissipate the least power. That's R1 .

And by the way, it's not "drawing" the most power. It's dissipating it.

Way #2:
Another expression for the power dissipated by a resistance is

(voltage across the resistance)² / (resistance) .

In a series loop, the voltage across each resistor is

[ (individual resistance) / (total resistance ] x battery voltage.

So the power dissipated by each resistor is

(individual resistance)² x [(battery voltage) / (total resistance)²]

This expression is smallest for the smallest individual resistance.
(The other two quantities are the same for each individual resistor.)
So again, the least power is dissipated by the smallest individual resistance.
That's R1 .

Way #3: (Einstein's way)
If we sat back and relaxed for a minute, stared at the ceiling, let our minds
wander, puffed gently on our pipe, and just daydreamed about this question
for a minute or two, we might have easily guessed at the answer.

===> When you wire up a battery and a light bulb in series, the part
that dissipates power, and gets so hot that it radiates heat and light, is
the light bulb (some resistance), not the wire (very small resistance).

3 0

3 years ago

A 5.00-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. the expanding gases behind it exert what force on the bull

muminat

Answer;

= 312 Newtons

Explanation;

The bullet has a mass of 0.005 kg, and a velocity of 320 m/s, so we need to find it's final kinetic energy.

KE = 1/2*m*v^2

= 1/2*0.005*320^2

= 256 Joules.

Divide this by the distance over which this energy was received and you have the force that provided that energy.

= 256/0.820 = 312.195 Newtons

Rounded off, this is 312 N

3 0

3 years ago

A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of

Lynna [10]

We have the relation

$\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}$

where $v_{A \mid B}$ denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

$v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}$

$v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}$

Let's assume the river flows South-to-North, so that

$\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath$

and let $-90^\circ < \theta < 90^\circ$ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

$\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)$

Then the velocity of the boat relative to the Earth is

$\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath$

The crossing is 153.0 m wide, so that for some time $t$ we have

$153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s$

which is minimized when $\theta=0^\circ$ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

$\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}$

The boat's position $\vec x$ at time $t$ is

$\vec x = \vec v_{B\mid E} t$

so that after 30.0 s, the boat's final position on the other side of the river is

$\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath$

and the boat would have traveled a total distance of

$\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}$

3 0

2 years ago

A block has a volume of 0.09 m3 and a density of 4,000 kg/m3. What's the force of gravity acting on the

SCORPION-xisa [38]

Density = (mass) / (volume)
4,000 kg/m³ = (mass) / (0.09 m³)
(4,000 kg/m³) x (0.09 m³) = mass
mass = 360 kg
force of gravity = (mass) x (acceleration of gravity) = (360 kg) x (9.8 m/s²) = (360 x 9.8) kg-m<span>/s² </span><span>= </span>3,528 newtons .

5 0

4 years ago