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3 years ago

HELP ASAP PLEASE PHYSICS

Physics

2 answers:

LiRa [457]3 years ago

6 0

Answer:

1. 45.65

2. 1.19x10^19

3. 514OHM

4. 2.45 ( NOT 1.86 OR 1.36)

6. 20.06A

7. 206 ohms

8. 13 Ohms

9. 2 ohms

10. 0.051 ( NOT 0.036 OR 0.089)

Explanation:

I've struggled with both problems 4 and 10

Nataly_w [17]3 years ago

5 0

Answer: did you get the answers?

Explanation:

You might be interested in

9. A Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table. How long does it take to hit the floor if no one sneezes? Wh

Paha777 [63]

By Considering the vertical distance and both vertical and horizontal final velocity, the time t = 0.45 s and Velocity V = 6.7 m/s

Given that a Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table.

Height h = 1.0 m

As the ball rolls off the table, it will be fallen under gravity. Where

g = 9.8 m/ $s^{2}$

Initial vertical velocity $u_{y}$ = 0

Initial horizontal velocity $u_{x}$ = 5 m/s

Considering the vertical distance, the formula to use to calculate the time will be;

h = ut + 1/2g $t^{2}$

1 = 0 + 1/2 x 9.8 $t^{2}$

1 = 4.9 $t^{2}$

$t^{2}$ = 1/4.9

t = $\sqrt{0.204}$

t = 0.45 seconds

It takes 0.45 seconds to hit the floor if no one sneezes.

To calculate its velocity when it hits the floor, we will need to calculate for both vertical and horizontal final velocity and find the resultant velocity of the two.

Vertical component

$V_{y}$ = $U_{y}$ + gt

$V_{y}$ = 0 + 9.8(0.45)

$V_{y}$ = 4.41 m/s

Horizontal component

$V_{x}$ = $u_{x}$ + at

but a = 0

$V_{x}$ = 5 m/s

Final velocity V = $\sqrt{5^{2} + 4.41^{2} }$

V = 6.67 m/s

Therefore, it will hit the floor at a velocity of 6.7 m/s

Learn more here: brainly.com/question/5063616

8 0

2 years ago

The human body has an average density of 979 kg/m3 , what fraction of a person is submerged when floating gently in fresh water?

FromTheMoon [43]

A person is submerged of about 97.9%.

The average density of the human body is given as 979 kg/ m³.

<h3>Define Law of floatation.</h3>

Law of floatation can be defined as the volume of the liquid displaced when a body floats on the liquid surface is equal to the body submerged in the water.

As body has the stable equilibrium state, the buoyancy of the fluid will be equal to the weight.

Weight of the body floating = Weight of the body immersed in fluid

Law of floatation = Density of the floating object / density of fluid

As fluid is the freshwater here, the density of fluid will be 1000 kg/ m³.

= (979 kg/ m³) / ( 1000 kg/ m³)

= 97.9 %

A person is submerged when floating gently in fresh water about 97.9%.

Learn more about law of floatation,

brainly.com/question/17032479

#SPJ4

4 0

5 months ago

Calculate the diffraction limit of the human eye, assuming a wide-open pupil so that your eye acts like a lens with diameter 0.8

guapka [62]

Hi, thank you for posting your question here at Brainly.

This problem could be solved using this equation:

Diffraction limit = 1.22*wavelength/diameter

diameter = 0.8 cm = 0.008 m
wavelength = 500E-9 m

Diffraction limit = 1.22(500E-9)/0.008
Diffraction limit = 0.00007625

6 0

2 years ago

A sample of gas with a volume of 750 ml exerts a pressure of 98 kpa at 30◦c. What pressure will the sample exert when it is comp

Tanzania [10]

Answer:

241 kPa

Explanation:

The ideal gas law states that:

$pV=nRT$

where

p is the gas pressure

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature of the gas

We can rewrite the equation as

$\frac{pV}{T}=nR$

For a fixed amount of gas, n is constant, so we can write

$\frac{pV}{T}=const.$

Therefore, for a gas which undergoes a transformation we have

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

where the labels 1 and 2 refer to the initial and final conditions of the gas.

For the sample of gas in this problem we have

$p_1 = 98 kPa=9.8\cdot 10^4 Pa\\V_1 = 750 mL=0.75 L=7.5\cdot 10^{-4}m^3\\T_1 = 30^{\circ}C+273=303 K\\p_2 =?\\V_2 = 250 mL=0.25 L=2.5\cdot 10^{-4} m^3\\T_2 = -25^{\circ}C+273=248 K$

So we can solve the formula for $p_2$ , the final pressure:

$p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}=\frac{(9.8\cdot 10^4 Pa)(7.5\cdot 10^{-4} m^3)(248 K)}{(303 K)(2.5\cdot 10^{-4} m^3)}=2.41\cdot 10^5 Pa = 241 kPa$

4 0

2 years ago

Suppose the Sun appeared to you 900 times dimmer than it does now. How far away from the Sun would you be? (A) 1/9 AU (B) 3 AU

Harrizon [31]

Answer:

The correct answer is option 'c': 30 AUs

Explanation:

For a spherical wave front emitted by sun with total energy 'E' the energy density over the surface when it is at a distance 'r' from the sun is given by

$e=\frac{E}{4\pi r^{2}}$

This energy per unit area is sensed by observer as intensity of the sun.

Let the initial intensity of sun at a distance $r_{1}$ be $e_{1}$

Thus if the sun becomes 900 times dimmer we have

$e'=\frac{e_{1}}{900}\\\\\frac{E}{4\pi r_{2}^{2}}=\frac{1}{900}\times \frac{E}{4\pi r_{1}^{2}}\\\\\Rightarrow r_{2}^{2}={r_{1}^{2}}\times 900\\\\\therefore r_{2}={r_{1}}\times {30}$

Thus the distance increases 30 times.

5 0

3 years ago