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3 years ago

Plzzzz helppppppppppp

Physics

1 answer:

pashok25 [27]3 years ago

7 0

1. Explain, using Kinetic theory, why when you have a hot room, the room cools down when you open a window to the cold outside.

Answer:

According to kinetic theory of gases, at higher temperature, the molecules in air have higher kinetic energy and they are moving at a faster speed. Hence the pressure is greater. And Air moves from higher pressure to lower pressure.

In a hot room, the air pressure is greater. The cool air outside has lower pressure. On opening the window, Air moves from higher pressure to lower pressure, from outside to inside.

2. If an atom is 1*10^(-10) m wide, and the nucleus is 1*10^(-15) m wide, how wide would the atom be if the nucleus was scaled up to be 1 cm wide?

Answer:

When nucleus was scaled upto 1 cm, atom would be 10^(5) cm.

The question is pretty simple, we just have to scale the size of nucleus to larger value and get the size of atom accordingly.

We know that

when nucleus is 10^(-15) m => atom is 10^(-10) m

When nucleus becomes 1 m => atom is 10^(-10) * 10^(15) m = 10^(5) m

When nucleus is 1 * 10^(-2) => atom is 10^(5) * 10^(-2) m

When nucleus is 1 cm => atom is 10^(5) cm

You might be interested in

A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 330 N/m. The object is droppe

pentagon [3]

Answer:

(A) l = 0.39 m

(B) l =0.38 m

(C) l = 0.14 m

Explanation:

Answer:

Explanation:

Answer:

Explanation:

from the question we are given the following values:

mass (m) = 1.6 kg

height (h) = 1.05 m

compression of spring (l) = ?

spring constant (k) = 330 N/m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) initial potential energy of the object = final potential energy of the spring

potential energy of the object = mg(1.05 + l)

potential energy of the spring = 0.5 x k x l^{2} (k= spring constant)

therefore we now have

mg(1.05 + l) = 0.5 x k x l^{2}

1.6 x 9.8 x (1.05 + l) = 0.5 x 300 x l^{2}

15.68 (1.05 + l) = 150 x l^{2}

16.5 + 15.68l = 150l^{2}

l = 0.39 m

(B) with constant air resistance the equation applied in part A above becomes

initial P.E of the object - air resistance = final P.E of the spring

mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}

1.6 x 9.8 x (1.05 + l) - 0.750(1.05 + l) = 0.5 x 300 x l^{2}

(16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}

16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

15.71 + 14.93l = 150^{2}

l =0.38 m

(C) where g = 1.63 m/s^{2} and neglecting air resistance

the equation mg(1.05 + l) = 0.5 x k x l^{2} now becomes

1.6 x 1.63 x (1.05 + l) = 0.5 x 300 x l^{2}

2.608 (1.05 +l) = 0.5 x 300 x l^{2}

2.74 + 2.608l = 150 x l^{2}

l = 0.14 m

6 0

3 years ago

A 1-megabit computer memory chip contains many 27 fF capacitors. Each capacitor has a plate area of 3.09 × 10−11 m 2 . Determine

valentina_108 [34]

Answer:

Plate separation of each capacitor is 101.132 °A

Explanation:

The formula to calculate the capacitance in empty space as a function of distance (square parallel plates) is:

$C=\epsilon_{0}\frac{Area}{distance}$

clearing for distance:

$distance=\epsilon_0 \frac{Area}{Capacitance} \\\\\epsilon_0=8.8542(10)^{-12}C^2/Nm\\\\Area=3.09(10)^{-11}m^2\\Capacitance=27(10)^{-15}F\\\\Replacing\\\\distance=\frac{8.8542(10)^{-12}*3.09(10)^{-11}}{27(10)^{-15}} =1.0133(10)^{-8}m\\\\In A\\distance= 101.132 A$

8 0

3 years ago

which planet should punch travel to if his goal is to weigh in at 118 lb? refer to the table of planetary masses and radii given

Harrizon [31]

The planet that Punch should travel to in order to weigh 118 lb is Pentune.

<h3 /><h3 /><h3>The given parameters:</h3>

Weight of Punch on Earth = 236 lb
Desired weight = 118 lb

The mass of Punch will be constant in every planet;

$W = mg\\\\m = \frac{W}{g}\\\\m = \frac{236}{g}$

The acceleration due to gravity of each planet with respect to Earth is calculated by using the following relationship;

$F = mg = \frac{GmM}{R^2} \\\\g = \frac{GM}{R^2}$

where;

M is the mass of Earth = 5.972 x 10²⁴ kg
R is the Radius of Earth = 6,371 km

For Planet Tehar;

$g_T =\frac{G \times 2.1M}{(0.8R)^2} \\\\g_T = 3.28(\frac{GM}{R^2} )\\\\g_T = 3.28 g$

For planet Loput:

$g_L =\frac{G \times 5.6M}{(1.7R)^2} \\\\g_L = 1.94(\frac{GM}{R^2} )\\\\g_L = 1.94g$

For planet Cremury:

$g_C =\frac{G \times 0.36M}{(0.3R)^2} \\\\g_C = 4(\frac{GM}{R^2} )\\\\g_C = 4 g$

For Planet Suven:

$g_s =\frac{G \times 12M}{(2.8R)^2} \\\\g_s = 1.53(\frac{GM}{R^2} )\\\\g_s = 1.53 g$

For Planet Pentune;

$g_P =\frac{G \times 8.3 }{(4.1R)^2} \\\\g_P = 0.5(\frac{GM}{R^2} )\\\\g_P = 0.5 g$

For Planet Rams;

$g_R =\frac{G \times 9.3M}{(4R)^2} \\\\g_R = 0.58(\frac{GM}{R^2} )\\\\g_R = 0.58 g$

The weight Punch on Each Planet at a constant mass is calculated as follows;

$W = mg\\\\W_T = mg_T\\\\W_T = \frac{236}{g} \times 3.28g = 774.08 \ lb\\\\W_L = \frac{236}{g} \times 1.94g =457.84 \ lb\\\\ W_C = \frac{236}{g}\times 4g = 944 \ lb \\\\ W_S = \frac{236}{g} \times 1.53g = 361.08 \ lb\\\\W_P = \frac{236}{g} \times 0.5 g = 118 \ lb\\\\W_R = \frac{236}{g} \times 0.58 g = 136.88 \ lb$

Thus, the planet that Punch should travel to in order to weigh 118 lb is Pentune.

<u>The </u><u>complete question</u><u> is below</u>:

Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer.

Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system (<em>find the image attached</em>).

<em>In the table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii.</em>

Learn more about effect of gravity on weight here: brainly.com/question/3908593

5 0

2 years ago

How can i calculate distances between objects by using the concepts of echo location

dangina [55]

Send wave from your location to the object and wait until echo is back.
Measure the time taken.

If you know the speed of wave (say sound wave), than just multiply by half time taken wave to return

5 0

3 years ago

Jenny pushes a 40 N crate down the hall 2m. How much work did she do?

Shalnov [3]

Work done = force x distance = 40 x 2 = 80 Joules.

8 0

3 years ago