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2 years ago

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Determine the acceleration of Earth due to its motion around the Sun. Assume the Earth's orbit to be circular with a radius of 1

.5 x 10^8 km.

1 answer:

Pie2 years ago

8 0

Answer:

Explanation:

Given

Radius of orbit $r=1.5\times 10^8\ km$

As earth moves around sun it experiences a centripetal force on it which is given by

$a_c=\frac{v^2}{r}$

where v=velocity of earth

Time period of Revolution

$T=\frac{2\pi \cdot r}{v}$

time period is 365 days i.e. $3.153\times 10^7\ s$

$v=\frac{2\pi \times 1.5\times 10^{11}}{3.153\times 10^7}$

$v=2.98\times 10^4\ m/s$

$a_c=\frac{v^2}{r}$

$a_c=\frac{9\times 10^8}{1.5\times 10^{11}}$

$a_c=0.006\ m/s^2$

acceleration due to gravity is $9.8\ m/s^2$

$a_c$ is less than gravity

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I have a set of calipers that can measure thickness of a few inches with an uncertainty of ±0.005 inches. I measure the thicknes

Shtirlitz [24]

Answer:

(a) 0.0113 ±0.0001 inches

(b) 5 decks

Explanation:

<u>Given information</u>

52 cards, thickness is 0.590 ±0.005 inches

Since this is thickness of all the cards, to get the thickness of a single card we divide the total thickness (plus uncertainty) by the number of cards hence

1 card= $\frac {0.590}{52}$ ± $\frac {0.005}{52}$ = 0.0113461 ±0.000096 inches

Considering that thickness is given to 3 significant figures while uncertainty is to 1 significant figure, the final answer should also conform to these hence giving the first part of the answer to 3 significant figures while second part to 1 significant figure yields 0.0113 ±0.0001 inches

(b)

Considering that the cards have uncertainty of 0.0001 inches and the number of decks, n required to create uncertainty of 0.00002 inch is given by

$n=\frac {0.0001}{0.00002}=5$

We need 5 decks for the given uncertainty

4 0

3 years ago

The method of heat transfer by which the sun energy reaches the earth is

snow_lady [41]

Answer:

Radiation

Explanation:

The heat transfer take place by three modes

1. Conduction

This type of heat transfer take place in the solid materials like metals .This heat transfer take place due to vibration of electron.

2.Convection

This type of heat transfer take place in the moving fluid like water ,air etc.

3.Radiation

Every materials which having temperature above 0 K are emitting radiation.This type of heat transfer dominate at the high temperature.

We know that temperature of sun is so high so the heat transfer from sun to the earth is take place by radiation mainly.

8 0

3 years ago

Read 2 more answers

An object of 4 cm length is placed at a distance of 18 cm in front of a convex mirror of radius of curvature 30 cm. Find the pos

erica [24]

Answer:

The position is 8.18cm from the mirror.

Nature is b=virtual

Size is 1.82cm

Explanation:

Note that for a convex mirror, the image distance and the focal length are negative;

Given

Object height H0 = 4cm

object distance u = 18cm

Radius of curvature R = 30cm

Since f = R/2

f = 30/2

f = -15cm

Recall that:

$\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v} \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm$

Since the image distance is negative, this shows that the image is a virtual image.

To get the size:

$\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm$

3 0

3 years ago

Hey can someone please help answer this question?

yan [13]

Answer:

yes ............................................ks

Explanation: is good

4 0

2 years ago

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)

Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

The power rating of the bulb is $P = 100 \ W$

The resistance is $R = 143 \ \Omega$

The voltage is $V = V_o sin [2 \pi ft]$

The energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

The voltage $V_o = 110 \ V$

The frequency is $f = 60 \ Hz$

The time considered is $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

$P = \frac{V^2}{ R}$

=> $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=> $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

So

$U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=> $U = 7.563 *10^{5} \ J$

7 0

3 years ago