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2 years ago

How many minutes of daylight do we gain after winter solstice

Physics

1 answer:

serg [7]2 years ago

4 0

Answer:

Depends on which hemisphere you are belong to and how much distance you are away from Ecuador.

Explanation:

Minutes of daylight is equal on everywhere only on the equinox days (21 March and 23 September). On other days it depends on the place that you are belong to. On winter solstice, places on Ecuador have 12 hours daylight. North side of Ecuador have less, south side of Ecuador have more hour of daylight.

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A bag of cement weighing 325 N hangs in equilibrium from three wires. Two of the wires make angles of theta1=60.0 degrees and th

Murljashka [212]

The sketch of the system is: two strings, 1 and 2, are attached to the ceiling and to a third string, 3.The third string holds the bag of cement.

The free body diagram of the weight with the string 3, drives to the tension T3 = weihgt => T3 = 325 N

The other free body diagram is around the joint of the three strings.

In this case, you can do the horizontal forces equilibrium equation as:

T1* cos(60) - T2*cos(40) = 0

And the vertical forces equilibrium equation:

Ti sin(60) + T2 sin(40) = T3 = 325 N

Then you have two equations with two unknown variables, T1 and T2

0.5 T1 - 0.766 T2 = 0

0.866 T1 + 0.643T2 = 325

When you solve it you get, T1 = 252.8 N and T2 = 165 N

Answer: T1 = 252.8 N, T2 = 165N, and T3 = 325N

7 0

3 years ago

A builder drops a brick from a height of 15 m above the ground. The gravitational field strength g is 10 N/ kg. What is the spee

Basile [38]

The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

Initial Velocity; $u = 0$

Height from which it has dropped; $h = 15m$

Gravitational field strength; $g = 10N/kg = 10 \frac{kg.m/s^2}{kg} = 10m/s^2$

Final speed of brick as it hits the ground; $v = \ ?$

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

$v^2 = u^2 + 2gh$

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

$v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s$

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Learn more about equations of motion: brainly.com/question/18486505

8 0

2 years ago

¡¡¡AYUDA CON ESTOS EJERCICIOS DE FÍSICA!!!

asambeis [7]

Answer:

(a) 8 V, (b) 144000 V, (c) 2 x 10^(-8) C

Explanation:

(a) charge, q = 5 μC , Work, W = 40 x 10-^(-6) J

The electric potential is given by

W = q V

$40\times10^{-6}=5 \times10^{-6}\times V\\\\V = 8 V$

(b)

charge, q = 8 x 10^(-6) C, distance, r = 50 cm = 0.5 m

Let the potential is V.

$V =\frac{k q}{r}\\\\V =\frac{9\times 10^{9}\times 8\times 10^{-6}}{0.5}\\\\V =144000 V$

(c)

Work, W = 8 x 10^(-5) J, Potential difference, V = 4000 V

Let the charge is q.

W= q V

$8\times10^{-5}= q\times 4000\\\\q =2\times 10^{-8} C$

3 0

2 years ago

light spring of force constant k = 158 N/m rests vertically on the bottom of a large beaker of water (Figure a). A 4.36-kg block

Fantom [35]

Answer:

0.146 m

Explanation:

f = -KΔL according to Hooke's law

volume of water displaced = mass / density of block since a body will displace equal volume of its own

weight of water displaced = mass of water × acceleration due to gravity

and mass of water = volume of water / density of water

weight of water displaced = Vw × dw × g = mg (dw / dblock)

net force = mg - mg (dw / dblock) = 42.728 - 65.74 = -23.00

it will be balanced by a restoring force of 23 N

ΔL = F / k = 23 / 158 = 0.146 m

4 0

2 years ago

A ball of radius R and mass m is magically put inside a thin shell of the same mass and radius 2R. The system is at rest on a ho

dlinn [17]

Answer:

$x =\frac{-R}{2}$

Explanation:

From the question we are told that mass

Thin layer radius $= 2R$

Generally the expression for ths solution is given as

Xcm =(m*0 =m(-2R))/2m =-mR/(2m)=-R/2

the center of mass will not move at initial state

Considering the center of mass of both bodies

$xcm=\frac{m*x+m*x)}{2m} =x$

$x =\frac{-R}{2}$

Therefore the enclosing layer moves $x =\frac{-R}{2}$

7 0

2 years ago