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3 years ago

When are the days in the northern hemisphere the shortest?

2 answers:

7 0

In the colder seasons. The winter solstice is the shortest day of the year which is usually in the third week of December

kolezko [41]3 years ago

3 0

The Winter...? I think...? Ya, it's the Winter. :)

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An object is transported to three different planets in our solar system. Which of the following is true about that object?

Digiron [165]

Answer;

D. The object’s weight changes, but its mass stays the same.

Explanation;

Mass is the amount of matter in a object, which is measured in kilograms. Mass of an object is measured using a beam balance. It is important to note that the mass of an object or a body remains constant, and does not vary from one place to another. For instance the mass of a person on the moon will be the same as when the person is on the earth surface.
Weight on the other hand is the measurement of gravitational pull of an object. weight is measured using a spring balance and measured in Newtons. Weight varies from one place to another depending on the gravitational pull of a given surface.

8 0

3 years ago

Read 2 more answers

I need help with 2 questions . Please help c:

dezoksy [38]

A) According to the nebular theory, the Solar System formed from a huge gaseous nebula which at a certain point was perturbated. Atoms and molecules started colliding, forming planetesimals (a sort of big rocks). The planetesimals were attracted to each other by gravity, forming bigger warm almost spherical objects called protoplanets, which at the end cooled down forming planets.
Therefore the correct answer is "all of the above".

b) The planets closer to the Sun were (and still are) subject to higher temperatures, due to their close distance to the Sun. In these conditions, rocky materials undergo condensation, while iced gaseous materials undergo vaporization. In the outer parts of the Solar System temperatures are too low to allow these transformations.
The correct answer is again "all of the above".

8 0

3 years ago

A shopper does 157 J of work pushing a cart with 10.9 N force

Tanzania [10]

The cart travelled a distance of 14.4 m

Explanation:

The work done by a force when pushing an object is given by:

$W=Fd cos \theta$

where:

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and the displacement

In this problem we have:

W = 157 J is the work done on the cart

F = 10.9 N is the magnitude of the force

$\theta=0^{\circ}$ , assuming the force is applied parallel to the motion of the cart

Therefore we can solve for d to find the distance travelled by the cart:

$d=\frac{W}{F cos \theta}=\frac{157}{(10.9)(cos 0)}=14.4 m$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0

3 years ago

A woman is running around a playground while keeping an eye on her child who is on a swing set. The child is going back and fort

Nutka1998 [239]

Answer:

(a) 1500 m

(b) 2827.43m

Explanation:

Given that the time for one cycle of the swing is 1 s

The radius of the swing, R= 3.0m

The angle covered, \theta, by each swing is a quarter of the circle. i.e.

$\Theta=\frac {\pi}{2}$

Speed of running of women =5 m/s.

Time of running = 5 minutes= 5 x 60 secondes= 300 s.

(a) As distance= (speed) x (time)

So, the required distance= 5 x 300 m= 1500 m.

(b) As there are two swings in one cycle, so, the distance covered in one swing is the length of the circular shown in the figure.

As arc length, $l= \theta R$ , where \theta is the angle, in radian, subtended by the arc at the center, and R is the radius of curvature of the arc.

So, the distance covered by the child in 1 swing $= \frac {\pi}{2}\times 3 m=\frac{3\pi}{2}m$ .

In 1 cycle, there are 2 swings, so distance covered in 1 cycle $= 3 \pi m$ .

Now, in 1 second there is 1 cycle, so in 5 minutes there will be 300 cycles.

So, the total distance covered by the child

$= 3\pi\times 300 m= 2827.43 m$ .

3 0

3 years ago

A solid conducting sphere with radius R carries a positive total charge Q. The sphere is surrounded by an insulating shell with

Illusion [34]

Answer:

Explanation:

Volume of the insulating shell is,

$V_{shell}=\frac{4}{3}\pi(R^3_2-R^3_1)$

Charge density of the shell is,

$\rho=\frac{Q_{shell}}{\frac{4}{3}\pi(R^3_2-R^3_1)}$

Here, $R_2 =2R, R_1 =R \,and\, Q_{shell} =-Q$

$\rho=\frac{Q_{shell}}{\frac{4}{3}\pi((2R)^3-R^3)}=\frac{-3Q}{28\piR^3}$

The electric field is $E=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}$

For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.

For R <r <2R According to gauss law

$E(4\pi r^2)=\frac{Q}{\epsilon_0}+\frac{4\pi\rho}{3\epsilon_0}(r^3-R^3)$

substitute $\rho=\frac{-3Q}{28\piR^3}$

$E=\frac{2}{7\pi\epsilon_0}\frac{Q}{r^2}-\frac{Qr}{28\piR^3}$

The net charge enclosed for each r in this range is positive and the electric field is outward

For r>2R

Charge enclosed is zero, so electric field is zero

8 0

3 years ago