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nasty-shy [4]
2 years ago
9

ASAP need physics help please:)

Physics
1 answer:
boyakko [2]2 years ago
3 0

Answer:

plantation of force of the earth acting on 15 kg of object on free fall the acceleration of free fall said about it now it is year 15 kg then it becomes 1.53 Newton

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a liquid with a specific heat of 1900J/Kg° C has 4750 joules of heat energy is added to it. if the temperature increases from 20
GREYUIT [131]
E=mcθ
where E is the energy added,m is the mass,c is the specfic heat capacity and θ is the change in temprature.
making m subject u get E/c<span>θ=m
</span><span>θ=(30-20)=10</span><span>
plugging the values  we get :
</span>\frac{4750}{1900*10}
<span>
solving it we get the answer that is 0.25kg or 250 grams

</span>
8 0
2 years ago
Photographs of many young stars show long jets of material apparently being ejected from their poles.
Rudiy27

Answer:

A) True

Explanation:

Researchers have detected numerous jets of gas ejected from poles of young stars and planetary nebulae.

By examining images of hydrogen molecules excited at infrared wavelengths, scientists have been able to see through the gas and dust in the Milky Way, in order to observe the most distant targets. These goals are normally hidden from view and many of them have never been seen before.

The entire study area covers approximately 1,450 times the size of the full moon, or the equivalent of an image of 95 gigapixels. The survey reveals jets emanating from proto-stars and planetary nebulas, as well as remnants of supernovae, the illuminated edges of vast clouds of gas and dust, and the warm regions that surround massive stars and their associated groups of smaller stars.

7 0
3 years ago
Decreasing a telescope's eyepiece focal length will _____.
Novay_Z [31]
I believe the correct answer from the choices listed above is the first option. Decreasing a telescope's eyepiece focal length will increase magnification. <span>The magnification of the </span>telescope<span> image is (</span>focal length<span> of the objective) divided by (</span>focal length <span>of the </span>eyepiece<span>). Hope this answers the question.</span>
7 0
2 years ago
Read 2 more answers
4. Which will have a higher atomic radius Mg or P? Why?
storchak [24]

Answer:

Magnesium has a larger atomic radius due to more electron levels.

6 0
2 years ago
A ball of mass m = 0.1kg is connected to a rope of length L = 1.2 m. The ball is swung around in a vertical circle and ball is m
bonufazy [111]

Answer:

The speed of the ball is approximately 5.94 m/s

The Tension of the string at the bottom is 3.92 N

Explanation:

We need to find the speed of the ball, which is constant due to the fact that we are in a uniform circular motion. Notice as well that the speed of the ball is the magnitude of the tangential velocity "v_t" (vector that changes direction with the position of the ball but doesn't change magnitude in this case).

We analyze first the top position of the circular motion, for which information on the tension of the string is given (see first free body diagram in the attached picture).  We are told that the tension at the top of the movement equals twice the force of gravity on the ball's mass: T - 2*m*g = 1.96 N. And we know that there are two forces acting on the ball in that position (illustrated with the green arrows pointing down): one is the ball's weight due to gravity, and the other is the string's tension. So we can write Newton's second law for this situation:

F_{net}= T_{top}+W\\F_{net}=2\,W+W\\F_{net}=3\,W\\F_{net}=2.94 N\\

Newton's second law tells us that the net force should equal the mass of the ball times its acceleration (F = m * a), and in this motion, the acceleration is the centripetal acceleration. Therefore weuse this equation to solve for the centripetal acceleration of the ball:

m\,a_c=2.94\,N\\a_c=\frac{2.94\,N}{0.1\,kg} \\a_c=29.4\,\frac{m}{s^2}

The centripetal acceleration is defined as the square of the tangential velocity divided the radius of the circular motion. Then we use it to derive the magnitude of the tangential velocity (speed of the ball):

a_c=\frac{v^2}{R} \\29.4\,\frac{m}{s^2} =\frac{v_t^2}{R} \\v_t^2=29.4\,(1.2)\,\frac{m^2}{s^2} \\v_t=5.94\,\frac{m}{s}

So we have found the speed of the ball.

Now we focus our attention to the bottom of the motion, and again use Newton's second law to solve for the string tension (see second free body diagram in the attached picture).

We notice here that the tension and the weight are acting in opposite directions, so we have such into account when finding the net force on the ball, and then solve for the tension knowing the value of the centripetal acceleration (recall that the magnitude of the tangential velocity is the same because of the uniform circular motion).

F_{net}= T_{bot}-W\\m\,a_c=T_{bot}-0.98\,N\\2.94\,N=T_{bot}-0.98\,N\\T_{bot}=(2.94+0.98)N\\T_{bot}=3.92\,N

4 0
2 years ago
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