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Doss [256]
4 years ago
8

StarsA. begin as protostars, which fire up when they collapse and become denser and hotter.B. create elements by splitting the n

uclei of small atoms into the nuclei of larger atoms.C. have unlimited amounts of fuel and therefore exist indefinitely.D. all explode cataclysmically when they die and contribute their matter to future star generations
Physics
1 answer:
Colt1911 [192]4 years ago
4 0

Answer:

A

Explanation:

Begin as protostars, which fire up when they collapse and become denser and hotter.

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: Consider a transformer used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 4 turns in
Aneli [31]

Answer:

a) 096V  b) 0.0288A  c) 0.3456W

Explanation:

a) Vp/Vs= Np/Ns

 120/Vs= 500/4

Vs= 096V

b) Np/Ns= Is/Ip

500/4= 3.6/Ip

Ip= 0.0288A

c) P= VI

P=(120)(0.0288)

P= 0.3456W

8 0
3 years ago
An ambulance is rapidly approaching you at a stop light. What happens to the frequency and pitch of the sound as the ambulance d
vagabundo [1.1K]

Answer:

A )

Explanation:

This change in frequency observation occur due to doppler effect

if the wave source moves,In the time between one wave peak being emitted and the next, the source will have moved so that the shells will no longer be concentric. The wavefronts will get closer together in front of the source as it travels and will be further apart behind it.  (see the graph)

when the person standing still in front of the ambulance, he will observe a <em>higher frequency </em>than before as the source travels towards them.

f_(observed)=\frac{V_(wave)}{V_(wave)-V_(source)} *f_(original)

The pitch we hear depends on the frequency of the sound wave.

A high frequency corresponds to a high pitch

as we hear a higher frequency , it makes the <em>pitch higher</em> too

5 0
3 years ago
Read 2 more answers
A ballast is dropped from a stationary hot-air balloon that is at an altitude of 576 ft. Find (a) an expression for the altitude
Nadya [2.5K]

Answer:

<h2>a) S = \frac{1}{2}gt^2\\</h2><h2>b) 6secs</h2><h2>c) 192ft</h2>

Explanation:

If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;

S = ut + \frac{1}{2}at^{2}

S is the altitude of the ballest

u is the initial velocity

a is the acceleration of the body

t is the time taken to strike the ground

Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s

Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g

Substituting this values into the equation of the motion;

S = 0 + \frac{1}{2}gt^2\\ S = \frac{1}{2}gt^2\\

a) An expression for the altitude of the ballast after t seconds is therefore

S = \frac{1}{2}gt^2\\

b) Given S = 576ft and g = 32ft/s², substituting this into the formula in (a);

576 = \frac{1}{2}(32)t^2\\\\\\576*2 = 32t^2\\1152 = 32t^2\\t^2 = \frac{1152}{32} \\t^2 = 36\\t = \sqrt{36}\\ t = 6.0secs

This means that the ballast strikes the ground after 6secs

c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.

v = 0 + 32(6)

v = 192ft

7 0
4 years ago
A 14-kg object moving with a constant velocity
natta225 [31]

The final velocity of the 14 kg object is 1.6 m/s in the same direction

Explanation:

We can solve this problem by using the law of conservation of momentum: the total momentum of the system must be conserved before and after the collision. Therefore, we can write

p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

m_1 = 14 kg is the mass of the first object

u_1 = 5.0 m/s is the initial velocity of the first object

v_1 is the final velocity of the first object

m_2 = 8.0 kg is the mass of the second object

u_2 = 3.0 m/s is the initial velocity of the second object

v_2 = 9.0 m/s is the final velocity of the second object

Re-arranging the equation and substituting the values, we find:

v_1 = \frac{m_1 u_1 + m_2 u_2 - m_2 v_2}{m_1}=\frac{(14)(5.0)+(8.0)(3.0)-(8.0)(9.0)}{14}=1.6 m/s

And the direction is the same as the initial direction, since it has the same sign.

Learn more about conservation of momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

4 0
3 years ago
How far can a mother push a 20.0 kg baby carriage, using a force of 62 N, if she can only do 2920 J of work? (Round to include t
alexgriva [62]
The mathematical definition of work (W) is force (F) multiplied by distance (x). In order to determine the distance for fixed force and work the above equation needs to be rearranged to make x the subject. The work divided by the force is equal to the distance. In this case the mother can push the baby carriage by a distance equal to 2920 divided by 62, which is 47.1 metres. 
5 0
4 years ago
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