PHYSICS COLLEGE

A spectroscope:measures light from distant objects
makes object far away look closer
receives radio signals from objects in space

Answers

Answer 1
Answer:

Answer:

Option A

Measures light from distant objects

Explanation:

A spectroscope is used to measure the use of light from a distant object to work out the object is made of.

It could be the single-most powerful tool astronomers use.

Professor Fred Watson from the Australian Astronomical Observatory says that "It lets you see the chemicals being absorbed or emitted by the light source"

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Answers

ANSWER:

D. Fission reactions can occur cheaply enough, but fusion requires very high temperatures

STEP-BY-STEP EXPLANATION:

One of the main reasons fusion power cannot be harnessed is that its power requirements are incredibly high. For fusion to occur, a temperature of at least 100,000,000°C is needed.

Therefore, the correct answer is D. Fission reactions can occur cheaply enough, but fusion requires very high temperatures

A 0.500-kg mass suspended from a spring oscillates with a period of 1.18 s. How much mass must be added to the object to change the period to 2.07 s?

Answers

Answer:

The add mass = 5.465 kg

Explanation:

Note: Since the spring is the same, the length and Tension are constant.

f ∝ √(1/m)........................ Equation 1  (length and Tension are constant.)

Where f = frequency, m = mass of the spring.

But f = 1/T ..................... Equation 2

Substituting Equation 2 into equation 1.

1/T ∝ √(1/m)

Therefore,

T ∝ √(m)

Therefore,

T₁/√m₁ = k

where k = Constant of proportionality.

T₁/√m₁ = T₂/√m₂ ........................ Equation 3

making m₂ the subject of the equation

m₂ = T₂²(m₁)/T₁²........................... Equation 4

Where T₁ = initial, m₁ = initial mass, T₂ = final period, m₂ = final mass.

Given: T₁ = 1.18 s m₁ = 0.50 kg, T₂ = 2.07 s.

Substituting into equation 4

m₂ = (2.07)²(0.5)/(1.18)²

m₂ = 4.285(1.392)

m₂ = 5.965 kg.

Added mass = m₂ - m₁

Added mass = 5.965 - 0.5

Added mass = 5.465 kg.

Thus the add mass = 5.465 kg

Is the magnet in a compass a permanent magnet or an electromagnet?

Answers

the needle of a compass is a permanent magnet and the north indicator of the compass is a magnetic north pole. the north pole of a magnet lines up with the magnetic field so a suspended compass needle will rotate it lines up with the magnetic field. Answer permanent magnet

A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. the average velocity for the entire trip is 26.5 mi/h. (a) what is the constant speed with which the car moved during the second distance d?

Answers

A distance of d is covered with 53 mile/hr initially.Time taken to cover this distance t1 = d/53 hourNext distance of d is covered with x mile hours.Time taken to cover this distance t2 = d/x hours.We have average speed = 26.5 mile / hour          

                                         = Total distance traveled/ total time taken                  

                                         = (2d)/((d)/(53)+(d)/(x)) = (2)/((1)/(53)+(1)/(x) )  = (106x)/(x+53)

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A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to move it 4.00 m to the side.(a) What is the magnitude of F when the crate is in this final position? During the crate's displacement, what are
(b) the total work done on it,
(c) the work done by the gravitational force on the crate, and
(d) the work done by the pull on the crate from the rope?
(e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force F does on the crate.
(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)?

Answers

Answer:

(a) magnitude of F = 797 N

(b)the total work done  W = 0

(c)work done by the gravitational force =  -1.55 kJ

(d)the work done by the pull  = 0

(e) work your force F does on the crate = 1.55 kJ

Explanation:

Given:

Mass of the crate, m =  220 kg

Length of the rope, L = 14.0m

Distance, d =  4.00m

(a) What is the magnitude of F when the crate is in this final position

Let us first determine vertical angle as follows

=>Sin \theta = (d )/(L)

=> \theta = Sin^(-1) (d)/(L) =

Now substituting thje values

=> \theta = Sin^(-1) (4)/(12) =

=> \theta = Sin^(-1) (1)/(3)

=> \theta = Sin^(-1)(0.333)

=> \theta = 19.5^(\circ)

Now the tension in the string resolve into components

The vertical component supports the weight

=>Tcos\theta = mg

=>T = (mg)/(cos\theta)

=>T = (230 * 9.8 )/(cos(19.5))

=>T = (2254 )/(cos(19.5))

=>T = (2254 )/(0.9426)

=>T =2391N

Therefore the horizontal force

F = TSin(19.5)

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

c) The work done by the gravitational force on the crate

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values                                                            

= -230 * 9.8* 12 ( 1 - cos(19.5) )

= -230 * 9.8* 12 ( 1 - 0.9426) )

= -230 * 9.8* 12 (0.0574)

= -230 * 9.8* 0.6888

=  -230 * 6.750

= -1552.55 J

The work done by gravity = -1.55 kJ

d) the work done by the pull on the crate from the rope

Since the pull  is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg  

WF = -(-1.55)

WF = 1.55 kJ

(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)

Here the work done by force is not equal to F*d  

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)      

The digital exchange of structured data is called ?

Answers

Answer:

Electronic data interchange

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