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Kisachek [45]
3 years ago
11

A copper rod is 0.450 m long. Heat is added to the rod until it expands by 0.00130 m. How much did the temperature change? (Unit

=C)
Physics
1 answer:
AnnyKZ [126]3 years ago
3 0

Answer:

170 ⁰C

Explanation:

α= ΔL/(LO×ΔT)

HERE α is coefficient of linear expansion which is constant for different materials , its value for copper is 17×10^-6( ⁰C)^-1

ΔL is change in length , Lo is original length , ΔT is change in temperature

re arranging the equation for ΔT we get

ΔT = ΔL/(LO×α)

    = (0.00130)/(0.450× 17×10^-6)

    = 169.9⁰C

    = 170 ⁰C

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A seagull flies at a velocity of 9.00 m/s straight into the wind.
RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity,\rm v_{SA}  = 9 \ m/sec

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

a)

The seagull's relative velocity with reference to the ground as;

\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec

Air velocity with reference to the ground is;

\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec

b)

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec

The time the bird takes;

\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7  \ min \  and  \ 42  \ sec

c)\

The total round-trip time compared to what it would be with no wind. is;

\rm  t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec

The time for the round trip is;

\rm  t = \frac{12 \times 10^ 3 }{ 9 \ m/sec }  \\\\ t  = 1333.33 \ sec

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be  4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

4 0
2 years ago
A 2.3 kg cart is rolling across a frictionless, horizontal track towards a 1.5 kg cart that is initially held at rest. The carts
Inga [223]

Answer:

total momentum = 8.42 kgm/s

velocity of the first cart is 3.660 m/s

Explanation:

Given data

mass m1 = 2.3 kg

mass m2 = 1.5 kg

final velocity V2 = 4.9 m/s

final velocity V3 = - 1.9 m/s

to find out

total momentum  and velocity of the first cart

solution

we know mass and final velocty

and initial velocity of second cart V1 = 0

so now we can calculate total momentum that is m1 v2 + m2 v2

total momentum =  2.3 ×4.9 + 1.5 ×(-1.9)

total momentum = 8.42 kgm/s

and

conservation of momentum  is

m1 V + m2 v1  = m1 v2  + m2 v3

put all value and find V

2.3 V + 1.5 ( 0) = 2.3 ( 4.9 ) + 1.5 ( -1.9)

V = 8.42 / 2.3

V = 3.660 m/s

so velocity of the first cart is 3.660 m/s

8 0
3 years ago
I'm not sure what equation to use.
Lelechka [254]
I would think that you would multiply then divide
7 0
3 years ago
A uniform 2.50m ladder of mass 7.30kg is leaning against a vertical wall while making an angle of 51.0degree with the floor. A w
Zigmanuir [339]

Answer:

19.95 J

Explanation:

The center of mass of the ladder is initially at a height of:

h_1=\frac{L}{2}sin\theta

The center of mass of the ladder ends at a height of:

h_2= \frac{L}{2}sin90 =L/2

So, the work done is equal to the change in potential energy which is:

W = PE = mg(h_2-h_1)

now h_2-h_1= 1-sin\theta

therefore

W = [mgL/2]×[1 - sin(theta)]

W = [(7.30)(9.81)(2.50)/2]×[1-sin(51°)]

solving this we get

W = 19.95 J

8 0
3 years ago
What is the mass of an object with a density of 18 g/cm3 and a volume of 25 cm3
DiKsa [7]

Answer: 0.72 grams

Explanation: Mass can be extracted from the formula of density. D=M/V where D is density and V is volume. Therefore:

18 g/cm^3 = M(25 cm^3) --> Divide by 18g/cm^3 by 25 cm^3 to isolate mass. --> <u>0.72 =M </u> --> Now, to find out which unit you need to use for mass, just look at the density. You can see it is in g/cm^3, and cm^3 was already used for the volume. Thus, gram units are left, so that will be the unit needed, making the final answer 0.72 grams. Hope this helps :)

5 0
1 year ago
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