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scoray [572]
3 years ago
15

A car traveled at an average speed of 60 MPhil for two hours how far did it travel

Physics
2 answers:
masha68 [24]3 years ago
8 0
<span>120 miles is the answer

</span>
astra-53 [7]3 years ago
5 0
Here, Speed = 60 Miles/hour
Time = 2 hour

Substitute it in the expression: Distance = Speed * Time
Distance = 60*2 = 120 Km

So, your final answer is 120 Km

Hope this helps!
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Listed following are the names and mirror diameters for six of the world’s greatest reflecting telescopes used to gather visible
ziro4ka [17]

Answer:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope

Explanation:

How much light a telescope can collect depends on its diameter, since in a bigger area more photons will be collected.    

Remember that in a circle the area is defined as:

A = \pi r^{2}  (1)

Where A is the area and r is its radius.

However, the radius can be determined by means of its diameter.

     

d = 2r

r = \frac{d}{2} (1)

Where d is its diameter.

An example of this is when a person is collecting raindrops with a bucket and with a cup. Since the bucket has a bigger area than the cup, it will collect more raindrops by unit of time. In this scenario the raindrops represent the photons.  

   

To determine the light collecting area of each telescope, equation 2 will be replaced in equation 1.

A = \pi (\frac{d}{2})^{2}  (3)

Case for Large binocular telescope:

A_{mirror1} = \pi (\frac{8.4m}{2})^{2}    

A_{mirror1} = 55.41m        

For the second mirror will be the same value

A = A_{mirror1}+A_{mirror2}  

A = 55.41m+55.41m

A= 110.82m

Case for Keck 1 telescope:

A = \pi (\frac{10m}{2})^{2}    

A = 78.53m  

Case for Hobby-Ebberly telescope:

A = \pi (\frac{9.2m}{2})^{2}    

A = 66.47m  

Case for Subaru telescope:

A = \pi (\frac{8.3m}{2})^{2}    

A = 54.10m  

Case for Gemini North telescope:

A = \pi (\frac{8m}{2})^{2}    

A = 50.26m  

Case for Magellan 2 telescope:

A = \pi (\frac{6.5m}{2})^{2}    

A = 33.18m  

Hence, they may be rank in the following way:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope.

<em>Key term:</em>

<em>Photons: particles that constitute light. </em>

3 0
3 years ago
A tensile test specimen has a gage length = 50 mm and its cross-sectional area = 100 mm2. The specimen yields at 48,000 N, and t
Shalnov [3]

Answer:

a) yield strength

   \sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa

b) modulus of elasticity

strain calculation

\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046

strain for offset yield point

\varepsilon_{new} = \varepsilon_0 -0.002

                              =0.0046-0.002 = 0.0026

now, modulus of elasticity

 E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}

    = 184615.28 MPa = 184.615 GPa

c) tensile strength

 \sigma_u =\dfrac{F_{max}}{A}=\dfrac{87000}{100}=870MPa

d) percentage elongation

\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%

e) percentage of area reduction

\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%                            

7 0
4 years ago
Is it possible that a speed of 254 and a speed of 100 could be the same speed?
EastWind [94]
Not if both speeds are in the same units.

However, if the 254 is 'centimeters per time' and the 100 is 'inches per time',
then the speeds are equal.
4 0
3 years ago
Consult Interactive Solution 10.37 to explore a model for solving this problem. A spring is compressed by 0.0647 m and is used t
padilas [110]

Answer:

\omega=32.14\ rad/s

Explanation:

Given that,

The compression in the spring, x = 0.0647 m

Speed of the object, v = 2.08 m/s

To find,

Angular frequency of the object.

Solution,

We know that the elation between the amplitude and the angular frequency in SHM is given by :

v=\omega\times A

A is the amplitude

In case of spring the compression in the spring is equal to its amplitude

\omega=\dfrac{v}{A}

\omega=\dfrac{2.08\ m/s}{0.0647\ m}

\omega=32.14\ rad/s

So, the angular frequency of the spring is 32.14 rad/s.

4 0
3 years ago
Which statement correctly describes a characteristic that a scientific measuring tool should have
Reil [10]

Explanation:

To be accurate, it must be able to make measurements that are close to the actual value.

6 0
3 years ago
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