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Scrat [10]
4 years ago
9

What must be your car’s average speed in order to travel 235km in 3.25 hours?

Physics
1 answer:
Serhud [2]4 years ago
3 0
235/ 3.25 = 72km/hr

Answer: 72km/hr
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Determine if a sentence is an argument?​
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The tone really matters and if there are any exclamation marks also.

Explanation:

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Que es electromagnetico?
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Could someone please explain to me Archimede's principle without using the term displace (or at least explaining what it means)?
disa [49]
Does this help?

When an object is immersed in a fluid (in this case water, but may include both liquids and gases) the fluid exerts an upward force on the object which is called buoyancy force or <span>up-thrust.  Archimedes’ Principle states that the buoyant force (upward push or force) applied to an object is equal to the weight of the fluid that the object takes the space of by that object.  Thus when an object is placed in water the rise in the water level is dictated by the mass of that object.</span>

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4 0
3 years ago
Refractive index of water is 1.3. calculate the speed of light in water.​
wolverine [178]

Answer:

The speed of light is faster in water. The Refractive index of water is 1.3 and the refractive index of glass is 1.5. From the equation n = c/v, we know that the refractive index of a medium is inversely proportional to the velocity of light in that medium. Hence, light travels faster in water.

5 0
3 years ago
Read 2 more answers
A yet-to-be-built spacecraft starts from Earth moving at constant speed to the yet-tobe-discovered planet Retah, which is 20 lig
Gre4nikov [31]

Answer:

The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

Explanation:

From the question we are told that

The distance between earth and Retah is  d = 20 \ light \ hours =  20 * 3600 *  c =  72000c \ m

Here c is the peed of light with value c =  3.0*10^8 m/s

The time taken to reach Retah from earth is  t =  25 \ hours  =  25 * 3600 =90000 \ sec

The velocity of the spacecraft is mathematically evaluated  as

     v_s =  \frac{d }{t}

substituting values

   v_s =  \frac{72000 * 3.0*10^{8} }{90000}

    v_s =  2.40*10^{8} \ m/s

The time elapsed in the spacecraft’s frame is mathematically evaluated as

      T  =  t *  \sqrt{ 1 -  \frac{v^2}{c^2} }

substituting value

       T  =  90000 *  \sqrt{ 1 -  \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }

        T = 54000 \ s

=>    T  = 15 \ hours

So  The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

       

7 0
3 years ago
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