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Zielflug [23.3K]
3 years ago
14

The distance traveled, in feet, of a ball dropped from a tall building is modeled by the equation d(t) = 16t2 where d equals the

distance traveled at time t seconds and t equals the time in seconds. What does the average rate of change of d(t) from t = 2 to t = 5 represent?
A. The ball travels an average distance of 112 feet from 2 seconds to 5 seconds.

B. The ball falls down with an average speed of 48 feet per second from 2 seconds to 5 seconds.

C. The ball falls down with an average speed of 112 feet per second from 2 seconds to 5 seconds.

D. The ball travels an average distance of 48 feet from 2 seconds to 5 seconds.
Physics
1 answer:
Ainat [17]3 years ago
6 0
When  t=2, the ball has fallen     d(2) = 16 (2²) = 64 feet .

When  t=5, the ball has fallen     d(5) = 16 (5²) = 400 feet .

Distance fallen from  t=2  until  t=5  is  (400 - 64) = 336 feet.

Time period between  t=2  until  t=5  is  (5 - 2) = 3 seconds.

Average speed of the ball from  t=2  until  t=5  is

                 (distance covered) / (time to cover the distance)

             =            336 feet        /        3 seconds       =  112 feet per second.

That's what choice-C says.        
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A top-fuel dragster starts from rest and has a constant acceleration of 42.0 m/s2. What are (a) the final velocity of the dragst
disa [49]

Answer:

a)  Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) The displacement of the dragster at the end of 1.8 s = 68.04 m

d) The displacement of the dragster at the end of 3.6 s = 272.16 m

Explanation:

a) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

  v = u + at

  v  = 0 + 42 x 1.8 = 75.6 m/s

Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

  v = u + at

  v  = 0 + 42 x 3.6 = 75.6 m/s

Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 1.8 + 0.5 x 42 x 1.8²

    s = 68.04 m

The displacement of the dragster at the end of 1.8 s = 68.04 m

d) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 3.6 + 0.5 x 42 x 3.6²

    s = 272.16 m

The displacement of the dragster at the end of 3.6 s = 272.16 m

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How are meteors and meteorites different?
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If any part of a meteoroid survives the fall through the atmosphere and lands on Earth, it is called a meteorite.

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3 years ago
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The weight of an object is measured in air to be 7.0 N. The
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Answer:

Buoyant force = 3.0 N

The object will not float.

Explanation:

Apparent weight of a body immersed in water is the actual weight of object minus buoyant force

Given in the question that;

Weight of object in air = 7.0 N

Apparent weight of object = 4.0 N

4.0 N = 7.0 N - Buoyant force

Buoyant force = 7.0 - 4.0 = 3.0 N

In this case, the buoyant force is less than weight of the object thus the object will sink.

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One watt is equal to which of these values?
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1 W = 1 J/s 
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3 years ago
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On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi
Lostsunrise [7]

Answer:

0.06\Omega/m

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire (I=\frac{V}{R}). But this is not the case.

Let the resistance of the ammeter be r

Hence, using Ohm's law we get the following 2 equations:

\frac{13}{20z+r} =7.6   .......(1)

\frac{13}{40z+r} =4.5     ......(2)

Substituting the value of r from (2) in (1), we have,

13=152z+7.6\times\frac{13-180z}{4.5}

which simplifying gives us, z=0.0589\Omega/m\approx0.06\Omega/m (which is our required solution)

putting the value of z in either (1) or (2) gives us, r = 0.5325 \Omega

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3 years ago
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