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Shalnov [3]
3 years ago
8

If we have less power, we most likely have

Physics
1 answer:
boyakko [2]3 years ago
6 0
The same amount of work being done over a long period of time!
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A 950-kg car strikes a huge spring at a speed of 22m/s (fig. 11-54), compressing the spring 5.0m. (a) what is the spring stiffne
alukav5142 [94]

(a) The spring stiffness constant of the spring is 18,392 N/m.

(b) The time the car was in contact with the spring before it bounces off in the opposite direction is 0.23 s.

<h3>Kinetic energy of the car</h3>

The kinetic energy of the car is calculated as follows;

K.E = ¹/₂mv²

K.E = ¹/₂ x 950 x 22²

K.E = 229,900 J

<h3>Stiffness constant of the spring</h3>

The stiffness constant of the spring is calculated as follows;

K.E =  U = ¹/₂kx²

k = 2U/x²

k = (2 x 229,900)/(5)²

k = 18,392 N/m

<h3>Force exerted on the spring</h3>

F = kx

F = 18,392 x 5

F = 91,960 N

<h3>Time of impact</h3>

F = mv/t

t = mv/F

t = (950 x 22)/(91960)

t = 0.23 s

Learn more about spring constant here: brainly.com/question/1968517

#SPJ4

3 0
1 year ago
How many atoms of oxygen in the chemical formula 2 Ca(ClO2)2?
alexgriva [62]

Answer:

8

Explanation:

Ca(ClO2)2  -    2*2 = 4 Oxygen atoms

2 Ca(ClO2)2  - 2*4 = 8 Oxygen atoms

8 0
3 years ago
A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 430 kg. Before a trip, the fl
fiasKO [112]

Answer:

a

  KE  =  7.17 *10^{7} \ J

b

 t = 6411.09 \ s

Explanation:

From the question we are told that

    The radius of the flywheel is  r =  1.50 \ m

      The mass of the flywheel is m  = 430 \ kg

          The rotational speed of the flywheel is w  =  5,200 \ rev/min = 5200 *  \frac{2 \pi }{60} =544.61 \ rad/sec

      The power supplied by the motor is  P  =  15.0 hp =  15 * 746 =  11190 \ W

         

     Generally the moment of inertia of the flywheel is  mathematically represented as

       I  = \frac{1}{2} mr^2

substituting values

       I  = \frac{1}{2}  ( 430)(1.50)^2

       I  = 483.75 \  kgm^2

The kinetic energy that is been stored is  

       KE  =  \frac{1}{2} * I * w^2

substituting values

        KE  =  \frac{1}{2} * 483.75 * (544.61)^2

        KE  =  7.17 *10^{7} \ J

Generally power is mathematically represented as

          P =  \frac{KE}{t}

=>      t =  \frac{KE}{P}

substituting the value

        t = \frac{7.17 *10^{7}}{11190}

        t = 6411.09 \ s

3 0
3 years ago
Determine the weight of an average physical science textbook whose mass is 3.1 kilograms. The acceleration due to gravity is 9.8
Mrrafil [7]

Given:

mass is 3.1 kilograms

The acceleration due to gravity is 9.8m/s2

 

Required:

Weight

 

Solution:

W = mg

W = (3.1 kilograms)( 9.8m/s2)

W = 30.38 Newtons

5 0
2 years ago
Consider the following examples of homeostatic regulation: In response to an increase in plasma K concentrations, secretion of t
TiliK225 [7]

Answer:

Both are examples of negative feedback regulation.

Explanation:

The maintenance of the homeostasis in the body is controlled by the the feedback regulation of the body. Two main types of feedback regulation are positive regulation and negative regulation.

The negative regulation occurs when the final product of the reactions inhibits the further secretion of that product. In the given examples of aldosterone and calcium mechanism, the secretion of aldosterone and calcium decreases as the normal levels are acheived in the body.

Thus, the answer is both are examples of negative feedback regulation.

5 0
3 years ago
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