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tekilochka [14]
4 years ago
9

A solid block, with a mass of 0.15kg, on a frictionless surface is pushed directly onto a horizontal spring, with a spring const

ant value of 316N/m, until the spring is compressed by 35cm. The block is then released. What is the block's final speed, in m/s?
Physics
1 answer:
iren [92.7K]4 years ago
4 0

Answer:

16.1 m/s

Explanation:

We can solve the problem by using the law of conservation of energy.

At the beginning, the spring is compressed by x = 35 cm = 0.35 m, and it stores an elastic potential energy given by

U=\frac{1}{2}kx^2

where k = 316 N/m is the spring constant. Once the block is released, the spring returns to its natural length and all its elastic potential energy is converted into kinetic energy of the block (which starts moving). This kinetic energy is equal to

K=\frac{1}{2}mv^2

where m = 0.15 kg is the mass of the block and v is its speed.

Since the energy must be conserved, we can equate the initial elastic energy of the spring to the final kinetic energy of the block, and from the equation we obtain we can find the speed of the block:

\frac{1}{2}kx^2=\frac{1}{2}mv^2\\v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(316 N/m)(0.35 m)^2}{0.15 kg}}=16.1 m/s

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A hypothesis can be accepted as true after ??? repeated trials. *ill give you 10 points
elena55 [62]

Answer:

true

Explanation:

I'm assuming this is a true or false question. if u get the same results after each repeated trial that is the only time a hypothesis can be used to support evidence

7 0
3 years ago
the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes fr
Dmitry [639]

Answer:

The approximate number of decays  this represent  is  N= 23*10^{10}  

Explanation:

 From the question we are told that

    The amount of Radiation received by an average american is I_a = 2.28 \ mSv

     The source of the radiation is S = 5.49 MeV \ alpha \ particle

 Generally

            1 \  J/kg = 1000 mSv

   Therefore  2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg

Also  1eV = 1.602 *10^{-19}J

  Therefore  2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg}  * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg

           An Average american weighs 88.7 kg

      The total energy received is mathematically evaluated as

        1 kg ------> 1.423*10^{16}ev \\88.7kg  --------> x

Cross-multiplying and making x the subject

           x = 88.7 * 1.423*10^{16} eV

              x = 126.2*10^{16}eV

Therefore the total  energy  deposited is x = 126.2*10^{16}eV

The approximate number of decays  this represent  is mathematically evaluated as

            N = \frac{x}{S}

Where n is the approximate number of decay

   Substituting values

                             N = \frac{126 .2*10^{16}}{5.49*10^6}  

                                  N= 23*10^{10}  

                     

             

7 0
4 years ago
In 1994, Leroy Burrell of the United States set what was then a new world record for the men’s 100 m run. He ran the 1.00  102
vovikov84 [41]

Answer:

61.33 Kg

Explanation:

From the question given above, the following data were obtained:

Distance = 1×10² m

Time = 9.5 s

Kinetic energy (KE) = 3.40×10³ J

Mass (m) =?

Next, we shall determine the velocity Leroy Burrell. This can be obtained as follow:

Distance = 1×10² m

Time = 9.5 s

Velocity =?

Velocity = Distance / time

Velocity = 1×10² / 9.5

Velocity = 10.53 m/s

Finally, we shall determine the mass of Leroy Burrell. This can be obtained as follow:

Kinetic energy (KE) = 3.40×10³ J

Velocity (v) = 10.53 m/s

Mass (m) =?

KE = ½mv²

3.40×10³ = ½ × m × 10.53²

3.40×10³ = ½ × m × 110.8809

3.40×10³ = m × 55.44045

Divide both side by 55.44045

m = 3.40×10³ / 55.44045

m = 61.33 Kg

Thus, the mass of Leroy Burrell is 61.33 Kg

5 0
3 years ago
Mr Jones launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, how far south does the arrow
DanielleElmas [232]

Answer:c

Explanation:its the answer because its the answer

4 0
3 years ago
With 51 gallons of fuel in its tank, the airplane has a weight of 2390.7 pounds. What is the weight of the plane with 81 gallons
Shtirlitz [24]

Answer: 2561.7 pounds

Explanation:

If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:

m=\frac{Y-Y_{1}}{X-X_{1}}  (1)

Where:

m=5.7 is the slope of the line

Y_{1}=2390.7pounds is the airplane weight with  51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)

X_{1}=51gallons is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)

This means we already have one point of the graph, which coordinate is:

(X_{1},Y_{1})=(51,2390.7)

Rewritting (1):

Y=m(X-X_{1})+Y_{1}  (2)

As Y is a function of X:

Y=f_{(X)}=m(X-X_{1})+Y_{1}  (3)

Substituting the known values:

f_{(X)}=5.7(X-51)+2390.7  (4)

f_{(X)}=5.7X-290.7+2390.7  (5)

f_{(X)}=5.7X+2100  (6)

Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):

f_{(81)}=5.7(81)+2100  (7)

f_{(81)}=2561.7  (8)   This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.

3 0
3 years ago
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