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2 years ago

A worker drives a 0.500 kg spike into a rail tie with a 2.50 kg sledgeham-

Mathematics

1 answer:

Marianna [84]2 years ago

8 0

Total internal energy increases by 1760 J

Step-by-step explanation:

The kinetic energy of an object is the energy possessed by the object due to its motion.

It is calculated as

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

For the hammer in this problem:

m = 2.50 kg

v = 65.0 m/s

So its kinetic energy is

$KE=\frac{1}{2}(2.50)(65)^2=5281 J$

Then the problem says that 1/3 of the hammer's kinetic energy is converted into internal energy: therefore, the total internal energy increases by

$\frac{1}{3}KE=\frac{1}{3}(5281)=1760 J$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

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Compute the cost of driving for two different job options.4 $Driving costs 0.50 dollars per mile. Driving occurs 250 days per ye

ohaa [14]

Answer:

Compute the cost of driving for two different job options.

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Job 1

Miles each way 5

What is the cost of driving?

Job 2

Miles each way 50

What is the cost of driving?

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2 years ago

Which city has a minmmum value that is lower arcadia or oakdale

spayn [35]

Oakdale is the answer I think

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2 years ago

Which angles are linear pairs? Check all that apply.

KengaRu [80]

Answer:

srtand trv

Step-by-step explanation:

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2 years ago

Read 2 more answers

A runner increases her speed from 3.1 m/s to 3.5 m/s during the last 15 seconds of her run, what was her acceleration during her

bezimeni [28]

We know, acceleration = change in speed / time
a = (3.5 - 3.1) / 15
a = 0.4 / 15
a = 0.026 m/s²

In short, Your Answer would be: 0.026 m/s²

Hope this helps!

4 0

2 years ago

One method of slowing the growth of an insect population without using pesticides is to introduce into the population a number o

Zielflug [23.3K]

To be clear, the given relation between time and female population is an integral:
 $t = \int { \frac{P+S}{P[(r - 1)P - S]} } \,
dP$

The problem says that r = 1.2 and S = 400, therefore substituting:
 $t = \int { \frac{P+400}{P[(1.2 - 1)P - 400]}
} \, dP$ 

= $
\int { \frac{P+400}{P(0.2P - 400)} } \, dP$

In order to evaluate this integral, we need to write this rational function in a simpler way:
 $\frac{P+400}{P(0.2P - 400)} = \frac{A}{P} +
\frac{B}{(0.2P - 400)}$ 

where we need to evaluate A and B. In order to do so, let's calculate the LCD:
 $\frac{P+400}{P(0.2P - 400)} = \frac{A(0.2P -
400)}{P(0.2P - 400)} + \frac{BP}{P(0.2P - 400)}$ 

the denominators cancel out and we get:
P + 400 = 0.2AP - 400A + BP
 = P(0.2A + B) - 400A

The two sides must be equal to each other, bringing the system:
 $\left \{ {{0.2A + B = 1} \atop {-400A =
400}} \right.$ 

Which can be easily solved:
 $\left \{ {{B=1.2} \atop {A=-1}} \right.$ 

Therefore, our integral can be written as:
 $t = \int { \frac{P+400}{P(0.2P - 400)} } \,
dP = \int {( \frac{-1}{P} + \frac{1.2}{0.2P-400} )} \, dP$ 
= $- \int { \frac{1}{P} \, dP +
1.2\int { \frac{1}{0.2P-400} } \, dP$ 
= $- \int { \frac{1}{P} \, dP +
6\int { \frac{0.2}{0.2P-400} } \, dP$ 
= - ln |P| + 6 ln |0.2P - 400| + C

Now, let’s evaluate C by considering that at t = 0 P = 10000:
0 = - ln |10000| + 6 ln |0.2(10000) - 400| + C
C = ln |10000| - 6 ln |1600|
C = ln (10⁴) - 6 ln (2⁶·5²)
C = 4 ln (10) - 36 ln (2) - 12 ln (5) 
 Therefore, the equation relating female population with time requested is:
t = - ln |P| + 6 ln |0.2P - 400| + 4 ln (10) - 36 ln (2) - 12 ln (5)

8 0

2 years ago