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m_a_m_a [10]
3 years ago
11

How is work related to force and displacement?

Physics
1 answer:
VashaNatasha [74]3 years ago
3 0

Answer:

Displacement is the distance and direction an object is moved. Force and work are directly proportional to each other, while force and displacement are indirectly propotional. The equation showing the relationship is W= Fd. OR Work is a scalar. The rate at which work is done or at which energy is transferred. Work is equal to the force parallel to the displacement.

Explanation:

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10. Suppose the tern travels 1.70*10^ km south, only to encounter bad weatherInstead of trying to fly around the stormthe tem tu
Ghella [55]

We can define average speed as the quotient between the whole distance traveled and the time it takes to travel that distance, and the velocity as the quotient between the displacement and the time it takes to displace that quantity.

We will find that:

Average speed =  2.6×10^2 km/day

Velocity = 2.52×10^2 km/day

So we know that:

The tent travels:

1.70×10^4 km to the south

6.00×10^2 km to the north

1.44×10^4 km to the south.

The total distance traveled is just the sum of the 3 above distances:

total distance = 1.70×10^4 km +  6.00×10^2 km + 1.44×10^4 km

Notice that the second term has a different exponent than the others, so to have the same exponent we can write:

6.00×10^2 km = 0.06×10^4 km

Now we get:

total distance = 1.70×10^4 km + 0.06×10^4 km + 1.44×10^4 km

total distance = (1.70 + 0.06 + 1.44)×10^4 km

<u>total distance = 3.2×10^4 km</u>

And it travels that distance in 122 days, then the average speed is:

AS = (3.2×10^4 km)/(122 days) = 2.6×10^2 km/day

Now we need to find the displacement, which is the difference between the final position and the initial position, the displacement will just be:

d =  1.70×10^4 km - 0.06×10^4 km + 1.44×10^4 km

(We add the distance traveled to south and subtract the distance that it travels due north)

d = (1.70 - 0.06 + 1.44)×10^4 km = 3.08×10^4 km

Then the velocity will be:

v = (3.08×10^4 km)/(122 days) = 2.52×10^2 km/day

Because we defined south as positive and north as negative, having a positive velocity means that the direction of the <u>velocity is due south.</u>

If you want to learn more, you can read:

brainly.com/question/12322912

 

3 0
3 years ago
Manganese-52 has a half-life of 6 days. How many days would a scientist have to wait for the radioactivity to be 12.5% the start
sineoko [7]

Answer:

Let N = N0 where N0 is the number of atoms originally present.

In 6 days    N = N0 / 2

In 12 days   N = N0 / 4

In 18 days   N = N0 / 8    = .125 N0

So it would take 18 days.

4 0
3 years ago
Read 2 more answers
What determines the quality of a conductor
aksik [14]
<span>number of free electrons present.</span>
8 0
3 years ago
A stone is dropped from the edge of a roof, and hits the ground with a velocity of -180 feet per second. How high (in feet) is t
Neko [114]

Answer:

d = 506.25 ft

Explanation:

As we know by kinematics that

v_f^2 - v_i^2 = 2 a d

here we know that initially the stone is dropped from rest from the edge of the roof

so here initial speed will be zero

now we have

v_i = 0

also the acceleration of the stone is due to gravity which is given as

g = 32 ft/s^2

now we have

v_f = 180 ft/s

so from above equation

180^2 - 0 = 2(32)d

d = 506.25 ft

6 0
3 years ago
You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac
dusya [7]

Answer:

20.62361 rad/s

489.81804 J

Explanation:

I_i = Initial moment of inertia = 9.3 kgm²

I_f = Final moment of inertia = 5.1 kgm²

\omega_i = Initial angular speed = 1.8 rev/s

\omega_f = Final angular speed

As the angular momentum of the system is conserved

I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

6 0
3 years ago
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