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olga_2 [115]
3 years ago
12

How does distance affect force when the amount of work remains the same?

Physics
1 answer:
devlian [24]3 years ago
5 0
As distance increases, the amount of force needed decreases.
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An object is said to move from a position of 10m East to a position of 5m west. Determine the object's distance travelled.
motikmotik

Answer:

5 i think

Explanation:

4 0
3 years ago
One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side
anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

L1^2=L2^2=L^2=2^2+(H/2)^2  Replacing this value in the previous equation:

\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34  Solving for H:

H=\sqrt{52}

We can now, calculate the angle between L1 and the 2m segment:

\alpha = atan(\frac{H/2}{2})=60.98°

If we make a sum of forces in the midpoint of the rope we get:

-2*T*cos(\alpha ) + F = 0  where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N

7 0
3 years ago
What is the name of the ratio of the voltage applied to a circuit and the current in a circuit
malfutka [58]
Resistance is the name of the ratio of the voltage applied to a circuit and the current in a circuit. Goes under <span>Ohm's Law</span>
4 0
3 years ago
Space Station Suppose a space-station is designed in s shape of a torus such as the one depicted in Stanley Kubrick's "2001: A s
yaroslaw [1]

Answer:

w = 3.2 rev / min

Explanation:

For this exercise we will use the centrine acceleration equal to the acceleration of gravity

      a = v² / r

Angular and linear variables are related.

     v = w r

Let's replace

     a = w² r = g

     w = √ g / r

     r = d / 2

     r = 175/2 = 87.5 m

    w = √( 9.8 / 87.5)

    w = 0.3347 rad / s

Let's reduce to rotations per min

     w = 0.3347 rad / s (1 rov / 2pi rad) (60 s / 1 min)

     w = 3.2 rev / min

Suppose the space station rotates counterclockwise, we have two possibilities for the car

The first car turns counterclockwise (same direction of the station

     v_{c} =  w_{c} r

     [texwv_{c}[/tex] =  v_{c} / r

     [texwv_{c}[/tex] = 25.0 / 87.5

     [texwv_{c}[/tex] = 0.286 rad / s

When the two rotate in the same direction their angular speeds are subtracted

     w total = w -[texwv_{c}[/tex]

     w total = 0.3347 - 0.286

    w  total= 0.487 rad / s

The car goes in the opposite direction of the station the speeds add up

    w = 0.3347 + 0.286

    w = 0.62 rad / s

From this values ​​we can see that the person feels a variation of the acceleration of gravity, feels that he has less weight when he goes in the same direction of the season and that his weight increases when he goes in the opposite direction to the season.

3 0
3 years ago
In the future, a spaceship has traveled three percent of the distance to a space station. If the ship has traveled 2.9 x 10^7 mi
leva [86]

Answer:

9.38 * 10^8 miles

Explanation:

The spaceship has traveled 3% of the distance to a space station and it has traveled 2.9 * 10^7 miles.

Let the total distance from the ship's starting point to the space station be x.

This means that:

\frac{3}{100}  * x = 2.9 * 10^7\\\\=> x = \frac{2.9 * 10^7 * 100 }{3} \\\\x = 9.67 * 10^8 miles

The total distance to be traveled is 9.67 * 10^8 miles.

Therefore, the distance left to travel is:

9.67 * 10^8 - 2.9 * 19^7\\\\= 9.38 * 10^8 miles

6 0
3 years ago
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