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3 years ago

A............... pulley helps us by changing the direction of the applied effort

Physics

2 answers:

DerKrebs [107]3 years ago

8 0

Explanation:

Malai thaxai. na

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garik1379 [7]3 years ago

8 0

Answer/Explanation:

True

A fixed pulley changes the direction of the force you exert by pulling, so you can pull down to move an object up. In this fixed pulley system, you pull down on one side of the rope and the other side goes up. You use the same amount of force to pull down as you would to lift the toy yourself.

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Read the following excerpt about water availability to living organisms.

lapo4ka [179]

Answer:

1 percent

Explanation:

It says that only 3 percent of the water is fresh. So it can be 1 percent or 3 percent. But then it says that most of the water is locked up in glaciers and polar ice caps. So the animals would have a hard time getting to this water. So the rest is available for them. Approximately 1 percent is most reasonable.

5 0

3 years ago

Read 2 more answers

Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is

victus00 [196]

Answer:

3.15 N towards the positive x-axis

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = $\frac{-kQq}{r^2}$

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = $\frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2}$ = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = $\frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2}$ = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = 3.15 N towards the positive x-axis

3 0

3 years ago

The direction of an electric field is always in the direction _______________ would naturally move.

Masja [62]

Answer:

Im not really sure but Id say weather .

Explanation:

6 0

3 years ago

What is the kinetic energy of a baseball moving at a speed of 40 m/s if the baseball has mass of 0.15 kg?

bonufazy [111]

0.5mv^2 > 0.5*0.15*40*40=120J

4 0

3 years ago

Read 2 more answers

4 This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive

zubka84 [21]

Answer:at 21.6 min they were separated by 12 km

Explanation:

We can consider the next diagram

B2------15km/h------->Dock

B1 at 20km/h

So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.

Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.

3 0

3 years ago

A............... pulley helps us by changing the direction of the applied effort​

A............... pulley helps us by changing the direction of the applied effort