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seraphim [82]
3 years ago
9

Which situation is an example of increasing potential energy? Question 4 options: A. a cat jumping from a tree B. pulling a wago

n uphill B. a bicyclist stopping at a stop sign C.\emptying a bucket of water
Physics
2 answers:
enyata [817]3 years ago
5 0
The answer to this  is a
jeka943 years ago
4 0
Pulling an wagon uphill I believe.
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The universal law of gravitation states that the force of attraction between two objects depends on which quantities?
Ronch [10]

Answer:

D. the masses of the objects and the distance between them

Explanation:

Gravitation is a force, a force doesn't care about the shape or density of objects, only about their masses... and distances.

And you can get it using the following equation:

f = \frac{Gm_{1}m_{2} }{d^{2} }

Where :

G is the universal gravitational constant : G = 6.6726 x 10-11N-m2/kg2

m represent the mass of each of the two objects

d is the distance between the centers of the objects.

4 0
3 years ago
Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of 10.0 m/s. At the start of a play, Matthews runs down
MA_775_DIABLO [31]

Answer:

a)   D_ total = 18.54 m,   b)        v = 6.55 m / s

Explanation:

In this exercise we must find the displacement of the player.

a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components

           sin 45 = y₁ / d

           cos 45 = x₁ / d

           y₁ = d sin 45

           x₁ = d sin 45

           y₁ = 8 sin 45 = 5,657 m

           x₁ = 8 cos 45 = 5,657 m

The second offset is d₂ = 12m at 90 of the 50 yard

            y₂ = 12 m

            x₂ = 0

total displacement

          y_total = y₁ + y₂

          y_total = 5,657 + 12

          y_total = 17,657 m

          x_total = x₁ + x₂

          x_total = 5,657 + 0

          x_total = 5,657 m

          D_total =   17.657 i^+ 5.657 j^  m

          D_total = Ra (17.657 2 + 5.657 2)

          D_ total = 18.54 m

b) the average speed is requested, which is the offset carried out in the time used

           v = Δx /Δt

the distance traveled using the pythagorean theorem is

         r = √ (d1² + d2²)

          r = √ (8² + 12²)

          r = 14.42 m

The time used for this shredding is

         t = t1 + t2

         t = 1 + 1.2

         t = 2.2 s

let's calculate the average speed

         v = 14.42 / 2.2

         v = 6.55 m / s

8 0
3 years ago
How does a compound differ from a mixture?
Mashcka [7]
A compound is the substances that are formed by combining two are more chemical elements. A mixture is a substance created from two or more matter that can be separate with the help of physical methods. ... Mainly pure water is part of the compound. Mixtures fall under impure water.

Hope this helps!

Have a great day!
4 0
3 years ago
Read 2 more answers
O que é multifásicos?
Oksana_A [137]

Answer:

i would like to help you but I can't help you i don't understand the question and i don't speak Spanish

4 0
3 years ago
A 1.50 µF capacitor and a 3.50 µF capacitor are connected in series across a 2.50 V battery. How much charge (in µC) is stored o
Nataly_w [17]

Explanation:

The given data is as follows.

      C_{1} = 1.50 \times 10^{-6} F

      C_{1} = 3.50 \times 10^{-6} F    

      Voltage = 2.50 V

Hence, calculate the equivalence capacitor as follows.

    \frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

    \frac{1}{C} = \frac{1}{1.50 \times 10^{-6} F} + \frac{1}{3.50 \times 10^{-6} F}

                 = 0.945 \times 10^{-6} F

          C = 1.06 \times 10^{-6} F

Now, we will calculate the charge across each capacitance as follows.

              Q = CV

                  = 1.06 \times 10^{-6} F \times 2.50 V

                  = 2.65 \times 10^{-6} C

                  = 2.65 \mu C

Thus, we can conclude that 2.65 \mu C is the charge stored on each given capacitor.

5 0
3 years ago
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