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nignag [31]
3 years ago
12

When net forces are not equal on an object, the object moves in the direction of: A) The lesser force B) It does not move C) The

greater force D) Towards the center of the Earth
Physics
2 answers:
Vilka [71]3 years ago
6 0

Answer:

The answer is C) The greater force and we will just ignore answer D because maybe the object is not even on the Earth. It could be floating in space.

Explanation:

It seems logical but let´s try to picture it to understand it better. Imagine a block sitting on the floor here on Earth. You are 10 years old and will compete against your 20 year old brother to see who can push harder and move the block in the direction of the other.

You are pushing as hard as you can and the block is not moving, Why? Because the net forces are equal. You have the floor pushing on the object as hard as the object is pushing on the floor due to the Earth´s gravity. Then you are pushing as hard as you can and your brother is pushing as hard as you, thus the block does not move. Next thing, your brother says he´s now going to push for real and you start moving backwards along with the block, Why? Because your 20 yo brother is now unbalancing the net forces on the block and accelerates the block in the direction of the greater force he is applying and the block will not stop moving until it finds enough resistance to do so.

Maurinko [17]3 years ago
5 0

C because bigger guys always push smaller guys more farther then vice versa plz mark me brainliest if you need more help let me know:)


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Emmy is standing on a moving sidewalk that moves at +2 m/s. Suddenly, she realizes she might miss her flight, so begins to speed
likoan [24]

Answer:

Refer to the attachment for the diagram.

3.53 m/s.

Explanation:

Acceleration is the first derivative of velocity relative to time. In other words, the acceleration is the same as the slope (gradient) of the velocity-time graph. Let t represents the time in seconds and v the speed in meters-per-second.

For 0 < x \le 1:

  • Initial value of v: \rm 2\;m\cdot s^{-1} at t = 0; Hence the point on the segment: (0, 2).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 2\; m\cdot s^{-2}.
  • Find the equation of this segment in slope-point form: v - 2 = 2 (t - 0) \implies v = 2t + 2, \quad 0 < t \le 1.

Similarly, for 1 < x \le 2:

  • Initial value of v is the same as the final value of v in the previous equation at t = 1: t = 2t + 2 = 4; Hence the point on the segment: (1, 4).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 1\; m\cdot s^{-2}.
  • Find the equation of this segment in slope-point form: v - 4 = (t - 1) \implies v = t + 3 \quad 1 < t \le 2.

For 2 < x \le 3:

  • Initial value of v is the same as the final value of v in the previous equation at t = 2: t = t + 3 = 5; Hence the point on the segment: (2, 5).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 0\; m\cdot s^{-2}. There's no acceleration. In other words, the velocity is constant.
  • Find the equation of this segment in slope-point form: v - 5 = 0 (t - 2) \implies v = 5 \quad 2 < t \le 3.

For 3 < x \le 4:

  • Initial value of v is the same as the final value of v in the previous equation at t = 3: t = 5; Hence the point on the segment: (3, 5).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm -3\; m\cdot s^{-2}. In other words, the velocity is decreasing.
  • Find the equation of this segment in slope-point form: v - 5 = -3 (t - 3) \implies v = -3t + 14 \quad 3 < t \le 4.

For 4 < x \le 5:

  • Initial value of v is the same as the final value of v in the previous equation at t = 4: t = -3t + 14; Hence the point on the segment: (4, 2).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 0\; m\cdot s^{-2}. In other words, the velocity is once again constant.
  • Find the equation of this segment in slope-point form: v - 2 = 0 (t - 4) \implies v = 2\quad 4 < t \le 5.

t = \rm 3.49\;s is in the interval 3 < x \le 4. Apply the equation for that interval: v = -3t +14 = \rm 3.53\; m \cdot s^{-1}.

4 0
2 years ago
PLEASE HELP ME WITH THIS ONE QUESTION
adoni [48]

Answer:

B) R1 = 6 V and R2 = 6V

Explanation:

In series, both resistors will carry the same current.

that current will be I = V/R = 12 / (10 + 10) = 0.6 A

The voltage drop across each resistor is V = IR = 0.6(10) = 6 V

6 0
3 years ago
Can someone tell me the answers??? thanks i need it asap!! i will give brainlist!!
Sloan [31]
Increases then decreases
4 0
3 years ago
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In everyday situations, it does seem that a force is needed to keep an object, for example the shopping trolley, moving at a con
Jobisdone [24]

Answer:

Because of friction

Explanation:

7 0
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Margarita [4]

To solve this problem we will apply the momentum conservation theorem, that is, the initial momentum of the bodies must be the same final momentum of the bodies. The value that will be obtained will be a vector value of the final speed of which the magnitude will be found later. Our values are given as,

m_1 = 1300kg

m_2 = 2400kg

u_1 = 12m/s i

u_2 = 18m/s j

Using conservation of momentum,

m_1u_1+m_2u_2 = (m_1+m_2)v_f

1300*12i-2400*18j = (1300+2400)v_f

Solving for v_f

v_f = 4.2162i-11.6756j

Using the properties of vectors to find the magnitude we have,

|v| = \sqrt{(4.2162^2)+(-11.6756)^2}

|v| = 12.4135m/s

Therefore the magnitude of the velocity of the wreckage of the two cars immediately after the collision is 12.4135m/s

6 0
2 years ago
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