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Jobisdone [24]
4 years ago
13

How much force is required to accelerate a 22Kg mass at 6 m/s?

Physics
2 answers:
GuDViN [60]4 years ago
7 0

Answer:

F = 132N

Explanation:

mel-nik [20]4 years ago
3 0
The correct one is this

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Radda [10]
A. 320 g
B. 160 g
C. 80 g
D. 40 g
6 0
3 years ago
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Assuming that voltage remains constant, what happens to the current in a filament if it is replaced by a longer wire?
Elena L [17]

Answer:

Choice B: The current decreases.

Explanation:

The resistance of a conductor is proportional to its length when radius and resistivity stays the same. The resistance of the longer filament will be greater than the resistance of the initial one.

For Ohmic conductors,

\displaystyle I = \frac{V}{R},

where

  • I is the current through the conductor,
  • V is the voltage across the conductor, and
  • R is the resistance of the conductor.

Voltage here stays the same. Increasing the resistance R of the conductor will reduce the current.

The filament might heat up over time. The filament might not be an ohmic conductor. Still, a similar trend shall exist. Resistance will be greater in the longer filament, and current will decrease.

4 0
4 years ago
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A body travels a distance of 20m in the 7th second and 24m in the 9th second. How much distance shall it travel in the 15th seco
DaniilM [7]

Answer:

<u>36 m</u>

Explanation:

We can consider this to be an AP.

Then,

  • a₇ = 20
  • a₉ = 24

<u>Subtract a₇ from a₉.</u>

  • a + 8d - a + 6d = 24 - 20
  • 2d = 4
  • d = 2

  • a + 6(2) = 20
  • a = 8

<u>Finding a₁₅</u>

  • a₁₅ = a + 14d
  • a₁₅ = 8 + 14(2)
  • a₁₅ = 8 + 28
  • a₁₅ = <u>36 m</u>
4 0
2 years ago
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Hi i think it 573 udaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
weqwewe [10]

Answer:

ummm I didn't understand the question

4 0
3 years ago
A truck drives with a constant linear speed v_iv i ​ v, start subscript, i, end subscript down a road with two curves. The first
Alex

Answer:

(C) Decreases by factor of 3

Explanation:

Centripetal acceleration is given by

a = \dfrac{v^2}{r}

where <em>v</em> is the linear velocity and <em>r</em> is the radius of the curve.

Let the centripetal acceleration on the curve of radius <em>R</em> be a_1.

Then

a_1 = \dfrac{v_i^2}{R}

Let the centripetal acceleration on the curve of radius 3<em>R</em> be a_2.

Then

a_2 = \dfrac{v_i^2}{3R} = \dfrac{1}{3}\dfrac{v_i^2}{R} = \dfrac{1}{3}a_1

Here, we see that the acceleration decreases by a factor of 3.

7 0
3 years ago
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