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777dan777 [17]
3 years ago
12

A 10 N force and a 15 N force are acting from a single point in opposite directions. What additional force must be added to prod

uce a net force of 0?
5 N acting in the same direction as the 10 N force
5 N acting in the same direction as the 15 N force
10 N acting in the same direction as the 10 N force
25 N acting in the same direction as the 15 N force
Physics
1 answer:
AleksAgata [21]3 years ago
7 0

Answer:

5 N acting in the same direction as the 10 N force

Explanation:

10+5=15

15=15

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How did scientists discover the Earth had a liquid outer core and solid inner core?
viktelen [127]
Dr. Inge discovered the make up of the earths inner core by studying how an earthquakes waves bounced off the core. And Inge Lehmann was studying the waves of a 1929 earthquake when she found them acting inconsistently with solid mantle crust 
hope it helps you


6 0
3 years ago
Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time (when they are not slee
user100 [1]

Answer:

I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s ,  I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s ,  I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s  and I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s

Explanation:

The impulse is equal to the variation of the moment, to apply this relationship to our case, we will assume that initially the mouse was at rest

    I = Δp = m v_{f} -m v₀

    I = m (v_{f}  -v₀)

Bold indicates vector quantities, let's calculate the momentum of each mouse in for the x and y axes

We recommend bringing all units to the SI system

Mouse 1.

It has a mass of 22.3 g = 22.3 10⁻³ kg, a final velocity of (v = 0.349 i ^ - 0.301 j ^) m / s with an initial velocity of zero

    Iₓ = m (v_{fx}  - v₀ₓ)

    Iₓ = 22.3 10⁻³ (0.349 -0)

    Iₓ = 7.78 10⁻³ J s

   I_{y} = m (v_{fy}  -v_{oy} )

   I_{y} = 22.3 10⁻³ (-0.301)

   I_{y} = -6.71 10⁻³ J s

   I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s

Mouse 2

Mass 17.9 g = 17.9 10⁻³ kg

Speed ​​(-0.699 i ^ - 0.815 j ^) m / s

    Iₓ = m (v_{fx}  - v₀ₓ)

    Iₓ = 17.9 10⁻³ (-0.699 -0)

    Iₓ = -12.5 10⁻³ J s

    I_{y} = 17.9 10⁻³ (-0.815 - 0)

    I_{y} = -14.6 10⁻³ J s

   I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s

Mouse 3

Mass 19.1 g = 19.1 10⁻³ kg

Speed ​​(0.745i ^ + 0.975 j ^) m / s

    Iₓ = 19.1 10⁻³ (0.745 -0)

    Iₓ = 14.2 10⁻³ J s

    I_{y} = 19.1 10⁻³(0.975 -0)

    I_{y} = 18.6 10⁻³ J s

    I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s

Mouse 4

Mass 10.1 g = 10.1 10⁻³ kg

Speed ​​(-0.905i ^ + 0.717j ^) m / s

    Iₓ = 10.1 10⁻³ (-0.905 -0)

    Iₓ = -9.14 10⁻³ J s

    I_{y} = 10.1 10⁻³ (0.717 -0)

    I_{y} = 7.24 10⁻³ J s

   I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s

8 0
3 years ago
The curved section of a speedway is a circular arc having a radius of 190 m. this curve is properly banked for racecars moving a
Anestetic [448]

The banking angle of the curved part of the speedway is determined as 32⁰.

<h3>Banking angle of the curved road</h3>

The banking angle of the curved part of the speedway is calculated as follows;

V(max) = √(rg tanθ)

where;

  • r is radius of the path
  • g is acceleration due to gravity

V² = rg tanθ

tanθ = V²/rg

tanθ = (34²)/(190 x 9.8)

tanθ = 0.62

θ = arc tan(0.62)

θ = 31.8

θ ≈ 32⁰

Learn more about banking angle here: brainly.com/question/8169892

#SPJ1

8 0
2 years ago
If an astronaut can throw a certain wrench 15.0 m vertically upward on earth, how high could he throw it on our moon if he gives
Sever21 [200]

Answer:

hm= 18.2 m

Explanation:

Solution is attached

5 0
3 years ago
Identical point charges (+50 x 10 power -6C) are placed at the corners of a square with sides of 2.0-m length. How much external
Gnesinka [82]

Answer:

636.4 J

Explanation:

The potential energy between one of the charges at the corner of the square and the fifth identical charge is U = kq²/r where q = charge = +50 × 10⁻⁶ C  and r = distance from center of square. = √2 m (since the midpoint of the sides = 1 m, so the distance from the charge at the corner to the center is thus √(1² + 1²) = √2)

Since we have four charges, the additional potential energy to move the charge to the centre of the square is U' = 4U = 4kq²/r

U' = 4kq²/r

= 4 × 9 × 10⁹ Nm²/C² (+50 × 10⁻⁶ C)²/√2 m

= 900 Nm²/√2 m

= 636.4 J

8 0
3 years ago
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