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3 years ago

On what principle does a bottle opener work

Physics

1 answer:

Genrish500 [490]3 years ago

4 0

Answer:

Bottle opener works on a fulcrum.

Explanation:

A bottle opener is a second-class lever because the pivot point is at one end of the opener and the load is in the middle.

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How do objects with the same charger interact

maxonik [38]

The interaction between two like-charged objects is repulsive. ... Positively charged objects and neutral objects attract each other; and negatively charged objects and neutral objects attract each other.

7 0

3 years ago

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Vector A → has magnitude 8.78 m at 37.0 ∘ from the + x axis. Vector B → has magnitude 8.26 m at 135.0 ∘ from the + x axis. Vecto

kodGreya [7K]

Answer:

R = (- 3.72î + 8.29j)

Magnitude of R = 9.09 m

Explanation:

Let î and j represent unit vectors along the x and y axis respectively.

Vector A --> magnitude 8.78 m, direction 37.0° from the +x-axis

Let the x and y components of this vector be Aₓ and Aᵧ

A = (Aₓî + Aᵧj) m

The components given magnitude and direction from the +x-axis are calculated as

Aₓ = A cos θ and Aᵧ = A sin θ

Aₓ = (8.78 cos 37°) = 7.01 m

Aᵧ = (8.78 sin 37°) = 5.28 m

A = (7.01î + 5.28j) m

Vector B has magnitude 8.26 m and direction 135° from the +x-axis

B = (Bₓî + Bᵧj) m

Bₓ = (8.26 cos 135°) = - 5.84 m

Bᵧ = (8.26 sin 135°) = 5.84 m

B = (-5.84î + 5.84j) m

Vector C has magnitude 5.65 m and direction 210° from the +x-axis

C = (Cₓî + Cᵧj) m

Cₓ = (5.65 cos 210°) = - 4.89 m

Cᵧ = (5.65 sin 210°) = - 2.83 m

C = (- 4.89î - 2.83j) m

The resultant force is a vector sum of all the forces. Let the resultant force be R

R = (Rₓî + Rᵧj) m

R = A + B + C = (7.01î + 5.28j) + (-5.84î + 5.84j) + (- 4.89î - 2.83j)

Summing the î and j components seperately,

R = (- 3.72î + 8.29j) m

To get its magnitude,

Magnitude of R = √(Rₓ² + Rᵧ²) = √((-3.72)² + (8.29)²) = 9.09 m

8 0

3 years ago

NEED ASAP PLEASE HELP

sveta [45]

1. GPE - 40 * 2 * 10 = 800j

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3 years ago

What's earths most precious element?

atroni [7]

I believe the answer is californium because it is used in most metals and is very strong and expensive like gold and silver

3 0

3 years ago

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As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte

Klio2033 [76]

Answer:

$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

Explanation:

For this exercise we must use the principle of conservation of energy

starting point. The proton very far from the nucleus

Em₀ = K = ½ m v²

final point. The point where the proton is stopped (v = 0)

Em_f = U = q V

where the potential is

V = k Ze / r²

Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

Energy is conserved

Em₀ = Em_f

½ m v² = e ( $k \frac{Ze}{r^2}$ )

$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

with this expression we can find the closest approach distance (r)

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3 years ago