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Fynjy0 [20]
3 years ago
12

18. When considering a magnet's force of attraction over a known distance, you would expect the strength of attraction 4 inches

away from the magnet to be _______ weaker than the strength of attraction at 1 inch from the magnet.
A. 8 times
B. 4 times
C. 16 times
D. 6 times

Physics
2 answers:
trapecia [35]3 years ago
4 0

When considering a magnet's force of attraction over a known distance, you would expect the strength of attraction 4 inches away from the magnet to be 16 times weaker than the strength of attraction at 1 inch from the magnet.

Answer: option C

<u>Explanation: </u>

The magnetic force of attraction is inversely dependent on the distance between the object and the magnet. That is the magnetic force of attraction is inverse proportional relation with the square of the distance between the object and the magnet.

Therefore, when the distance is 4 inches away, the magnetic force of attraction would be square of the distance that is 16 times. And as it is away, the force will be weaker.

sineoko [7]3 years ago
3 0
C. 16 times
Because depending on this ... (4)^2= 16
And (1)^2 =1

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At what angle torque is half of max
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Which value represents accuracy: the mean, median, mode, range from center, or distance between farthest beanbags?
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Answer:

the mean

Explanation:

The mean (informally, the “average“) is found by adding all of the numbers together and dividing by the number of items in the set: 10 + 10 + 20 + 40 + 70 / 5 = 30. The median is found by ordering the set from lowest to highest and finding the exact middle. The median is just the middle number: 20.

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3 years ago
A 332 kg mako shark is moving in the positive direction at a constant velocity of 2.30 m/s along the bottom of a sea when it enc
Digiron [165]

To solve this problem we will apply the concepts related to the conservation of momentum. By definition we know that the initial moment must be equivalent to the final moment of the two objects therefore

p_1 = p_2

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Here,

m_{1,2} = Mass of each object

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v_{1,2}= Final velocity of each object

Since the initial velocity relative to the metal tank is at rest, that velocity will be zero. And considering that in the end, the speed of the two bodies is the same, the equation would become

m_1u_1 = (m_1+m_2)v_f

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4 0
3 years ago
An 8.75 kg point mass and a 14.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be
marysya [2.9K]

Answer

given,

mass of the = m₁ = 8.75 Kg

another mass of the object = m₂ = 14 Kg

distance between them = 50 cm

R₁ = 17 cm

R₂ = 50 -17 = 33 cm

a) Force applied due to the Mass 8.75  in +ve x- direction

F_1 = \dfrac{GM_1 m}{R_1^2}

F_1 = \dfrac{6.67\times 10^{-11} \times 8.75\ m}{0.17^2}

F_1 = 2.019\times 10^{8}\ m

Force applied due to mass 14 Kg in -ve x-direction

F_2 = \dfrac{GM_2 m}{R_2^2}

F_2 = \dfrac{6.67\times 10^{-11} \times 14\ m}{0.33^2}

F_2 = 0.857\times 10^{8}\ m

net force

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F = 2.019\times 10^{8}\ m - 0.857\times 10^{8}\ m

F = 1.162\times 10^{8}\ m

Using newton second law

a = \dfrac{F}{m}

a = \dfrac{ 1.162\times 10^{8}\ m}{m}

a =1.162\times 10^{8} \ m/s^2

b) As the acceleration of mass comes out to be  +ve hence, the direction will be toward the mass of 8.75 Kg

6 0
3 years ago
In a lesson about the behavior of gases, Genaris and her classmates learn that the volume of a gas is affected by its temperatur
Liula [17]
<span>To know if there were other factors that affected the volume of a gas, Genaris and her classmates should: </span>"formulate a new hypothesis with the same dependent variable but a different independent variable as the original hypothesis." In this case, the dependent variable is the volume of the gas and the new independent variable is a factor they think will affect the volume of the gas.
7 0
3 years ago
Read 2 more answers
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