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1 year ago

A mass attached to a spring vibrates back and forth. At maximum displacement, the spring force and the

Physics

1 answer:

Svetradugi [14.3K]1 year ago

8 0

ANSWER

Velocity of the mass reaches zero

EXPLANATION

We want to identify what hapens to a mass attached toa a spring at maximum displacement.

When a mass attached to a spring is at its maximum position of displacement, the direction of the mass begins to change. This implies that the velocity of the mass will reach zero.

Hence, at maximum displacement, the velocity of the mass reaches zero.

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If there is unequal sharing among electrons, which of the following

Dennis_Churaev [7]

Answer:

D. Fluorine.

Explanation:

For an atom to accept an electron, from another atom, it must have high electron affinity compared to the other elements.

Large atoms then to have high electron affinity compared to smaller atoms and most large atoms are non metals while elements with smaller atomic radius are metals. Tin(Sn), Arsenic(As) are metals and they have lower electrons affinity compared to Sulfur (S) and Fluorine (F).

Fluorine have a high electron affinity than Sulfur and hence has a larger atomic radius and can easily draw most of the electrons shared between the two atoms.

There's a trend to electron affinity and I.e it increases down the period and decreases down the group.

Most group (VII)A elements form electrovalent compounds due to their tendency to pull the shared electron towards them(high electron affinity) compared to Sulfur (a group VIA) element.

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4 years ago

How dose an atom become an ion with a -2 charge?

Aloiza [94]

A atom can become an ion with a -2 charge, if it gains 2 electrons during a chemical reaction.

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4 years ago

Please help!!! The radius of a circle is 5.5 cm, the circumference in meters is 34.56, the area in square meters is 95.03. What

olga nikolaevna [1]

Answer:

47.515 cm²

In meter square= 4.7515 *10^-3 m²

Explanation:

The radius of a circle = 5.5 cm

In meter= 5.5*0.01

= 0.055 m

the circumference= 34.56 cm

in meters = 34.56 cm*0.01

= 0.3456 meters

the area= 95.03 cm²

in square meters =95.03*(0.01)²

in square meters= 9.503*10^-3 m²

. area under the curve at the right

= Area of semi circle

= Area of circle/2

= 95.03 /2

= 47.515 cm²

In meter square= 4.7515 *10^-3 m²

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3 years ago

Gayle runs at a speed of 3.85 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After s

enot [183]

Answer:

Final velocity at the bottom of hill is 15.56 m/s.

Explanation:

The given problem can be divided into four parts:

1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)

From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:

$p_i = p_f$

$m_1u_1 + m_2v_2 = (m_1 + m_2)v$

$v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)}$

$v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s$

2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.

The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}$

Here, initial velocity is the final velocity from the first stage. Therefore:

$v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s$

3. Use conservation of momentum to find the combined speed of Gayle and her brother.

Given:

Initial velocity of Gayle and sled is, $u_1(i)=10.5$ m/s

Initial velocity of her brother is, $u_2(i)=0$ m/s

Mass of Gayle and sled is, $m_1=55.0$ kg

Mass of her brother is, $m_2=30.0$ kg

Final combined velocity is given as:

$v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}$

$v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79$ m/s

4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.

Using conservation of energy, the final velocity at the bottom of the hill is:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s$

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3 years ago

In electroplating, the object to be plated is which part of an electrolytic cell? cathode or anode

Bingel [31]

Since the anode dissolves and goes to the cathode. The object to be electroplated is kept at the cathode.

Please mark me as brainliest.

7 0

3 years ago

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