Which one is the answer please

A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of

Nastasia [14]

Answer:

Amplitude = 0.058m

Frequency = 6.25Hz

Explanation:

Given

Amplitude (A) = 8.26 x 10-2 m

Frequency (f) = 4.42Hz

Conversation of energy before split

½mv² = ½KA²

Make A the subject of formula

A = $v\sqrt{\frac{m}{k} }$

Conversation of energy after split

½(m/2)V'² = ½(m/2)V² = ½KA'²

½(m/2)V² = ½KA'²

Make A the subject of formula

First divide both sides by ½

(m/2)V² = KA'²

Divide both sides by K

$\frac{m}{2K}$ V² = A'²

$v\sqrt{\frac{m}{2k} }$ = A'

Substitute $v\sqrt{\frac{m}{k} }$ for A in the above equation

A' = A/√2

A' = 8.26 x 10^-2/√2

A' = 0.05840702012600882

Amplitude after split = 0.058 (Approximated)

Frequency (f') = f√2

f' = 4.42√2

f' = 6.25082394568908011

Frequency after split = 6.25Hz (approximated)

8 0

3 years ago

Which has the most momentum?

boyakko [2]

Answer:

Both objects have the same magnitude of momentum.

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum of that object would be $m\, v$ .

The $100\; {\rm g}$ ( $0.1\; {\rm kg}$ ) object is moving at a speed of $1\; {\rm m\cdot s^{-1}}$ . The magnitude of the momentum of this object would be $0.1\; {\rm kg} \times 1\; {\rm m\cdot s^{-1}} = 0.1\; {\rm kg \cdot m \cdot s^{-1}}$ .

Similarly, the momentum of the $1\; {\rm g}$ ( $10^{-3}\; {\rm kg}$ ) object moving at a speed of $100\; {\rm m\cdot s^{-1}}$ would be $10^{-3}\; {\rm kg} \times 100\; {\rm m\cdot s^{-1}} = 0.1\; {\rm kg \cdot m \cdot s^{-1}}$ .

Hence, the magnitude of momentum is the same for the two objects.

7 0

2 years ago

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the

algol13

Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa

<u>Explanation:</u>

Given -

Stress Direction, A = [1 0 0 ]

Slip plane = [ 1 1 1]

Normal to slip plane, B = [ 1 1 1 ]

Critical stress, Sc = 2.92 MPa

Let the direction of slip on = [ 1 1 0 ]

Let Ф be the angle between A and B

cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3

cos Ф = 1/√3

σ = Sc / cosФ cosλ

For slip along [ 1 1 0 ]

cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1

cos λ = 1/√2

Therefore,

σ = 2.92 / 1/√3 1/√2

σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa

Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa

4 0

3 years ago

James Joule (after whom the unit of energy is named) claimed that the water at the bottom of Niagara Falls should be warmer than

Molodets [167]

Answer:

0.12 K

Explanation:

height, h = 51 m

let the mass of water is m.

Specific heat of water, c = 4190 J/kg K

According to the transformation of energy

Potential energy of water = thermal energy of water

m x g x h = m x c x ΔT

Where, ΔT is the rise in temperature

g x h = c x ΔT

9.8 x 51 = 4190 x ΔT

ΔT = 0.12 K

Thus, the rise in temperature is 0.12 K.

7 0

3 years ago

What is the use of force to move an object over a distance?

steposvetlana [31]

Answer:

In physics, work is defined as the use of force to move an object. For work to be done, the force must be applied in the same direction that the object moves. Work is directly related to both the force applied to an object and the distance the object moves. <em>[I HOPE THIS HELPS* PLS MARK ME BRAINLIEST]</em>

6 0

3 years ago