All categories

3 years ago

Can someone answer this please

Physics

1 answer:

Ber [7]3 years ago

6 0

Answer:

ask the question again we cant really see it

Explanation:

You might be interested in

A spring with spring constant 13.1 N/m hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest.

Alenkinab [10]

Answer:

Explanation:

spring constant, K = 13.1 N/m

22 oscillations in 20 seconds

time taken to complete one oscillation is called time period.

T = 20 / 22 second = 0.909 seconds

(a) let m be the mass.

The formula for the time period is

$T = 2\pi \sqrt{\frac{m}{K}}$

$m = \frac{T^{2}K}{4\pi ^{2}}$

$m = \frac{0.909^{2}\times 13.1}{4\pi ^{2}}$

m = 0.275 kg

(b) maximum speed, v = ω A = 2π A / T

v = ( 2 x 31.4 x 0.1) / 0.909

v = 0.691 m/s

6 0

4 years ago

I just need someone to check my answer!

Crazy boy [7]

That’s actually correct!! Good luck with the test

7 0

3 years ago

Can someone please Help me with this? It’s Due today

kirza4 [7]

helium group no val elect

mg is reactive when activated. when burned, very intense

pot\asssium 1 valence elect ... KCl eg

theone with H and sodium in it

http://perendis.webs .com

6 0

4 years ago

A 0.48 kg pool cue moving at 3.39 m/s hits the 0.22 kg cue ball that was a rest and the pool cue stops after the impact. What is

Grace [21]

Answer:

7.39 m/s is the velocity of the cue ball after impact.

Explanation:

Given that,

$\text { Mass of the pool cue is } 0.48 \mathrm{kg} .\left(\mathrm{m}_{\mathrm{i}}\right)$

$\text { Moving at velocity } 3.39 \mathrm{m} / \mathrm{s} .(\mathrm{v_i})$

$\text { Mass of the cue ball that } 0.48 \text { mass ball hits is } 0.22 \mathrm{kg} .\left(\mathrm{m}_{\mathrm{f}}\right)$

$\text { We need to find the velocity ( } \mathrm{v}_{\mathrm{f}} \text { ) of the cue ball of mass } 0.22 \mathrm{kg} \text { . }$

We know that,

If two objects collide then the “total momentum” before is equal to the “total momentum” after collision.

$\mathrm{m}_{\mathrm{i}} \mathrm{v}_{\mathrm{i}}=\mathrm{m}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}$

Substitute the given values in the formula.

$0.48 \times 3.39=0.22 \times \mathrm{V}_{\mathrm{f}}$

$\frac{1.6272}{0.22}=\mathrm{V}_{\mathrm{f}}$

$\mathrm{v}_{\mathrm{f}}=7.39 \mathrm{m} / \mathrm{s}$

3 0

4 years ago

A piece of wire 26 m long is cut into two pieces. one piece is bent into a square and the other is bent into an equilateral tria

kiruha [24]

<span>a) All the 26m should be used to bent into a square to maximize the total area (= 42.25 m2) b) 1.4136m should be bent into a square and the rest (= 24.586m) should be bent into an equilateral triangle to minimize the total area (= 29.0835m2)</span>

8 0

3 years ago