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3 years ago

Help plz i need itttttttttttttt

Physics

1 answer:

OLga [1]3 years ago

4 0

Answer:

The answer is A or B. Sorry if incorrect. Hope this helps. :)

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A vertical spring stretches 9.6 cm when a 1.3 kg block is hung from its end. (a) Calculate the spring constant. This block is th

tatiyna

Answer:

F = - K x

a) K = 1.3 kg * 9.8 m/s^2 / .096 m = 133 kg/sec^2

b) ω = (K/m)^1/2 angular frequency of SHM

ω = (133 / 1.3)^1/2 = 10.1 / sec

f = 2 π ω = 6.28 * 10.1 / sec = 63.5 / sec

P = 1/f = .0157 sec

5 0

3 years ago

2. A can filled with sand has a mass of 0.65kg is swung overhead in a horizontal circle of radius 0.70m at a constant rate of 2.

Aliun [14]

<h3><u>Answer</u>;</h3>

≈ 5 Kgm²/sec

<h3><u>Explanation</u>;</h3>

Angular momentum is given by the formula

L = Iω, where I is the moment of inertia and ω is the angular speed.

I = mr², where m is the mass and r is the radius

= 0.65 × 0.7²

= 0.3185

Angular speed, ω = v/r

= (2 × 3.142 × r × 2.5) r

= 15.71

Therefore;

Angular momentum = Iω

= 0.3185 × 15.71

= 5.003635

<u>≈ 5 Kgm²/sec</u>

6 0

3 years ago

A parallel-plate capacitor is constructed from two aluminum foils of 1 square centimeter area each placedon both sides of a rubb

Svet_ta [14]

Answer:

The voltage will be 0.0125V

Explanation:

See the picture attached

4 0

4 years ago

What scientisists stated that animals, as well as plants, are made of cells.?

Andrews [41]

<span>The answer is Mathias Schleiden and <span><span>Theodor Schwann</span></span></span>

7 0

4 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

4 years ago