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3 years ago

Using the image provided, which moon phase is next?

Physics

2 answers:

ella [17]3 years ago

5 0

Answer:

Waning Crescent

Sergio [31]3 years ago

3 0

Answer: waxing crescent

Explanation:

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1) the strength of an electromagnet can be increased by

miss Akunina [59]

Using coils of fewer turns on the electromagnet

5 0

3 years ago

The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on

Neporo4naja [7]

Answer:

162.8 K

Explanation:

initial current = io

final current, i = io/8

Let the potential difference is V.

coefficient of resistivity, α = 43 x 10^-3 /K

Let the resistance is R and the final resistance is Ro.

The resistance varies with temperature

R = Ro ( 1 + α ΔT)

V/i = V/io (1 + α ΔT )

8 = 1 + 43 x 10^-3 x ΔT

7 = 43 x 10^-3 x ΔT

ΔT = 162.8 K

Thus, the rise in temperature is 162.8 K.

5 0

3 years ago

2) A skier stands at rest and begins to ski downhill with an acceleration of 3.0 m/s² {downhill). What is

olga2289 [7]

Answer:

her displacement <em>s=337.5m</em>

Explanation:

check out the above attachment ☝️

7 0

2 years ago

Read 2 more answers

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

3 years ago

Select the correct answer.

Serhud [2]

Answer:

b hope this helps

Explanation:

5 0

3 years ago

Read 2 more answers