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2 years ago

Please help me Why do objects in the sky appear to move and change so much?

Physics

1 answer:

bearhunter [10]2 years ago

6 0

Answer:

It depends on the climate and wind speed at the time.

If the climate is hot and no breeze it will move slower.

If it is cold and very windy it will move rapidly.

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PLEASE HELP WILL GIVE BRAINLIEST!!!!

zimovet [89]

Answer:

pelvic gridle

Explanation:

8 0

3 years ago

Landslides are most common in

nikitadnepr [17]

shorelines of the southeast U.S.

3 0

3 years ago

A hydraulic system for a dentist's chair is designed to be able to lift 3,112 newtons. The surface area over which this force is

In-s [12.5K]

Answer:

13.8 N

Explanation:

Pressure on the one end of the hydraulic system = Pressure on the other end

Pressure = Force / Area where Force is in Newton, area is in m²

so Force of one end (F1) / area of that end = force of the other end (F2) / area of that end

3112 / ( 707 /10000) in m² = F2 / ( 3.14 / 10000) in m²

cross multiply

44016.97 × 0.000314 = 13.82 N

5 0

2 years ago

Kelli weighs 425 N, and she is sitting on a playground swing that hangs 0.36 m above the ground. Her mom pulls the swing back an

kodGreya [7K]

Answer:

V = 3.54 m/s

Explanation:

Using the conservation of energy:

$E_i = E_f$

so:

$wh = \frac{1}{2}mV^2$

where w is te weigh of kelly, h the distance that kelly decends, m is the mass of kelly and V the velocity in the lowest position.

So, the mass of kelly is:

m = 425N/9.8 = 43.36 Kg

and h is:

h = 1m-0.36m =0.64m

then, replacing values, we get:

$(425N)(0.64m) = \frac{1}{2}(43.36kg)v^2$

Solving for v:

V = 3.54 m/s

7 0

3 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

3 years ago