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3 years ago

Can someone help me with universal gravation

1 answer:

6 0

States that particles are attracts with every other particle. wich force is directily proportional product of two masses and inversely proportional to the distance between the centers.

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A train has an initial velocity of 44m/s and an accelaration of _4m/s calculate its velocity

Kobotan [32]

Complete question:

A train has an initial velocity of 44m/s and an acceleration of -4m/s². calculate its velocity after 10s ?

Answer:

the final velocity of the train is 4 m/s.

Explanation:

Given;

initial velocity of the train, u = 44 m/s

acceleration of the train, a = -4m/s² (the negative sign shows that the train is decelerating)

time of motion, t = 10 s

let the final velocity of the train = v

The final velocity of the train is calculated using the following kinematic equation;

v = u + at

v = 44 + (-4 x 10)

v = 44 - 40

v = 4 m/s

Therefore, the final velocity of the train is 4 m/s.

7 0

3 years ago

Who was the first woman to compete and win a championship in the Olympics

natita [175]

Hélène de Pourtalès she was the first women

6 0

3 years ago

Kali left school and traveled toward her friend's house at an average speed of 40 km/h. Matt left one hour later and traveled in

zlopas [31]

Answer:

t = 5 hr

Explanation:

Let kali moves toward east with velocity= V₁= 40 km/ h

Mat moves toward west with velocity = V₂= 50 km/hr

As Klai left one hour earlier = t₁= 1 hr

distance traveled in 1st hour = s₁ = v * t = 40 * 1 = 40 km

Remaining distance = 400 - 40 = 360 km

As they move in the opposite directions:

Relative speed= 40 + 50 = 90 km/ h

s = v * t

⇒ t = s / v

⇒ t₂ = 360 / 90

⇒ t₂ = 4 hr

Total time = t = t₁ + t₂

t = 1 hr + 4 hr

t = 5 hr

5 0

3 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

All are examples of electric forces except _________.

pochemuha

The correct answer is A

6 0

3 years ago

Can someone help me with universal gravation ​

Can someone help me with universal gravation