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zloy xaker [14]

3 years ago

11

The leader of a bicycle race is traveling with a constant velocity of +12.60 m/s and is 11.4 m ahead of the second-place cyclist

. the second-place cyclist has a velocity of +9.20 m/s and an acceleration of +1.50 m/s2. how much time elapses before he catches the leader?

1 answer:

Svetllana [295]3 years ago

8 0

Let the time be t.

so,In time t , distance travelled by 1st cyclist = 12.6 t

distance travelled by 2nd cyclist = 9.2t + 0.5 (1.5) t^2

Now, cyclist 1st is already 11.4 m ahead of 2nd cyclist.

so, 9.2t + 0.5 (1.5) t^2 = 11.4 + 12.6t

find t :

t = 6.77 sec

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Uranium-235 decays to thorium-231 with a half-life of 700 million years. When a rock was formed, it contained 6400 million urani

Dahasolnce [82]

Answer:

proof in explanation

Explanation:

First, we will calculate the number of half-lives:

$n = \frac{t}{t_{1/2}}$

where,

n = no. of half-lives = ?

t = total time passed = 2100 million years

$t_{1/2}$ = half-life = 700 million years

Therefore,

$n = \frac{2100\ million\ years}{700\ million\ years}\\\\n = 3$

Now, we will calculate the number of uranium nuclei left ( $n_u$ ):

$n_u = \frac{1}{2^{n} }(total\ nuclei)\\\\n_u = \frac{1}{2^{3} }(6400\ million)\\\\n_u = \frac{1}{8}(6400\ million)\\\\n_u = 800\ million$

and the rest of the uranium nuclei will become thorium nuclei ( $u_{th}$ )

$n_{th} = total\ nuclei - n_u\\n_{th} = 6400\ million-800\ million\\n_{th} = 5600\ million$

dividing both:

$\frac{n_{th}}{n_u}=\frac{5600\ million}{800\ million} \\\\n_{th} = 7n_u$

<u>Hence, it is proven that after 2100 million years there are seven times more thorium nuclei than uranium nuclei in the rock.</u>

6 0

2 years ago

Two identical small charged spheres hang in equilibrium with equal masses (0.02kg). The length of the strings is equal (0.18m) a

Yuliya22 [10]

Answer:

The value is $q = 3.4 *10^{-6} \ C$

Explanation:

From the question we are told that

The mass of each sphere is $m_1 = m_2 = m = 0.020 \ kg$

The length of the string is $l = 0.18 \ m$

The angle of with the vertical is $\theta = 7^o$

The acceleration due to gravity is $g = 9.8 \ m/s^2$

Generally the force acting between the forces is mathematically represented as

$F = T cos \theta = \frac{k* q^2}{ r^2}$

=> $T cos \theta = \frac{k* q^2}{ r^2}$

Generally from Pythagoras theorem the radius of the circular curve created by the force is

$r = 2 L sin (\theta )$

=> $r = 2* 0.180 sin (7)$

=> $r = 0.043 \ m$

=> $q = tan \theta * \frac{m * g * r^2 }{k}$

=> $q = tan(7)* \frac{ 0.02 * 9.8 * 0.043^2 }{9*10^{9}}$

=> $q = 3.4 *10^{-6} \ C$

7 0

2 years ago

the pressure of gas in a cylinder is 70 kilopascals. if the volume of the cylinder is reduced from 8.0 liters to 4.0 liters, wha

Olegator [25]

By Boyle's law:

P₁V₁ = P₂V₂

70*8 = P<span>₂*4

</span>P<span>₂*4 = 70*8
</span>
P<span>₂ = 70*8/4 = 140
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P<span>₂ = 140 kiloPascals.</span>

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3 years ago

I NEED HELP PLEASE, THANKS! :)

mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

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