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2 years ago

11

A researcher measures 200 counts per minute coming from a radioactive source at noon. At 3:00pm, she finds that this has dropped

to 25 counts per minute. What percentage of the original source will remain at 6:00pm?
A) 3.125%
B) 4.5%
C) 5.8%
D) 6.25%

1 answer:

patriot [66]2 years ago

4 0

Answer:

UWU ForkNit Nub X3 Scrub get plied uWu x333333

Explanation:

idk lol I need the answer too

Ok I commited yolo the answer is 3.125%

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A charge Q is distributed uniformly along the x axis from x1 to x2. What would be the magnitude of the electric field at x0 on t

Lena [83]

Answer:

E = k Q 1 / (x₀-x₂) (x₀-x₁)

Explanation:

The electric field is given by

dE = k dq / r²

In this case as we have a continuous load distribution we can use the concept of linear density

λ= Q / x = dq / dx

dq = λ dx

We substitute in the equation

∫ dE = k ∫ λ dx / x²

We integrate

E = k λ (-1 / x)

We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁

E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)

E = k λ (x₂ -x₁) / (x₀-x₂) (x₀-x₁)

We replace the density

E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]

E = k Q 1 / (x₀-x₂) (x₀-x₁)

8 0

2 years ago

An 80-g particle moving with an initial speed of 50 m/s in the positive x direction strikes and sticks to a 60-g particle moving

liubo4ka [24]

The collision is a form of inelastic collision because the it forms a single mass after is collides. So it can be solve by momentum balance

( 0.08 kg * 50 m/s ) + ( 0.06 kg * 50 m/s) = ( 0.08 + 0.06 kg ) v

V = 50 m/s

So the kinetic energy lost is

KE = 0.5 (50 m/s)^2) *( 0.14 – 0.08kg )

KE = 75 J

8 0

2 years ago

The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo

irina1246 [14]

Answer:

$p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

Explanation:

first write the newtons second law:

F $_{s}$ =δma $_{s}$

Applying bernoulli,s equation as follows:

∑ $δp+\frac{1}{2} ρδV^{2} +δγz=0\\$

Where, $δp$ is the pressure change across the streamline and $V$ is the fluid particle velocity

substitute $ρg$ for {tex]γ[/tex] and $g_{0}-cz$ for $g$

$dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0$

integrating the above equation using limits 1 and 2.

$\int\limits^2_1 \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

there the bernoulli equation for this flow is $p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

note: $ρ$ =density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

4 0

2 years ago

irina [24]

Answer:

v' = -18 m/s

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

$p_{o} = p_{f} (1)$

The initial momentum can be expressed as follows (taking as positive the initial direction of the ball):

$m_{b} * v_{b} -M_{c}*V_{c} = m_{b} * 18 m/s + (-M_{c}* 20 m/s) (2)$

The final momentum can be expressed as follows (since we know that v'b is opposite to the initial vb):

$-(m_{b} * v'_{b}) + M_{c}*V'_{c} (3)$

If we assume that Mc >> mb, we can assume that the car doesn't change its speed at all as a result of the collision, so we can replace V'c by Vc in (3).
So, we can write again (3) as follows:

$-(m_{b} * v'_{b}) +(- M_{c}*V_{c}) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (4)$

Replacing (2) and (4) in (1), we get:

$m_{b} * 18 m/s + (-M_{c}* 20 m/s) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (5)$

Simplifying, and rearranging, we can solve for v'b, as follows:
$v'_{b} = -18 m/s (6)$ , which is reasonable, because everything happens as if the ball had hit a wall, and the ball simply had inverted its speed after the collision.

3 0

1 year ago

A 2.70 kg cat is sitting on a windowsill. The cat is sleeping peacefully until a dog barks at him. Startled, the cat falls from

Alchen [17]

Answer:

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat ( $U_{g}$ ), in joules, is equal to the sum of the final translational kinetic energy ( $K$ ), in joules, and work losses due to air resistance ( $W_{l}$ ), in joules:

$U_{g} = K +W_{l}$ (1)

By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:

$m \cdot g\cdot h = \frac{1}{2}\cdot m \cdot v^{2}+W_{l}$ (2)

Where:

$m$ - Mass of the cat, in kilograms.

$g$ - Gravitational acceleration, in meters per square second.

$h$ - Initial height of the cat, in meters.

$v$ - Final speed of the cat, in meters per second.

If we know that $m = 2.70\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h = 5.20\,m$ and $W_{l} = 120\,J$ , then the final speed of the cat is:

$v = \sqrt{\frac{2\cdot (m\cdot g\cdot h-W_{l})}{m} }$

$v = \sqrt{2\cdot g\cdot h-\frac{W_{l}}{m} }$

$v \approx 7.586\,\frac{m}{s}$

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

4 0

2 years ago