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3 years ago

Heather and Matthew take 45 s to walk eastward along a straight road to a store 72 m away. What is their average velocity?

Physics

1 answer:

vladimir1956 [14]3 years ago

5 0

Answer:

v = 1.6 m/s

Explanation:

Given that,

Distance, d = 72 m

Time taken, t = 45 s

We need to find their average velocity. Average velocity of an object is given by total distance divided by total time taken.

$v=\dfrac{d}{t}\\\\v=\dfrac{72\ m}{45\ s}\\\\v=1.6\ m/s$

So, their average velocity is 1.6 m/s.

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Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co

patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of $^{3890}Sr\, t_{\frac{1}{2}}$ = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

$\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

where

N = No. of nuclei left after time t

$N_{o}$ = No. of nuclei initially started with

$\frac{N}{N_{o}} = 1\times 10^{- 4}$

(Since, 100% - 99.99% = 0.01%)

Thus

$1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}$

Taking log on both the sides:

$- 4 = \frac{t}{28.5}log\frac{1}{2}$

$t = \frac{-4\times 28.5}{log\frac{1}{2}}$

t = 387.69 yrs

5 0

3 years ago

In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e

Zanzabum

Answer:

The velocity is $v_t = 0.02175 \ m/s$

Explanation:

From the question we are told that

The mass of the bullet is $m_b = 0.024 \ kg$

The initial speed of the bullet is $u_b = 1200 \ m/s$

The mass of the target is $m_t = 320 \ kg$

The initial velocity of target is $u_t = 0 \ m/s$

The final velocity of the bullet is is $v_b = 910 \ m/s$

Generally according to the law of momentum conservation we have that

$m_b * u_b + m_t * u_t = m_b * v_b + m_t * v_t$

=> $0.024 * 1200 + 320 * 0 = 0.024 * 910 + 320 * v_t$

=> $v_t = 0.02175 \ m/s$

3 0

3 years ago

A 65 kg person jumps from a window to a fire net 18 m below, which stretches the net 1.1 m. Assume the net behaves as a simple s

posledela

Answer:

0.03167 m

1.52 m

Explanation:

x = Compression of net

h = Height of jump

g = Acceleration due to gravity = 9.81 m/s²

The potential energy and the kinetic energy of the system is conserved

$P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m$

The spring constant of the net is 20130.76 N

From Hooke's Law

$F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m$

The net would strech 0.03167 m

If h = 35 m

From energy conservation

$65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0$

Solving the above equation we get

$x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45$