Ask question

Ask question

All categories

3 years ago

5

The human tibia breaks under an impulse of 55 Ns. If after falling a person typically comes to eat in a time span of 0.005 s how

fast would a 75 kg person need to be falling to break their leg?

1 answer:

ozzi3 years ago

5 0

Answer:

0.73 m/s

Explanation:

From Newton second law of motion,

I = m(v-u)...................... Equation 1

Where I = Impulse, m = mass of the person, v = final velocity, u = Initial velocity.

make v the subject of the equation

v =(I/m)+u................. Equation 2

Note: u = 0 m/s as the person is falling from an height.

Given: I = 55 Ns, m = 75 kg, u = 0 m/s

Substitute into equation 2

v = 55/75

v = 0.73 m/s

You might be interested in

Which option lists a form of kinetic energy followed by a form of potential

Artemon [7]

Answer:

D. Sound Energy, Magnetic energy

Explanation:

Sound energy is in motion, and Magnetic energy is about to be in motion.

6 0

3 years ago

An car is brought to rest in a distance of 652 m using a constant acceleration of -2.7 m/s2. What was the

Goshia [24]

Answer:

I’m so sorry I tried solving it but I don’t understand it can you explain the question a little bit more ty

Explanation:

7 0

3 years ago

What do you mean by Parallel Universe ?

BartSMP [9]

Answer:

Parallel universe, or alternate reality, is a hypothetical self-contained plane of existence, co-existing with one's own

3 0

3 years ago

Read 2 more answers

The energy emitted from the sun is a product of ________.

Tamiku [17]

Q. The energy emitted from the sun is a product of ________.

A. Fusion

3 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