Ask question

Ask question

All categories

3 years ago

15

A pole-vaulter is nearly motionless as he clears the bar, set 4.2 m above the ground. he then falls onto a thick pad. the top of

the pad is 80 cm above the ground, and it compresses by 50 cm as he comes to rest. what is the magnitude of his accelerations he comes to rest on the pad

2 answers:

scZoUnD [109]3 years ago

8 0

Refer to the diagram shown below.

Neglect wind resistance, and use g = 9.8 m/s².

The pole vaulter falls with an initial vertical velocity of u = 0.
If the velocity upon hitting the pad is v, then
v² = 2*(9.8 m/s²)*(4.2 m) = 82.32 (m/s)²
v = 9.037 m/s

The pole vaulter comes to res after the pad compresses by 50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
0 = (9.037 m/s)² + 2*(a m/s²)*(0.5 m)
a = - 82.32/(2*0.5) = - 82 m/s²

Answer: - 82 m/s² (or a deceleration of 82 m/s²)

sertanlavr [38]3 years ago

8 0

The magnitude of his accelerations on the pad is 66.64 m/s²

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

height of pole vaulter above the pad = H = 4.2 - 0.8 = 3.4 m

distance of compression of pad = d = 50 cm = 0.5 m

<u>Unknown:</u>

a = ?

<u>Solution:</u>

Firstly , we calculate the velocity of pole-vaulter before hitting the pad.

$v^2 = u^2 + 2gH$

$v^2 = 0^2 + 2 \times 9.8 \times 3.4$

$v^2 = 1666/25$

$v = \frac{7 \sqrt{34}}{5} ~ m/s$

Next , we can calculate the magnitude of his accelerations on the pad.

$v^2 = u^2 + 2gH$

$0^2 = 1666/25 + 2 \times a \times 0.5$

$a = -1666/25$

$\large {\boxed {a = -66.64 ~ m/s^2} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

You might be interested in

Homeostasis is necessary for these organisms to survive.

alexandr1967 [171]

Answer:

B, unicellular

Explanation:

8 0

3 years ago

What is magnification

il63 [147K]

The action or process of magnifying something or being magnified, especially visually. Hope this helped

5 0

4 years ago

A rectangular plate has a length of (21.7 ± 0.2) cm and a width of (8.2 ± 0.1) cm. Calculate the area of the plate, including it

Grace [21]

Answer:

(177.94 ± 3.81) cm^2

Explanation:

l + Δl = 21.7 ± 0.2 cm

b + Δb = 8.2 ± 0.1 cm

Area, A = l x b = 21.7 x 8.2 = 177.94 cm^2

Now use error propagation

$\frac{\Delta A}{A}=\frac{\Delta l}{l}+\frac{\Delta b}{b}$

$\frac{\Delta A}{A}=\frac{0.2}{21.7}+\frac{0.1}{8.2}$

$\Delta A=177.94 \times \left ( 0.0092 + 0.0122 \right )=3.81$

So, the area with the error limits is written as

A + ΔA = (177.94 ± 3.81) cm^2

8 0

4 years ago

Describe how elements are formed?<br>

kenny6666 [7]

Elements are identified by the nuclei of the atoms of which they are made. For example, an atom having six protons in its nucleus is carbon, and one having 26 protons is iron. ... Heavy elements can be formed from light ones by nuclear fusion reactions; these are nuclear reactions in which atomic nuclei merge together. This is how they are formed

3 0

4 years ago

Assume the average value of the vertical component of Earth's magnetic field is 42 μT (downward) in some region that has an area

Oliga [24]

Answer:

The net magnetic flux through the rest of Earth's surface is $5.782\times10^{-2}\ T$

Explanation:

Given that,

Magnetic field = 42 μT

Area $A=3.71\times10^{5}\ km^2$

$A=3.71\times10^{11}\ m^2$

We need to calculate the flux per unit area

$flux\ per\ unit\ area=\dfrac{42\times10^{-6}}{3.71\times10^{11}}$

$flux\ per\ unit\ area=1.132\times10^{-16}\ T/m^2$

We need to calculate the total earth's surface area

$A'=4\pi r^2$

$A'=4\times\pi\times(6.3781\times10^{6})^2$

$A'=5.1120\times10^{14}\ m^2$

We need to calculate the rest of earth's area

$A''=A-A'$

Put the value into the formula

$A''=5.1120\times10^{14}-3.71\times10^{11}$

$A''=5.10829\times10^{14}\ m^2$

We need to calculate the net magnetic flux through the rest of Earth's surface

$B'=5.10829\times10^{14}\times1.132\times10^{-16}$

$B'=5.782\times10^{-2}\ T$

Hence, The net magnetic flux through the rest of Earth's surface is $5.782\times10^{-2}\ T$

3 0

4 years ago