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Yakvenalex [24]

3 years ago

9

In a typical badminton swing the racket is in contact with the birdy for about 0.0010 seconds. If the 0.045kg birdy acquires a s

peed of 67 m/s, what is the force exerted by the racket on the ball

1 answer:

zepelin [54]3 years ago

3 0

Answer:

3015 N

Explanation:

From Newton's second law, we know that;

F.t = mv

F = force on the ball= ?

m= mass of the ball

v= velocity

F= mv/t

F= 0.045 × 67/0.0010

F= 3.015/0.0010

F= 3015 N

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A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of

ANEK [815]

Answer:

The distance the piece travel in horizontally axis is

L=3.55m

Explanation:

$a=2 \frac{rev}{s^{2}} \\h=0.820m\\r = 0.125 m
\\d=150rev$

$d= 155 rev = 155(2\pi ) = 310\pi rad$

$a= 2.0 \frac{rev}{s^{2} } = 2.0(2\pi ) = 4.0\pi \frac{rev}{s^{2} }$

$d=d_{i}+vo*t+\frac{1}{2}*a*t^{2} \\ di=0\\vo=0\\d=\frac{1}{2}*a*t^{2}\\t=\sqrt{\frac{2*d}{a}}\\t=\sqrt{\frac{2*310 rad}{4\frac{rad}{s^{2}}}} \\t=12.449$

$w=a*t\\w=4\frac{rad}{s^{2}}*12.449s\\ w=49.79 \frac{rad}{s}$

Now the angular velocity is the blade speed so:

$V=w*r\\V=49.79 \frac{rad}{s}*0.175m\\V=8.7 \frac{m}{s}$

assuming no air friction effects affect blade piece:

time for blade piece to fall to floor

$t=\sqrt{\frac{2*h}{g}}\\t=\sqrt{\frac{2*0.820m}{9.8\frac{m}{s^{2} } }}\\t=0.409s$

Now is the same time the piece travel horizontally

$L=t*V\\L=0.409s*8.7\frac{m}{s}\\L=3.55m$

blade piece travels HORIZONTALLY = (24.5)(0.397) = 9.73 m ANS

6 0

3 years ago

A rock is launched vertically upward by a slingshot. The elastic band of the slingshot accelerates the rock from rest to 40 m/s

eimsori [14]

Answer:

0.016 s

Explanation:

Initial velocity, u = 0

use the third equation of motion to find the acceleration of the rock.

$2as=v^2-u^2\\a = \frac{40^2-0}{2\times0.1} =8000 m/s^2$

Net displacement in the vertical direction would be zero. y = 0

Use second equation:

$s=ut+\frac{1}{2}at^2\\0=0+\frac{1}{2}(8000)t^2\\t= 0.016 s$

Thus, the rock will return to its initial release point in 0.016 s.

5 0

3 years ago

A swimmer swims north at 0.10 m/s relative to still water across a river that flows at 0.25 m/s from east to west. Calculate the

ValentinkaMS [17]

0.15 m/s East
If you follow the equation a=vf-vi/t, you'll discover that subtracting the final velocity, 0.25 m/s, by the initial velocity, 0.10 m/s, and divide by zero, (bc there was no given time) the answer is 0.15 m/s East

7 0

3 years ago

What fraction of 5 MeV α particles will be scattered through angles greater than 8.5° from a gold foil (Z = 79, density = 19.3 g

aalyn [17]

Answer:

$f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)$

And after do all the operations we got:

$f(8) = 1.96 x10^{-4] atoms/m^2$

And that would be the final answer for this case.

Explanation:

For this case we can use the fomrula for the fraction of incident particles scattered by an angle $\theta$ , given by:

$f(\theta) = \pi nt (\frac{Z_1 e Z_2 e}{8 \pi e_o K})^2 cos^2 (\theta/2)$

Where:

$Z_1 e$ represent the charge of the projectile (Z1=2)

$Z_2 e$ is the target charge (z2=79)

$K= 5x10^6 eV$ represent the kinetic energy of incident particle

n represent the density of target particles (we need to find it first)

$t= 10^{-8] m$ represent the thicknss of the foil

The first step would be calculate the density of target particles with the following formula:

$n =\frac{\rho N_A N_M}{M_g}$

Where:

$\rho = 19.3 g/m^3= 19300 Kg/m^3$

$N_A = 6.022 x10^{23} molecules/mol$ the Avogadro's number

$N_M = 1$ represent the atoms per molecule

$M_g = 197 g/mol = 0.197 Kg/mol$ represent the molecular weigth

If we replace we got:

$n = \frac{19300 kg/m^3 *6.022x10^{23} molecu/mol * 1 atom/mole}{0.197 Kg/mol}= 5.90 x106{28] atoms/m^3$

Now we can calculate the fraction of 5 MeV alpha particles that would be scatteres with angle higher than 8 degrees in a piece of thickness $t=10^{-8}m$

And using the first formula we got:

$f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)$

And after do all the operations we got:

$f(8) = 1.96 x10^{-4] atoms/m^2$

And that would be the final answer for this case.

4 0

3 years ago

How does friction affect object acceleration down an incline?

Colt1911 [192]

Friction is the force that acts on the opposite side of direction of force, thus it manages to decelerate an object, so it acts upward along the plane

4 0

2 years ago