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3 years ago

12

You are standing on a log and a friend is trying to knock

1 answer:

WINSTONCH [101]3 years ago

3 0

Answer: catching the ball is a better choice.

Explanation:

The collision of 2 objects involves involves large impact force since the force is inversely proportional to the time in which the momentum of the object changes.

Mathematically

$F=\frac{\Delta p}{\Delta t}$

If we catch the ball we increase the time in which the momentum of the ball is decreased thus the impact force that acts on us is lower as larger time is allowed for the ball to decrease it's momentum.

If we allow the ball to hit us the momentum of the ball changes in a short period of time thus applying a large impact force on our body thus increasing the chances of toppling.

You might be interested in

Period of a 60Hz circuit

4vir4ik [10]

Period is T=1/f
T=1/60 = 16.67 milliseconds

3 0

2 years ago

The water in a river flows uniformly at a constant speed of 2.27 m/s between parallel banks 69.3 m apart. You are to deliver a p

Tomtit [17]

Answer: Got it!

Explanation: The water in a river flows uniformly at a constant speed of 2.50m/s between two parallel banks 80.0m apart. You are to deliver a package directly across the river, but you can only swim at 1.5m/s.

3 0

3 years ago

An emu moving with constant acceleration covers the distance between two points that are 92 m

nasty-shy [4]

Answer:

a) $V_{o}=14.30 m/s$

b) $a=-0.046 m/s^{2}$

Explanation:

The complete question is written below:

An emu moving with constant acceleration covers the distance between two points that are 92 m apart in 6.5s. Its speed as it passes the second point is 14 m/s. What are (a) its speed at the first point and (b) its acceleration?

Since we are talking about constant acceleration, we can use the following equations:

$d=x-x_{o}=(\frac{V_{o}-V}{2})t$ (1)

$V=V_{o}+at$ (2)

Where:

$d=92 m$ is the distance between the two points

$V_{o}$ is the velocity of the emu at the first point

$V=14 m/s$ is the velocity of the emu at the second point

$t=6.5 s$ is the time it takes to the emu to cover the distance $d$

$a$ is the emu's constant acceleration

Knowing this, let's begin with the answers:

<h2>a) Speed at the first point</h2>

In this situation wi will use equation (1):

$d=(\frac{V_{o}-V}{2})t$ (1)

Finding $V_{o}$ :

$V_{o}=\frac{2d}{t}-V$ (3)

$V_{o}=\frac{2(92 m)}{6.5 s}-14 m/s$ (4)

$V_{o}=14.30 m/s$ (5)

<h2>b) Emu's acceleration</h2>

Now we will substitute (5) in equation (2):

$14 m/s=14.30 m/s+a(6.5 s)$ (6)

Finding $a$ :

$a=-0.046 m/s^{2}$ (7) This means the emu is decreasing its speed at a constant rate.

8 0

3 years ago

A centrifuge in a medical laboratory rotates at an angular speed of 3,400 rev/min. When switched off, it rotates through 52.0 re

eduard

The constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².

<h3> Constant angular acceleration</h3>

Apply the following kinematic equation;

ωf² = ωi² - 2αθ

where;

ωf is the final angular velocity when the centrifuge stops = 0
ωi is the initial angular velocity
θ is angular displacement
α is angular acceleration

ωi = 3400 rev/min x 2π rad/rev x 1 min/60s = 356.05 rad/s

θ = 52 rev x 2π rad/rev = 326.7 rad

0 = ωi² - 2αθ

α = ωi²/2θ

α = ( 356.05²) / (2 x 326.7)

α = 194.02 rad/s²

Thus, the constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².

Learn more about angular acceleration here: brainly.com/question/25129606

#SPJ1

7 0

1 year ago

Problem 3) Bob stands at the edge of the swimming pool holding a laser 1.5m above the ground. He shines the red laser beam onto

Vadim26 [7]

Answer:

d = 5.75m

Explanation:

Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

n1= refractive index of 1st medium= 1

n2= refractive index of 2nd medium = 1.33

r= angle of reflection

therefore,

$r = \sin^{-1}\frac{n_1\sin i}{n_2}$

Here,

i = 90 - θ

$\theta = \tan^-^1(\frac{1.5}{3} )\\\\=26.56^\circ$

$r = \sin^{-1}\frac{n_1\sin i}{n_2}$

$r = \sin^{-1}\frac{(1)\sin (90-26.56)}{1.33}\\\\r = 42.26m$

$\tan r = \frac{2.5}{d_1}$

$d_1 = \frac{2.5}{\tan (42.26)} \\\\d_1 = 2.75m$

Therefore, the distance is

d = 3 + d₁

d = 3 + 2.75

d = 5.75m

6 0

3 years ago