Is there a correlation between moonrise and high tide?

A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object without damage to the body of th

poizon [28]

Answer:

The force bumper at 0.200m

F=2722.5 N

Explanation:

Using the energy theorem of work

$W=K_{f}- K_{i}$

W=ΔK

W=F*d

ΔK= $\frac{1}{2}*m*v_{f} ^{2} -\frac{1}{2}*m*v_{i} ^{2}$

$v_{i} =0$

ΔK=F*d= $\frac{1}{2}*m*v_{f} ^{2} -\frac{1}{2}*m*v_{i} ^{2}$

$F*d=\frac{1}{2}*m*0 ^{2}- \frac{1}{2}*m*v_{i} ^{2}\\F*d=-\frac{1}{2}*m*v_{i} ^{2}=\frac{1}{2}*900kg*(1.1\frac{m}{s})^{2}\\F*d=544.5\frac{kg*m}{s^{2}}\\ F=\frac{544.5\frac{kg*m}{s^{2}}}{0.2m} \\F=2722.5 N$

3 0

3 years ago

What is the definition of energy transformation?

mixas84 [53]

Energy Transformation is the process of changing one form of energy to another form...

4 0

4 years ago

Read 2 more answers

A whistle of frequency 564 Hz moves in a circle of radius 71.2 cm at an angular speed of 17.1 rad/s. What are (a) the lowest and

trapecia [35]

Answer:

a) $f'=544.66 \textup{Hz}$

b) $f'=584.75 \textup{Hz}$

Explanation:

Given:

Frequency of the whistle, f = 564 Hz

Radius of the circle, r = 71.2 cm = 0.712 m

Angular speed, ω = 17.1 rad/s

speed of source, $v_s$ = rω = 0.712 × 17.1 = 12.1752 m/s

speed of sound, v = 343 m/s

Now, applying the Doppler's effect formula, we have

$f'=f\frac{v\pm v_d}{v\pm v_s}$

where,

$v_d$ = relative speed of the detector with respect to medium = 0

a) for lowest frequency, we have the formula as:

$f'=f\frac{v}{v+v_s}$

on substituting the values, we get

$f'=564\times\frac{343}{343+12.1752}$

or

$f'=544.66 \textup{Hz}$

b) for maximum frequency, we have the formula as:

$f'=f\frac{v}{v-v_s}$

on substituting the values, we get

$f'=564\times\frac{343}{343-12.1752}$

or

$f'=584.75 \textup{Hz}$

3 0

3 years ago

A truck tire rotates at an initial angular speed of 21.5 rad/s. The driver steadily accelerates, and after 3.50 s the tire's ang

erma4kov [3.2K]

Given:

initial angular speed, $\omega _{i}$ = 21.5 rad/s

final angular speed, $\omega _{f}$ = 28.0 rad/s

time, t = 3.50 s

Solution:

Angular acceleration can be defined as the time rate of change of angular velocity and is given by:

$\alpha = \frac{\omega_{f} - \omega _{i}}{t}$

Now, putting the given values in the above formula:

$\alpha = \frac{28.0 - 21.5}{3.50}$

$\alpha = 1.86 m/s^{2}$

Therefore, angular acceleration is:

$\alpha = 1.86 m/s^{2}$

5 0

3 years ago

A woman takes her dog rover for a walk on a leash. She pulls on the leash with a force of 30.0 N at an angle of 29° above the ho

IrinaK [193]

19.8 N force is tending to lift Rover vertically off the ground.

<h3>What is horizontal and vertical component?</h3>

The horizontal velocity component ( $v_{x}$ ) describes the influence of the velocity in displacing the projectile horizontally. The vertical velocity component ( $v_{y}$ ) describes the influence of the velocity in displacing the projectile vertically.