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2 years ago

Does the car's speed alone determine whether the egg breaks?

Physics

2 answers:

otez555 [7]2 years ago

6 0

Answer:

Somehow you must have information about the 'stopping time' of the egg. Then, the force will be proportional to the momentum divided by the stopping time. Pretend both the eggshell and the surface it hits are perfectly rigid, and the force will be infinite from any height, and the egg will break from any height.

This will help you solve other question like these

Hope it helps PLS MARK ME AS BRAINLIST I BEG YOU thanks :)

blsea [12.9K]2 years ago

4 0

Answer:

yes it is

Explanation:

when the speed increases the car will vibration than egg will breaks

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Stan is driving north on his scooter at 8m/s, accelerates 11m/s (North) in 4s, drives a constant velocity for the next 15s, and

kow [346]

A) Acceleration: $a_1 = 0.75 m/s^2, a_2 = 0, a_3 = -1.57 m/s^2$

B) The total displacement is 209.5 m north

C) The average velocity is 8.06 m/s north

Explanation:

Acceleration is defined as:

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

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

Here we have:

- In the first segment,

u = 8 m/s north

v = 11 m/s north

t = 4 s

So the acceleration is

$a_1 = \frac{11-8}{4}=0.75 m/s^2$ (north)

- In the second segment, Stan drives at a constant velocity: so the final velocity is equal to the initial velocity,

u = v

Therefore, the acceleration is zero: $a_2 = 0$

- In the third segment,

u = 11 m/s (north)

v = 0 (he comes to a stop)

t = 7 s

So the acceleration is

$a=\frac{0-11}{7}=-1.57 m/s^2$

And the negative sign means the acceleration is south, opposite to the direction of motion.

In a uniformly accelerated motion, the displacement can be calculated as:

$s=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

t is the time

- For the first segment, we have

$u = 0\\a = 0.75 m/s^2\\t=4 s$

So the displacement is

$s_1 = 0+\frac{1}{2}(0.75)(4)^2=6 m$

- For the second segment, we have

$u = 11 m/s\\a = 0\\t=15 s$

So the displacement is

$s_2 = (11)(15)+0=165 m$

- For the third segment, we have

$u = 11\\a = -1.57 m/s^2\\t=7 s$

So the displacement is

$s_3 = (11)(7)+\frac{1}{2}(-1.57)(7)^2=38.5 m$

So the total displacement is:

s = 6 m + 165 m + 38.5 m = 209.5 m

In the north direction (positive direction)

The average velocity is given by:

$v=\frac{d}{t}$

where

d is the total displacement

t is the total time

Here we have:

d = 209.5 m

t = 26 s

Therefore, the average velocity is

$v=\frac{209.5}{26}=8.06 m/s$ (north)

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3 years ago

Need help ASAP!!!! here are the options: Amplitude, Compression, Rarefaction, and Wavelength

Strike441 [17]

Amplitude is the pair of vertical buttons, so to speak. Compressions are the bunched up vertical lines with the purple arrows pointing left and right. Rarefactions are purple arrows pointing down. Wavelength is crest to crest purple buttons. Associated LH and RH pointing arrows.

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3 years ago

Conveyor belts are often used to move packages around warehouses. The conveyor shown below moves packages at a steady 4.0 m/s. A

Alisiya [41]

Answer:

0 j

Explanation:

4 0

3 years ago

An ice skater starts with a velocity of 2.25 m/s in a 50.0 degree direction. After 8.33s, she is moving 4.65 m/s in a 120 degree

Mnenie [13.5K]

The y-component of the acceleration is $0.22 m/s^2$

Explanation:

The y-component of the acceleration is given by

$a_y = \frac{v_y-u_y}{t}$

where

$v_y$ is the y-component of the final velocity

$u_y$ is the y-component of the initial velocity

t is the time elapsed

For the ice skater in this problem, we have:

u = 2.25 m/s is the initial velocity, in a direction $\theta=50.0^{\circ}$

v = 4.65 m/s is the final velocity, in a direction $120^{\circ}$

t = 8.33 s is the time elapsed

The y-components of the initial and final velocity are:

$u_y = u sin \theta = (2.25)(sin 50^{\circ})=1.72 m/s\\v_y = v sin \theta = (4.65)(sin 50^{\circ})=3.56 m/s$

So the y-component of the acceleration is

$a_y = \frac{3.56-1.72}{8.33}=0.22 m/s^2$

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3 years ago

Is everyone in your class able to hear a quiet sound equally well?

Sergio039 [100]

Answer:

Explanation:

Loudness describes how people perceive sound (see loudness). ... If people could hear equally well at all frequencies, the contour lines would be flat because the same measured sound intensity would be perceived to be equally loud regardless of the sound frequency. In fact, people do not hear as well at low frequencies.

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3 years ago