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2 years ago

Why is it more comfortable to fall onto a big pillow than onto a wooden floor ?

1 answer:

6 0

I believe the answer to this question is C. The pillow distributes the decelerating force over a greater amount of time .

sudden stops are damaging, not the fall itself.

When falling on a mat it will take longer since a smaller force=longer falling time

when falling on a wooden floor it will take shorter since a larger force=smaller falling time

Hence, if the sudden fall is what causes impact and discomfort, and a smaller force equals longer falling time, and a larger force equals shorter falling time and smaller falling time equals less impact, then, C. must be the answer.

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At a baseball game, you are in the cheap seats and far from the batter. You see the ball moving before you hear the sound of the

bogdanovich [222]

Answer:

it is he sound of source it uses ur ears to hear something unectpected can u give me brainly plz

Explanation:

6 0

3 years ago

if the amplitude of sine curve is 4 and the frequency is 1 with no phase shift, what is the function that descirbes this curve?

Vlada [557]

Answer:

y = 4 Sin (2πt)

Explanation:

Amplitude, A = 4

frequency, f = 1

Wave function is given by

y = A sinωt

where, ω is angular frequency

ω = 2 π f = 2 π x 1 = 2π

So, the desired wave function

y = 4 Sin (2πt)

8 0

3 years ago

A shopper pushes a 5.32 kg grocery cart with a 12.7 N force directed at -28.7° below horizontal. A friction force of 8.33 N push

atroni [7]

Answer:

0.8214 m/s^2

Explanation:

Fnet= Fpushed - Ffriction

Fpushed = 12.7N Ffriction = 8.33N

Fnet = 12.7N - 8.33N = 4.37N

Fnet= mass(acceleration)

Fnet = 4.37N mass = 5.32 kg

4.37N = 5.32 kg(acceleration)

acceleration= 0.8214 m/s^2

3 0

2 years ago

Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g

ollegr [7]

Answer:

The beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

velocity of wave on the string with greater tension;

$v_1 = \sqrt{\frac{T_1}{\mu }$

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

$v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}$

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

$v_2^2 = \frac{T_2v_1^2}{T_1} \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s$

Fundamental frequency of wave on the string with greater tension;

$f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1} =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz$

Beat frequency = F₂ - F₁

= 270.6 - 258

= 12.6 Hz

Therefore, the beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

6 0

3 years ago

Mary starts from her house, walks 80 meters south, and stops to chat with her aunt on the sidewalk. After chatting for a few min

marin [14]

Mary walks:
d 1 = 80 m, d 2 = 125 m, d 3 = 45 m
t = 10 minutes = 600 seconds;
Average speed:
v = ( d 1 + d 2 + d 3 ) / t
v = ( 80 m + 125 m + 45 m ) / 600 s
v = 250 m / 600 s
v = 0.4167 m/s ≈ 0.42 m/s
Answer:
E ) 0.42 meters/second

5 0

2 years ago