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3 years ago

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Alicia is asked to calculate the change in momentum that occurs as she stops her bicycle. This is equivalent to the______ Applie

d to the bicycle for a certain amount of time.

1 answer:

emmainna [20.7K]3 years ago

3 0

It is the force applied to the bycicle

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How many Sig Figs (Significant Figures) are in each number?<br><br> 5070.0<br><br> 870.064080

nlexa [21]

∑ Hey, Lethality ⊃

Answer:

5070.0 has 5 significant figures

870.064080 has 9 significant figures

Explanation:

<u><em>Given:</em></u>

<em>How many Sig Figs (Significant Figures) are in each number?</em>

<em>5070.0</em>

<em>870.064080</em>

<u><em>Solution:</em></u>

<em>-------------------------------------------------------------------------------------------------------------</em>

<em>5070.0</em>

<em>5070.0 has 5 significant figures ( 5 , 0 , 7 , 0 , and 0 )</em>

<em>Number of significant figures: 5</em>

<em>-------------------------------------------------------------------------------------------------------------</em>

<em>870.064080</em>

<em>870.064080 has 9 significant figures ( 8, 7 ,0,0, 6,4,0,8 and 0 )</em>

<em>Number of significant figures: 9</em>

<em>-------------------------------------------------------------------------------------------------------------</em>

<em />

<u><em>xcookiex12</em></u>

<em>8/23/2022</em>

5 0

1 year ago

If a person believes that dreams have hidden meaning he or she would agree with Freud's ideas about _ content

uysha [10]

A.lucid because it makes more sense yo the answer how about you do it yourself

4 0

3 years ago

A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit

likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density $\mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m$

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 120=753.6rad/sec$

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by $P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt$

From the relation we can see that power $P\ \propto\ A^2$

So if amplitude is halved then power will be $\frac{1}{4}$ times

So power will be equal to $\frac{0.2023}{2}=0.0505watt$

4 0

3 years ago

When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6

Stella [2.4K]

Answer:

it is separated by 80 cm distance

Explanation:

As per Coulombs law we know that force between two point charges is given by

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1 = +8.4\mu C$

$q_2 = +5.6 \mu C$

force between two charges is given as

$F = 0.66 N$

now we have

$F = \frac{kq_1q_2}{r^2}$

$0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}$

$r = 0.8 m$

so it is separated by 80 cm distance

3 0

3 years ago

Read 2 more answers

A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the

mylen [45]

Answer:

Angular acceleration, is $708.07\ rad/s^2$

Explanation:

Given that,

Initial speed of the drill, $\omega_i=0$

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, $\omega_f=28940\ rev/min=3030.58\ rad/s$

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

$\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2$

So, the drill's angular acceleration is $708.07\ rad/s^2$ .

4 0

3 years ago