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2 years ago

Help someone :(((((((((((((((

Physics

2 answers:

d1i1m1o1n [39]2 years ago

6 0

Answer:

Explanation:

3500Hz*2.5m=9800m/s

Give Brillienst please!!!!

Dimas [21]2 years ago

3 0

Wave Speed (M/s) = Frequency (Hz) x Wavelength (M)

Wave Speed = 3500Hz x 2.8M = 9800M/s

Answer = Top one

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A particle moves along a straight line and its position at time t is given by

Alisiya [41]

(A) Use interval notation to indicate the time interval or union of time intervals when the particle is moving forward and backward

the particle is speeding up

if a(t) >0, it means 12t-48>0, and t >4, I = ]4, infinity[

the particle is speeding down

if a(t) < 0, it means 12t-48>0, and t < 4, I = ] -infinity, 4[

when the particle is moving forward and backward

that is depending of the sign of v(t)

moving forward if v(t)> 0

moving backward if v(t)< 0

but v(t) = 6t² - 48 t +90, this quadratic equation has exactly two solutions, t =1, and t =3

so after taking care the sign of 6t² - 48 t +90,

v(t)> o when t is in I = ] –inf, 1[ U ]4, inf[,the particle is moving forward

v(t) < o when t is in I = ]1, 4[,the particle is moving backward

3 0

3 years ago

How does natural selection produce change in a population of mice?

sergey [27]

Answer:

The population of the mice will decrease to the faster, stronger, and smarter mice because the weaker will die because of natural selection.

Explanation:

5 0

2 years ago

What is the minimum force require to move a 5kg wooden crate on a wooden floor?

kolbaska11 [484]

You need to know the coefficient of static friction between a wooden object and a wooden surface. I'll denote it with µ. If you're given a specific value you should obviously use that.

By Newton's second law, the horizontal and vertical net forces are

• net horizontal:

∑ F = p - f = 0

• net vertical:

∑ F = n - w = 0

where

p = magnitude of the pushing force

f = mag. of friction

n = mag. of the normal force

w = weight of the crate

The second equation gives

n = w = (5 kg) (9.8 m/s²) = 49 N

Friction is proportional to the normal force by a factor of µ, so

f = µ (49 N) = 49µ N

To overcome static friction, the push has to exceed this in magnitude, so that

p > 49µ N

For instance, if p = 0.25, then p would need to greater than 12.25 N. (This example isn't particularly helpful, though, since both possibly correct options are larger than 12.25 N...)

7 0

3 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

Given:

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

Note: The arguments of the sine is calculated in unit of radian and not in degree.

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

3 years ago

Which depends on your location weight or mass

Zepler [3.9K]

I believe it would be weight. mass never changes.

6 0

2 years ago

Read 2 more answers