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3 years ago

Big Bee has a mass of 5 kg and hits a newly cleaned windshield accelerating at a rate of 2 m/s2.

Physics

1 answer:

Juli2301 [7.4K]3 years ago

8 0

Answer

The Little Bee will have a greater force of 12N

<u>Explanation</u>

As we know that Force is equal to the product of mass and acceleration (F=ma)

The Force with which the Big Bee will hit the windshield = (5kg x 2m/s^2)

Force of Big Bee= 10 Newton (N)

The Force with which the Little Bee will hit the windshield = (3kg x 4m/s^2)

Force of Little Bee = 12N

Hence, the little bee will hit the windshield with a greater force.

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Losing or gaining electrons makes an atom a/an: A. Electron B. Charged atom. C. Proton. D. Ion

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3 years ago

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gavmur [86]

Answer:

the period of the physical pendulum is 0.498 s

Explanation:

Given the data in the question;

$T_{simple$ = 0.61 s

we know that, the relationship between T and angular frequency is;

T = 2π/ω ---------- let this be equation 1

Also, the angular frequency of physical pendulum is;

ω = √(mgL / $I$ ) ------ let this equation 2

where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and $I$ is moment of inertia of rod.

Now, moment of inertia of thin uniform rod D is;

$I$ = $\frac{1}{3}$ mD²

since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.

we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/ $I$ ) OR T = 2π√( $I$ /mgL)

so we can use $I$ = $\frac{1}{3}$ mD² for moment of inertia of the rod

Since center of gravity of the uniform rod lies at the center of rod

so that L = $\frac{1}{2}$ D.

now, substituting these equations, the period becomes;

T = 2π/√( $I$ /mgL) OR T = $2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } }$ OR T = 2π√(2D/3g ) ----- equation 3

length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

ω $_{simple$ = 2π/ $T_{simple$ OR ω $_{simple$ = √(g/D) OR ω $_{simple$ = 2π√( D/g )

so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) × $T_{simple$

we substitute in value of $T_{simple$

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s

8 0

3 years ago

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Luba_88 [7]

Answer:

Explanation:

4 0

3 years ago

c. A car was running with a velocity of 20m/s. what will be its velocity after 30s if it's acceleration is 5m/s2

weqwewe [10]

$\large \mathfrak{Solution : }$

let's use first equation of motion to solve this ;

$\boxed{ \boxed{ v = u + at}}$

$v = 20 + (5 \times 30)$

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Velocity after 30 seconds = 170 m/s

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3 years ago

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is

Lana71 [14]

Answer:

$m = 0.169 \ kg$

$|v_{max} |= 0.5653 \ m/s$

Explanation:

From the question we are told that

The spring constant is $k = 14 \ N/m$

The maximum extension of the spring is $A = 6.0 \ cm = 0.06 \ m$

The number of oscillation is $n = 30$

The time taken is $t = 20 \ s$

Generally the the angular speed of this oscillations is mathematically represented as

$w = \frac{2 \pi}{T}$

where T is the period which is mathematically represented as

$T = \frac{t}{n}$

substituting values

$T = \frac{20}{30 }$

$T = 0.667 \ s$

Thus

$w = \frac{2 * 3.142 }{ 0.667}$

$w = 9.421 \ rad/s$

this angular speed can also be represented mathematically as

$w = \sqrt{\frac{k}{m} }$

=> $m =\frac{k }{w^2}$

substituting values

$m =\frac{ 15 }{(9.421)^2}$

$m = 0.169 \ kg$

In SHM (simple harmonic motion )the equation for velocity is mathematically represented as

$v = - Awsin (wt)$

The velocity is maximum when $wt = \(90^o) \ or \ 1.5708\ rad$

$v_{max} = - A* w$

=> $|v_{max} |= A* w$

=> $|v_{max} |= 0.06 * 9.421$

=> $|v_{max} |= 0.5653 \ m/s$

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3 years ago