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2 years ago

PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP

Physics

2 answers:

aleksley [76]2 years ago

6 0

Answer:

red is not at the bottom. its density is 0.9 not 9!

also I don't need the brainlest thanks

boyakko [2]2 years ago

4 0

Explanation: That is not meant to be red, it‘s the bottom of the beaker. The star is at the very bottom of the beaker. it’s just the base of the beaker.

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A swinging pendulum has a total energy of <img src="https://tex.z-dn.net/?f=E_i" id="TexFormula1" title="E_i" alt="E_i" align="a

Zolol [24]

Answer:

$\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta}$ (for small oscillations)

Explanation:

The total energy of the pendulum is equal to:

$E_{1} = m\cdot g \cdot (1-\cos \theta)\cdot L$

For small oscillations, the equation can be re-arranged into the following form:

$E_{1} \approx m\cdot g \cdot (1-\theta) \cdot L$

Where:

$\theta = \frac{A}{L^{2}}$ , measured in radians.

If the amplitude of pendulum oscillations is increase by a factor of 4, the angle of oscillation is $4\theta$ and the total energy of the pendulum is:

$E_{2} \approx m\cdot g \cdot (1-4\theta)\cdot L$

The factor of change is:

$\frac{E_{2}}{E_{1}} \approx \frac{1 - 4\theta}{1-\theta}$

$\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta}$

3 0

3 years ago

You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 14.8m above the ground, it

VladimirAG [237]

Answer:

(A) The speed just as it left the ground is 30.25 m/s

(B) The maximum height of the rock is 46.69 m

Explanation:

Given;

weight of rock, w = mg = 20 N

speed of the rock at 14.8 m, u = 25 m/s

(a) Apply work energy theorem to find its speed just as it left the ground

work = Δ kinetic energy

F x d = ¹/₂mv² - ¹/₂mu²

mg x d = ¹/₂m(v² - u²)

g x d = ¹/₂(v² - u²)

gd = ¹/₂(v² - u²)

2gd = v² - u²

v² = 2gd + u²

v² = 2(9.8)(14.8) + (25)²

v² = 915.05

v = √915.05

v = 30.25 m/s

B) Use the work-energy theorem to find its maximum height

the initial velocity of the rock = 30.25 m/s

at maximum height, the final velocity = 0

- mg x H = ¹/₂mv² - ¹/₂mu²

- mg x H = ¹/₂m(0) - ¹/₂mu²

- mg x H = - ¹/₂mu²

2g x H = u²

H = u² / 2g

H = (30.25)² / 2(9.8)

H = 46.69 m

4 0

3 years ago

A mass on the end of a spring undergoes simple harmonic motion. At the instant when the mass is at its maximum displacement from

Mrac [35]

Answer:

C) True. At maximum displacement, its instantaneous velocity is zero.

Explanation:

The simple harmonic movement is given by

x = A cos wt

Speed

v = - A w sin wt

At the point of maximum displacement x = A

A = A cos wt

cos wt = 1

wt = 0

We replace the speed

v = -Aw sin 0 = A w

Speed is maximum

Let's review the claims

A) False. Speed is zero

B) False. It can be determined

C) True. Agree with our result

D) False. When one is maximum the other is minimum

4 0

3 years ago

Which is the best example of tropism in plants?

seraphim [82]

Answer:

Here are some examples of tropism in plants : Sunflowers are one of the best examples of positive phototropism , as they always grow facing the sun.

Explanation:

Although some experts consider that perhaps such a clear growth movement should not be considered tropism.

3 0

2 years ago

A particle with charge q = 10-9 C and mass m = 5.0 x 10-9 kg is moving in a magnetic field whose magnitude is 0.003 T. The speed

Marina86 [1]

Answer:

a=0.212 m/s²

Explanation:

Given that

q= 10⁻⁹ C

m = 5 x 10⁻⁹ kg

Magnetic filed ,B= 0.003 T

Speed ,V= 500 m/s

θ= 45°

Lets take acceleration of the mass is a m/s²

The force on the charge due to magnetic filed B

F= q V B sinθ

Also F= m a ( from Newton's law)

By balancing these above two forces

m a= q V B sinθ

$a=\dfrac{qVB\ sin\theta}{m}$

$a=\dfrac{10^{-9}\times 500\times 0.003\times\ sin45^{\circ}}{5\times 10^{-9}}\ m/s^2$

$a=\dfrac{10^{-9}\times 500\times 0.003\times\dfrac{1}{\sqrt2}}{5\times 10^{-9}}\ m/s^2$

a=0.212 m/s²

4 0

3 years ago