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2 years ago

In an electric motor, electrical energy is converted into

Physics

1 answer:

N76 [4]2 years ago

5 0

Answer:

idrk im jus here for the points $\lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n$ ∈ω≠≠

Explanation:

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A box with mass (m) it's sliding along on a friction-free surface at 9.87 m/s at a height of 1.81 meters. It travels down the hi

Rus_ich [418]

A) The answer is 11.53 m/s

The final kinetic energy (KEf) is the sum of initial kinetic energy (KEi) and initial potential energy (PEi).
KEf = KEi + PEi

Kinetic energy depends on mass (m) and velocity (v)
KEf = 1/2 m * vf²
KEi = 1/2 m * vi²

Potential energy depends on mass (m), acceleration (a), and height (h):
PEi = m * a * h

So:
KEf = KEi + <span>PEi
</span>1/2 m * vf² = 1/2 m * vi² + m * a * h
..
Divide all sides by m:
1/2 vf² = 1/2 vi² + a * h

We know:
vi = 9.87 m/s
a = 9.8 m/s²
h = 1.81 m

1/2 vf² = 1/2 * 9.87² + 9.8 * 1.81
1/2 vf² = 48.71 + 17.74
1/2 vf² = 66.45
vf² = 66.45 * 2
vf² = 132.9
vf = √132.9
vf = 11.53 m/s

b) The answer is 6.78 m

The kinetic energy at the bottom (KE) is equal to the potential energy at the highest point (PE)
KE = PE

Kinetic energy depends on mass (m) and velocity (v)
KE = 1/2 m * v²

Potential energy depends on mass (m), acceleration (a), and height (h):
PE = m * a * h

KE = PE
1/2 m * v² = m * a * h

Divide both sides by m:
1/2 * v² = a * h
v = 11.53 m/s
a = 9.8 m/s²
h = ?

1/2 * 11.53² = 9.8 * h
1/2 * 132.94 = 9.8 * h
66.47 = 9.8 * h
h = 66.47 / 9.8
h = 6.78 m

3 0

2 years ago

A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surf

Sati [7]

Answer:

a) The minimum thickness of the oil slick at the spot is 313 nm

b) the minimum thickness be now will be 125 nm

Explanation:

Given the data in the question;

a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?

t $_{min$ = λ/2n

given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20

we substitute

t $_{min$ = 750 / 2(1.20)

t $_{min$ = 750 / 2.4

t $_{min$ = 312.5 ≈ 313 nm

Therefore, The minimum thickness of the oil slick at the spot is 313 nm

Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?

minimum thickness of the oil slick at the spot will be;

t $_{min$ = λ/4n

given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50

we substitute

t $_{min$ = 750 / 4(1.50)

t $_{min$ = 750 / 6

t $_{min$ = 125 nm

Therefore, the minimum thickness be now will be 125 nm

4 0

2 years ago

what occurs when a swimmer pushes through the water to swim answers are (A) the water exerts a reaction force on the swimmer (B)

miv72 [106K]

When a swimmer pushes through water to swim they are propelled forward because of the water resistance against the hand and feet. It's A. The water doesn't automatically push the swimmer forward. It releases a reaction after the swimmer pushes through the water.

6 0

3 years ago

Read 2 more answers

You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend

Bogdan [553]

Answer:

a. $\tau=200J$ b. $\alpha=0.44 \frac{rad}{s^2}$ c. $\alpha=0.33\frac{rad}{s^2}$ d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by $\tau=rF\sin \theta$ , being r the radius, F the force aplied and $\theta$ the angle between the vector force and the vector radius. Since $\theta=90^{\circ}, \, \sin\theta=1$ and so $\tau=rF=2m100N=200Nm=200J$ .

b. Since the relation $\tau=I\alpha$ hols, being I the moment of inertia, the angular acceleration can be calculated by $\alpha=\frac{\tau}{I}$ . Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is $I=\frac{1}{2}Mr^2$ , being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by $I_p=m_pr_p^{2}$ , being $m_p$ the mass of the person and $r_p^{2}$ the distance from the person to the center. Given all of this, we have

$\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}$ .

c. Similar equation to b, but changing $r_p=2m$ , so

$alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}$ .

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

5 0

3 years ago

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

3 years ago