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Helen [10]
3 years ago
13

Which of the following is an example of potential energy? View Available Hint(s) Which of the following is an example of potenti

al energy? a hot cup of coffee the mercury rising within a thermometer a bowling ball placed on the top shelf of a closet waves crashing against the shoreline
Physics
1 answer:
Brums [2.3K]3 years ago
4 0

Answer:

a bowling ball placed on the top shelf of a closet.

Explanation:

Potential energy is the energy that is stored in an object due to its position relative to some zero position. An object possesses gravitational potential energy if it is positioned at a height above (or below) the zero height.

Mathematically,

Potential energy, PE = M × g × h

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A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it
Degger [83]

Answer:

a)\omega_1=8.168\,rad.s^{-1}

b)n_1=7.735 \,rev

c)\alpha_1 =0.6864\,rad.s^{-2}

d)\alpha_2=4.1454\,rad.s^{-2}

e)t_2=1.061\,s

Explanation:

Given that:

  • initial speed of turntable, N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}
  • full speed of rotation, N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}
  • time taken to reach full speed from rest, t_1=11.9\,s
  • final speed after the change,  N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}
  • no. of revolutions made to reach the new final speed,  n_2=11\,rev

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

\omega=\frac{2\pi.N}{60}\, rad.s^{-1}

where N = angular speed in rpm.

putting the respective values from case 1 we've

\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}

\omega_1=8.168\,rad.s^{-1}

(c)

using the equation of motion:

\omega_1=\omega_0+\alpha . t_1

here α is the angular acceleration

78=0+\alpha_1\times 11.9

\alpha_1 = \frac{8.168 }{11.9}

\alpha_1 =0.6864\,rad.s^{-2}

(b)

using the equation of motion:

\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1

8.168^2=0^2+2\times 0.6864\times n_1

n_1=48.6003\,rad

n_1=\frac{48.6003}{2\pi}

n_1=7.735\, rev

(d)

using equation of motion:

\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2

12.5664^2=8.168^2+2\alpha_2\times 11

\alpha_2=4.1454\,rad.s^{-2}

(e)

using the equation of motion:

\omega_2=\omega_1+\alpha_2 . t_2

12.5664=8.168+4.1454\times t_2

t_2=1.061\,s

4 0
3 years ago
A rabbit runs 28 m toward the left in 9 s; then the rabbit suddenly changes direction and runs 18 m toward the right in the next
Viefleur [7K]

Answer:

The answer is 3.111111.

Explanation:

It runs 28 m in the first 9 s, and 28 divided by 9 equals 3.1 and the one goes on forever.

8 0
3 years ago
are the properties of a specific element the same as the properties of a molecule made up of that element ? why or why not
konstantin123 [22]
Because it’s science
8 0
3 years ago
What does the negative sign in F = –kx mean?
Andru [333]
The force is opposite to the displacement
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3 years ago
You are working in cooperation with the Public Health department to design an electrostatic trap for particles from auto emissio
Mademuasel [1]

Answer:

Explanation:

Given that:

Charge (q) on the particle = 3 × 10⁻⁸ C

mass (m) of the particle = 6 × 10⁻⁹ kg

at a distance x = 15 cm , the velocity in the plate = 900 m/s²

For the square plate, the surface charged density σ = -8 × 10⁻⁶ C/m²

To start with calculating the electric field as a result of the square plate; we use the formula;

E = \dfrac{\sigma }{2 \varepsilon_o}

E = \dfrac{8 \times 10^{-6} }{2 \times  8.85 \times 10^{-12}}

E = 4.51977 \times 10^5 \ V/m

On the square plate; The electric force F = Eq

F = (4.51977 \times 10^5 \ V/m )(3\times 10^{-8} \ C)

F = 1.3559 \times 10^{-2} \ N

The acceleration a =\dfrac{ F}{m}{

a = \dfrac{1.3559\times 10^{-2} \ N}{6 \times 10^{-9} \ Kg}

a = 2.25988 \times 10^6 \ m/s^2

For the particle, the velocity at distance x = 7 m can be calculated by using the formula:

(\dfrac{1}{2}) mv^2 = \Delta Vq

v^2 = \dfrac{2 Eq}{dm}

v^2 = \dfrac{2 * 4.51977 \times 10^5 \times 3 \times 10^{-8} }{0.07 \times 6\times 10^{-9} }

v^2 = 64568142.86  \ m/s

v =\sqrt{ 64568142.86  \ m/s}

\mathbf{v = 8.035 \times 10^3 \ m/s}

From the calculation, we realize that the charge acting between the particle and the plate is said to be "opposite".

Hence, the force is an attractive force.

Similarly, there is a gradual increase exhibited by the velocity of the particle.

Therefore, the particles get to the detector, but the detector failed to get detect due to the velocity which is greater than 1000 m/s.

6 0
3 years ago
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