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3 years ago

What is kinetic energy related to?

Physics

1 answer:

mestny [16]3 years ago

6 0

Answer:

Motion, or Temperature.

Explanation:

Kinetic Energy refers to energy expelled from moving, such as if you are going down a giant hill in a rollercoaster. Potential Energy refers to stored energy, or energy collected when you don't move. An example of this is right at the top of a rollercoaster, before you go down the giant hill.

Similarly, temperature is determined by the energy of moving molecules, or how fast they move.

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What unit of measure would you use for mass?

creativ13 [48]

grams, pounds, kilos, etc

6 0

3 years ago

Please help me guys never mind the calculations

vlada-n [284]

The shape is connected in parallel so;

5.1) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{2} + \frac{1}{3} \\ \frac{1}{R} = \frac{3 + 2}{6} = \frac{5}{6} \\ R = \frac{6}{5} = 1.2 \: \: ohm$

5.2) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{8} + \frac{1}{10} \\ \frac{1}{R} = \frac{5 + 4}{40} = \frac{9}{40} \\ R = \frac{40}{9} = 4.4 \: \: ohm$

I hope I helped you^_^

7 0

2 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

8 0

2 years ago

Question 9 (28 points)

ladessa [460]

Im sure the answer is letter B

5 0

3 years ago

Read 2 more answers

In a certain two slit diffraction experiment, two slits 0.02mm wide are spaced0.2mm between centers.(a) How many fringes appear

Kamila [148]

Answer:

a) m = 10 and b) λ = 3.119 10⁻⁷ m

Explanation:

In the diffraction experiments the maximums appear due to the interference phenomenon modulated by the envelope of the diffraction phenomenon, for which to find the number of lines within the maximum diffraction center we must relate the equations of the two phenomena.

Interference equation d sin θ = m λ

Diffraction equation a sin θ = n λ

Where d is the width between slits (d = 0.2 mm), a is the width of each slit (a = 0.02 mm). θ is the angle, λ the wavelength, m and n are an integer.

Let's find the relationship of these two equations

d sin θ / a sin θ = m Lam / n Lam

The first maximum diffraction (envelope) occurs for n = 1, let's simplify

d / a = m

Let's calculate

m = 0.2 / 0.02

m = 10

This means that 10 interference lines appear within the first maximum diffraction.

b) let's use the interference equation, remember that the angles must be given in radians

θ = 0.17 ° (π rad / 180 °) = 2.97 10⁻³ rad

d sin θ = m λ

λ = d sin θ / m

λ = 0.2 10⁻³ sin (2.97 10⁻³) / 2

λ = 3.119 10⁻⁷ m

8 0

3 years ago