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3 years ago

14

Kim holds a pinwheel in the air and says it can be used to model a source of energy. Identify the type of energy Kim's pinwheel

models. Explain the type of energy source used.

1 answer:

Natali5045456 [20]3 years ago

7 0

Answer:

it is kenetic

Explanation: its in motion:D

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1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.

harkovskaia [24]

Answer:

(a) The net force is 80.394 N

The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, $F_y = Fsin \theta = 250sin45 = 176.78 \ N$

Applied force in x-direction, $F_x = Fcos \theta = 250cos45 = 176.78 \ N$

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

$F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N$

Apply Newton's second law of motion;

F = ma

$a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2$

(b) the velocity of the crate after 5.0 s

$F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s$

7 0

3 years ago

What are the conditions necessary for a terrestrial planet to have a strong magnetic field?.

Stells [14]

Both a molten metallic core and reasonably fast rotation.

6 0

2 years ago

If you wish to observe features that are around the size of atoms, say 1 .5 x 100 m, with electromagnetic radiation, the radiati

chubhunter [2.5K]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

If you wish to observe features that around the size of atoms, say 1.5×10⁻¹⁰ m, with electromagnetic radiation, the radiation must have a wavelength about the size of the atom itself.

a) If you had a microscope which was capable of doing this, what would the frequency of electromagnetic radiation be, in hertz that you would have to use?

b) What type of electromagnetic radiation would this be?

Given Information:

Wavelength = λ = 1.5×10⁻¹⁰ m

Required Information:

a) Frequency = f = ?

b) Type of electromagnetic radiation = ?

Answer:

a) Frequency = f = 2×10¹⁸ Hz

b) Type of electromagnetic radiation = X-rays

Explanation:

a) The frequency of the electromagnetic radiation is given by

f = c/ λ

Where λ is the wavelength of the electromagnetic radiation and c is the speed of light and its value is 3×10⁸ m/s

f = 3×10⁸/1.5×10⁻¹⁰

f = 2×10¹⁸ Hz

Therefore, the frequency of the electromagnetic radiation would be 2×10¹⁸ Hz.

b)

The frequency range of X-rays is 3×10¹⁶ Hz to 3×10¹⁹ Hz

The frequency 2×10¹⁸ lies in that range, therefore, the type of electromagnetic radiation is X-rays

5 0

3 years ago

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm

andrew11 [14]

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}$

For second order

$\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

8 0

2 years ago

In a lab, a student drags a shoe across the floor at constant speed. If the coefficient of static friction between the floor and

Svetllana [295]

<span>B) 0.6 N
I suspect you have a minor error in your question. Claiming a coefficient of static friction of 0.30N is nonsensical. Putting the Newton there is incorrect. The figure of 0.25 for the coefficient of kinetic friction looks OK. So with that correction in mind, let's solve the problem. The coefficient of static friction is the multiplier to apply to the normal force in order to start the object moving. And the coefficient of kinetic friction (which is usually smaller than the coefficient of static friction) is the multiplied to the normal force in order to keep the object moving. You've been given a normal force of 2N, so you need to multiply the coefficient of static friction by that in order to get the amount of force it takes to start the shoe moving. So: 0.30 * 2N = 0.6N And if you look at your options, you'll see that option "B" matches exactly.</span>

7 0

3 years ago

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