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Delicious77 [7]
3 years ago
15

A boy is exerting a force of 70 N at a 50-degree angle on a lawnmower. He is accelerating at 1.8 m/s2. Round the answers to the

nearest whole number.
What is the mass of the lawnmower?
What is the normal force exerted on the lawnmower?
Physics
1 answer:
Marysya12 [62]3 years ago
5 0

The lawnmower accelerates in the positive horizontal direction, so that the net horizontal force is, by Newton's second law,

(70 N) cos(-50°) = <em>m</em> (1.8 m/s²)

where <em>m</em> is the mass of the lawnmower. Solve for <em>m</em> :

<em>m</em> = ((70 N) cos(-50°)) / (1.8 m/s²)

<em>m</em> ≈ 25 kg

The lawnmower presumably doesn't get lifted off the ground, so that the net vertical force is 0. If <em>n</em> is the magnitude of the normal force, then by Newton's second law,

<em>n</em> - <em>m g</em> + (70 N) sin(-50°) = 0

<em>n</em> = <em>m g</em> + (70 N) sin(50°)

<em>n</em> = (25 kg) (9.8 m/s²) + (70 N) sin(50°)

<em>n</em> ≈ 300 N

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When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic ener
vodomira [7]

This question is incomplete, the complete question is;

When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.28 × 10⁻¹⁹ J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.

Answer:

the required wavelength of light is 161.9 nm

Explanation:

Given the data in the question;

Let us represent work function of the metal by W.

Now, using Einstein photoelectric effect equation;

E_{proton = W + K_{max

hc/λ = W + K_{max ------- let this be equation 1

we solve for W

W = hc/λ - K_{max

given that; λ = 221 nm = 2.21 × 10⁻⁷ m, K_{max= 3.28 × 10⁻¹⁹ J

we know that speed of light c = 3 × 10⁸ m/s and Planck's constant h = 6.626 × 10⁻³⁴ Js.

so we substitute

W = [( (6.626 × 10⁻³⁴)(3 × 10⁸) )/2.21 × 10⁻⁷ ] -  3.28 × 10⁻¹⁹

W =  8.99457 × 10⁻¹⁹ -   3.28 × 10⁻¹⁹

W = 5.71457 × 10⁻¹⁹ J

Now, to determine λ for which maximum kinetic energy is double

so;

K'_{max = double = 2( 3.28 × 10⁻¹⁹ J ) = 6.56 × 10⁻¹⁹ J

from from equation 1

we solve for λ'

λ' = hc / W + K'_{max

we substitute

λ' = ( (6.626 × 10⁻³⁴)(3 × 10⁸) ) / ( (5.71457 × 10⁻¹⁹ J) + ( 6.56 × 10⁻¹⁹ J ))

λ' = 1.9878 × 10⁻²⁴ /  1.227457 × 10⁻¹⁸

λ' =  1.619 × 10⁻⁷ m

λ' =  161.9 nm

Therefore, the required wavelength of light is 161.9 nm

4 0
3 years ago
The magnetic circuit below is excited by a 100-turn coil wound over the central leg. The mean length of the central leg is 5.5cm
Dafna11 [192]

Answer:

Hello your question is incomplete attached below is the complete question and solution

<em>answer; </em>

attached below

Explanation:

<em>Given data:</em>

100-turn coil

mean length of central leg = 5.5 cm

mean length of outer paths = 15.5 cm

relative permeability = 2000

cross sectional area ( A ) = 1 cm^2

distance x = 1 cm  

attached below is a detailed solution

3 0
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