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3 years ago

Consider three scenarios in which a particular box moves downward under the pull of gravity :

Physics

1 answer:

luda_lava [24]3 years ago

5 0

Answer:

1. True WA > WB > WC

Explanation:

In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.

A) Work is the product of force by distance and the cosine of the angle between them

WA = W h cos 0

WA = mg h

B) On a ramp without rubbing

Sin30 = h / L

L = h / sin 30

WB = F d cos θ

WB = F L cos 30

WB = mf (h / sin30) cos 30

WB = mg h ctan 30

C) Ramp with rubbing

W sin 30 - fr = ma

N- Wcos30 = 0

W sin 30 - μ W cos 30 = ma

F = W (sin30 - μ cos30)

WC = mg (sin30 - μ cos30) h / sin30

Wc = mg (1 - μ ctan30) h

When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum

Let's review the claims

1. True The work of gravity is the greatest and the work where there is friction is the least

2 False. The job where there is friction is the least

3 False work with rubbing is the least

4 False work with rubbing is the least

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Choose the pair of elements that will most likely have the least ionic character.

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I think the answer is c h and o

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2 years ago

Read 2 more answers

A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 4.0 m/s, skat

Cerrena [4.2K]

Answer:

a) $t=59.817652833125294s \approx 59.8s$

b) $D_1=894.01m$

Explanation:

From the question we are told that

Speed of opposing player $V_2=4.0m/s$

First player chase his opponent after $t=2.80t$

Acceleration of first player $a=0.14 m/s2$

Let time of catch be $t_c$

a)Generally the Equation for distance covered is mathematically given as follows

Distance to catch First opponent

$D_1=\frac{1}{2}(0.14)t^2$

$D_1=0.07t^2$

Distance to covered Second opponent

$D_2=(2.8+t)*4$

Generally when first opponent catch the second opponent it is represented mathematically as

$D_2=D_1$

$0.07t^2=(2.8+t)*4$

$0.07t^2=11.2+4t$

$0.07t^2-4t-11.2$

$t=59.817652833125294s \approx 59.8s$

b)Generally the the total time traveled by the first opponent is mathematically given as

$D_1=\frac{t^2}{4}$

$D_1=\frac{59.8^2}{4}$

$D_1=894.01m$

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2 years ago

Two long cylinders of radii 5 cm and 10 cm form a coaxial cylindrical capacitor, the space between them being filled with a diel

V125BC [204]

Answer:

Explanation:

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2 years ago

Let's assume you use green light (λ = 550 nm) to look at an electron. What is the uncertainty in determining the electron's velo

Rudiy27

By uncertainty principle

λ = h / p.

Where λ = wavelength, h = Planck's constant = 6.63 * 10⁻³⁴ Js

λ = wavelength = 550 nm = 550 * 10⁻⁹ m, Mass of electron = 9.1 * 10 ⁻³¹ kg

p = Momentum = mv

λ = h / mv

v = h / mλ

v = 6.63 * 10⁻³⁴ / (9.1 * 10⁻³¹ * 550 * 10⁻⁹)

v = 1 324 675.325 m/s

v ≈ 1.325 *10⁶ m/s

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2 years ago

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olga_2 [115]

The three scales are fahrenheit, celsius, and kelvin.

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Celsius (C or o)(C * 9/5) + 32C
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One degree Celsius is a bigger unit of measurement than one degree Fahrenheit. One degree C equals 1.8 degrees F.

Normal body temp is 98.6 F, which converts to 37 C. If your temp increased 2 degrees F, it would be 100.6, which is a fairly mild fever. An increase of 2 degrees C is 39, or 102.2 F. Obviously, 102.2 is a higher fever than 100.6.

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2 years ago