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RSB [31]
2 years ago
13

An angle marking is like a meter stick marking. The marking of 1.2m on a meter stick only means an object is 1.2m long if the ob

ject starts at zero meters. If it starts at 0.2 meters then the object is really 1.0 meters long. You will be making angle calculations from angle markings in a similar fashion.
Given a green marking of 223 degrees and a straight-on marking of 253 degrees, how much has the light bent?
Physics
1 answer:
grandymaker [24]2 years ago
6 0

Answer:

The angle of bend = 20°

Explanation:

Since the question describes that the angle marking is similar to the metre stick marking.

The length of the object when it begins at 0 meters on the meter rule

= 1.2 m

The length of the meter stick when it begins at 0.2 m on the meter rule

= 1.0 m

The difference in length of the object based on the markings

= 1.2 - 1.0 = 0.2 m

If the green marking is on 223°, i.e. θ₂ = 223°

and the straight marking is on 253°, i.e.  θ₁ = 253°

The angle of bend =  θ₁ -  θ₂

The angle of bend = 253° - 223°

The angle of bend = 20°

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A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a
UNO [17]

Answer:

P_{avg} = 6.283*10^{-9} \ W

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})

where;

E= \frac{d \phi}{dt}

E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t

E_{rms} =   \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}

Replacing our values into above equation; we have:

E_{rms} =   \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}

E_{rms} =   \frac {8.98909588*10^{-4} }{\sqrt{2}}

E_{rms} =   6.356*10^{-4} \ V

Then the P_{avg is calculated as:

P_{avg} = \frac{E^2}{R}

P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}

P_{avg} = 6.283*10^{-9} \ W

6 0
3 years ago
Too much skepticism can
VladimirAG [237]
<span>Contemporary skepticism (or scepticism) is loosely used to denote any questioning attitude,[1] or some degree of doubt regarding claims that are elsewhere taken for granted.[2] Usually meaning those who follow the evidence, [ [ versus those who are skeptical of the evidence (see:Denier) Skepticism is most controversial when it questions beliefs that are taken for granted by most of the

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5 0
2 years ago
A ball moving at positive 3.0 m per s along a table rolls off a table and lands on the ground 2.0 m away. How high was the table
MA_775_DIABLO [31]

consider the motion along the horizontal direction :

v₀ = initial velocity in horizontal direction as the ball rolls off the table = 3.0 m/s

X = horizontal displacement of the ball = 2.0 m

a = acceleration along the horizontal direction = 0 m/s²

t = time taken to land = ?

using the kinematics equation

X = v₀ t + (0.5) a t²

2.0 = 3.0 t + (0.5) (0) t²

t = 2/3


consider the motion of the ball along the vertical direction

v₀ = initial velocity in vertical direction as the ball rolls off the table = 0 m/s

Y = vertical displacement of the ball = height of the table = h

a = acceleration along the vertical direction = 9.8 m/s²

t = time taken to land = 2/3

using the kinematics equation

Y = v₀ t + (0.5) a t²

h = 0 t + (0.5) (9.8) (2/3)²

h = 2.2 m


C 2.2 m

3 0
3 years ago
A 55.0-g piece of copper wire is heated, and the temperature of the wire changes from 19.0°C to 86.0°C. The amount of heat absor
Alinara [238K]
The specific heat of a metal or any element or compound can be determined using the formula Cp = delta H / delta T / mass. delta pertains to change. That is change in enthalpy and change in temperature. From the given data, Cp is equal to 343 cal per (86-19) c per 55 grams. This is equal to 0.093 cal / g deg. Celsius
4 0
3 years ago
If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far
zavuch27 [327]

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d=\frac{N2}{mg}*l * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= \frac{241.75}{69.1*9.8} *l * tan 58

d= 0.57* l

So painter can be along the 57% of length before the ladder begins to slip

3 0
3 years ago
Read 2 more answers
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