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2 years ago

5

at the sewing store, ava bought a bag of buttons. 21 in all. 21 of the buttons were large. what percentage of the buttons wre la

rge

2 answers:

erma4kov [3.2K]2 years ago

6 0

If she only has 21 buttons and all 21 of them are large, then all of her buttons are large. so 100% of the buttons would be large.

lukranit [14]2 years ago

4 0

Answer: 100%

Explanation:

Given : The total number of buttons in the bag= 21

The number of larger buttons = 21

Then , the percentage of the buttons were large is given by :-

$\dfrac{\text{Number of large buttons}}{\text{Total buttons}}\times100\\\\\dfrac{21}{21}\times100=100\%$

Therefore, the percentage of the buttons were large = 100%

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A ray of light is incident on a plane surface separating two sheets of glass with refractive indexes 1.70 and 1.58. The angel of

dlinn [17]

Answer:

r = 71.8⁰

Explanation:

given,

refractive index of the glass 1 = 1.70

refractive index of glass 2 = 1.58

angle of incidence = 62°

angle of refraction =?

using Snell's law

$\dfrac{sin\ i}{sin\ r} = \dfrac{n_2}{n_1}$

$\dfrac{sin\ 62^0}{sin\ r} = \dfrac{1.58}{1.70}$

1.7 ×sin 62 ^0 = 1.58× sin r

$sinr = \dfrac{1.7\times sin 62^0}{1.58}$

sin r = 0.95

r = sin⁻¹(0.95)

r = 71.8⁰

angle of refraction =r = 71.8⁰

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2 years ago

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A 20 watt lightbulb uses 20 Joules of energy every second.A person expends 50 watts of energy per stair when climbing up stairs.

Pachacha [2.7K]

20W = 20 J/s

Energy expended during climbing stairs = 50 W of energy/stair = 50J/stair

For 20 stairs, Total energy = 50x20 = 1000 J

This can light bulbs for, T= 1000J/20 J/s =50 seconds

6 0

2 years ago

 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

we know that $y_{f}=0$ and that $v_{y0}=0$ as well, so the formula is simplified to:

$0=y_{0}+\frac{1}{2}at^{2}$

we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

we know that $y_{0}=1.20m$ and that $a=g=-9.8m/s^{2}$

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

$V_{x}=\frac{x}{t}$

so we can solve this for x, so we get:

$x=V_{x}t$

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

$a=\frac{v_{f}-v_{0}}{t}$

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

$a=\frac{v_{f}}{t}$

so we can solve this for the final velocity:

$V_{yf}=at$

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

which yields:

$V_{yf}=-4.851m/s$

so now we know the components of the final velocity, which are:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

so now we can find the speed by determining the magnitude of the vector, like this:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

so we get:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

so we get:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

which yields:

$\theta = -72.82^{o}$

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2 years ago

Write the equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and

yKpoI14uk [10]

Answer:

The newton’s second law is F=ma

The Gravitational force is $F=\dfrac{Gm_{1}m_{2}}{r^2}$

Explanation:

Given that,

The equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and Gravitation.

We know that,

Velocity :

The velocity is equal to the rate of position of the object.

$v=\dfrac{dx}{dt}$ ....(I)

Acceleration :

The acceleration is equal to the rate of velocity of the object.

$a=\dfrac{dv}{dt}$ ....(II)

Newton’s second Laws

The force is equal to the change in momentum.

In mathematically,

$F=\dfrac{d(p)}{dt}$

Put the value of p

$F=\dfrac{d(mv)}{dt}$

$F=m\dfrac{dv}{dt}$

Put the value from equation (II)

$F=ma$

This is newton’s second laws.

Gravitational force :

The force is equal to the product of mass of objects and divided by square of distance.

In mathematically,

$F=\dfrac{Gm_{1}m_{2}}{r^2}$

Where, m₁₂ = mass of first object

m= mass of second object

r = distance between both objects

Hence, The newton’s second law is F=ma

The Gravitational force is $F=\dfrac{Gm_{1}m_{2}}{r^2}$

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2 years ago

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nexus9112 [7]

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6 0

2 years ago

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